WEBVTT 00:00:00.000 --> 00:00:01.984 align:middle line:90% [SQUEAKING] 00:00:01.984 --> 00:00:03.968 align:middle line:90% [RUSTLING] 00:00:03.968 --> 00:00:04.960 align:middle line:90% [CLICKING] 00:00:04.960 --> 00:00:20.358 align:middle line:90% 00:00:20.358 --> 00:00:22.400 align:middle line:84% CASEY RODRIGUEZ: We defined the Lebesgue integral 00:00:22.400 --> 00:00:36.480 align:middle line:84% for simple functions, which have this canonical representation 00:00:36.480 --> 00:00:42.450 align:middle line:84% as a finite linear combination of indicator functions 00:00:42.450 --> 00:00:45.390 align:middle line:84% on sets which are pairwise disjoint 00:00:45.390 --> 00:00:51.220 align:middle line:84% and whose union gives me E. We defined their integral-- 00:00:51.220 --> 00:00:54.210 align:middle line:90% 00:00:54.210 --> 00:01:00.960 align:middle line:84% the integral of phi was defined to be the sum from j equals 1 00:01:00.960 --> 00:01:10.290 align:middle line:84% to n aj times the measure of capital Aj sum as j 00:01:10.290 --> 00:01:11.340 align:middle line:90% goes from 1 to n. 00:01:11.340 --> 00:01:13.230 align:middle line:84% And we proved a couple of properties of this. 00:01:13.230 --> 00:01:18.120 align:middle line:84% Namely, we proved that if I multiply a simple function 00:01:18.120 --> 00:01:19.950 align:middle line:84% by a non-negative scalar, then the scalar 00:01:19.950 --> 00:01:21.033 align:middle line:90% pulls out of the integral. 00:01:21.033 --> 00:01:22.920 align:middle line:84% The integral of the scalar multiple of phi 00:01:22.920 --> 00:01:26.490 align:middle line:84% is equal to scalar multiple of the integral. 00:01:26.490 --> 00:01:29.880 align:middle line:84% If I have two non-negative simple functions 00:01:29.880 --> 00:01:32.580 align:middle line:84% and I add them, I get another non-negative simple function 00:01:32.580 --> 00:01:36.220 align:middle line:84% and the integral of the sum is the sum of the integrals. 00:01:36.220 --> 00:01:40.560 align:middle line:84% And we also prove that if I have two simple functions, 00:01:40.560 --> 00:01:42.360 align:middle line:84% one less than or equal to the other, 00:01:42.360 --> 00:01:44.730 align:middle line:84% then the integrals respect this inequality. 00:01:44.730 --> 00:01:46.170 align:middle line:84% The integral of the smaller one is 00:01:46.170 --> 00:01:48.810 align:middle line:84% less than or equal to the integral of the bigger one. 00:01:48.810 --> 00:01:52.320 align:middle line:90% 00:01:52.320 --> 00:01:58.410 align:middle line:84% So now, we're going to define the Lebesgue 00:01:58.410 --> 00:02:02.520 align:middle line:84% integral for a general non-negative measurable 00:02:02.520 --> 00:02:04.440 align:middle line:90% function. 00:02:04.440 --> 00:02:12.180 align:middle line:84% In some sense, how one should view 00:02:12.180 --> 00:02:16.620 align:middle line:84% the Lebesgue is kind of how one should view the Riemann 00:02:16.620 --> 00:02:18.150 align:middle line:90% integral, I guess, if you-- 00:02:18.150 --> 00:02:20.370 align:middle line:84% or at least when you think about the Riemann integral 00:02:20.370 --> 00:02:26.130 align:middle line:84% as you build up these approximations to the integral 00:02:26.130 --> 00:02:30.180 align:middle line:84% by cutting up the domain for the Riemann integral. 00:02:30.180 --> 00:02:33.870 align:middle line:84% And you choose points in between and you form these boxes. 00:02:33.870 --> 00:02:37.950 align:middle line:84% Now, you have some freedom in how you choose these boxes that 00:02:37.950 --> 00:02:41.160 align:middle line:90% approximate the integral of f. 00:02:41.160 --> 00:02:42.660 align:middle line:84% But one way to choose the boxes-- 00:02:42.660 --> 00:02:44.285 align:middle line:84% or at least when I picture it, I always 00:02:44.285 --> 00:02:48.810 align:middle line:84% picture the boxes sitting below the graph of f. 00:02:48.810 --> 00:02:52.110 align:middle line:84% And as you dice up the domain smaller and smaller 00:02:52.110 --> 00:02:55.530 align:middle line:90% then these things are kind of-- 00:02:55.530 --> 00:02:59.290 align:middle line:84% these boxes are getting narrower and kind of your approximation 00:02:59.290 --> 00:03:04.510 align:middle line:84% is filling in the integral or the area underneath the curve 00:03:04.510 --> 00:03:05.140 align:middle line:90% from below. 00:03:05.140 --> 00:03:07.880 align:middle line:90% 00:03:07.880 --> 00:03:11.260 align:middle line:84% And we've already seen that for non-negative measurable 00:03:11.260 --> 00:03:11.800 align:middle line:90% functions. 00:03:11.800 --> 00:03:15.730 align:middle line:84% There always exists a sequence of simple functions 00:03:15.730 --> 00:03:17.200 align:middle line:90% that increase to f. 00:03:17.200 --> 00:03:21.533 align:middle line:84% So that for every x I stick into the sequence of functions-- 00:03:21.533 --> 00:03:23.950 align:middle line:84% phi 1 is less than or equal to phi 2 is less than or equal 00:03:23.950 --> 00:03:25.250 align:middle line:90% to phi 3 and so on-- 00:03:25.250 --> 00:03:28.330 align:middle line:84% and these phis are increasing pointwise to f. 00:03:28.330 --> 00:03:32.830 align:middle line:84% So if you're trying to build this on your own, 00:03:32.830 --> 00:03:35.440 align:middle line:84% you would think, OK, let me define the integral as 00:03:35.440 --> 00:03:38.140 align:middle line:84% of a function, of a non-negative measurable function, 00:03:38.140 --> 00:03:45.340 align:middle line:84% as the limit of the integrals of a sequence of simple functions 00:03:45.340 --> 00:03:46.240 align:middle line:90% increasing to f. 00:03:46.240 --> 00:03:49.030 align:middle line:84% Which I know exists, because we constructed one. 00:03:49.030 --> 00:03:52.000 align:middle line:84% Of course-- and that's one way to do it, 00:03:52.000 --> 00:03:54.640 align:middle line:90% and some textbooks do that. 00:03:54.640 --> 00:03:57.820 align:middle line:84% But you come across this problem that this number that you've 00:03:57.820 --> 00:04:00.640 align:middle line:84% defined to be the limit of these integrals-- maybe it 00:04:00.640 --> 00:04:05.950 align:middle line:84% depends on the sequence of measurable functions 00:04:05.950 --> 00:04:08.990 align:middle line:90% that you took in the beginning. 00:04:08.990 --> 00:04:11.170 align:middle line:84% So we're not going to quite do that. 00:04:11.170 --> 00:04:13.690 align:middle line:84% In the end, we'll see that this number we defined 00:04:13.690 --> 00:04:19.600 align:middle line:84% can be given as of the limit of the integrals 00:04:19.600 --> 00:04:20.870 align:middle line:90% of simple functions. 00:04:20.870 --> 00:04:31.630 align:middle line:84% So for a general non-negative measurable functions, 00:04:31.630 --> 00:04:37.680 align:middle line:84% we define the integral of f over E 00:04:37.680 --> 00:04:43.290 align:middle line:84% to be the sup of the integral over E of phi, 00:04:43.290 --> 00:04:53.950 align:middle line:84% where now, phi is a non-negative simple function 00:04:53.950 --> 00:04:56.515 align:middle line:90% and phi sits below f. 00:04:56.515 --> 00:05:00.688 align:middle line:90% 00:05:00.688 --> 00:05:05.130 align:middle line:84% So in some sense, this is kind of like taking the integral 00:05:05.130 --> 00:05:10.115 align:middle line:90% of f to be defined as kind of-- 00:05:10.115 --> 00:05:12.240 align:middle line:84% I don't want to say limit, because it's not exactly 00:05:12.240 --> 00:05:15.660 align:middle line:84% a limit, but in some sense, the thing that's getting filled up 00:05:15.660 --> 00:05:19.020 align:middle line:84% by all the integrals of the simple functions 00:05:19.020 --> 00:05:21.900 align:middle line:84% below the graph that sit below the graph of f. 00:05:21.900 --> 00:05:27.125 align:middle line:90% 00:05:27.125 --> 00:05:32.000 align:middle line:84% OK, so let me just prove a very simple theorem, 00:05:32.000 --> 00:05:34.250 align:middle line:90% which is useful. 00:05:34.250 --> 00:05:41.880 align:middle line:84% So if E is a subset of R is measurable-- 00:05:41.880 --> 00:05:50.180 align:middle line:84% in fact, it's a set of measures 0. 00:05:50.180 --> 00:05:53.120 align:middle line:84% So remember, all sets of outer measure 0 are measurable. 00:05:53.120 --> 00:05:57.140 align:middle line:84% So I don't have to say really if E is a subset of R 00:05:57.140 --> 00:05:57.890 align:middle line:90% with E measurable. 00:05:57.890 --> 00:06:00.182 align:middle line:84% When I say this, I'm kind of saying two things at once. 00:06:00.182 --> 00:06:03.620 align:middle line:84% It's outer measure of E is 0 or 1, 00:06:03.620 --> 00:06:05.330 align:middle line:90% and therefore it's measurable. 00:06:05.330 --> 00:06:09.490 align:middle line:90% 00:06:09.490 --> 00:06:15.100 align:middle line:84% So then for all f that are non-negative and measurable 00:06:15.100 --> 00:06:19.510 align:middle line:84% on E, the integral over E of f is 0. 00:06:19.510 --> 00:06:21.940 align:middle line:84% OK, so it's only kind of interesting to take 00:06:21.940 --> 00:06:27.250 align:middle line:84% the integral over sets that have positive measure. 00:06:27.250 --> 00:06:29.740 align:middle line:84% So no matter what function I take, 00:06:29.740 --> 00:06:32.860 align:middle line:84% the integral over a set of measure 0 is 0. 00:06:32.860 --> 00:06:37.070 align:middle line:84% OK, this is kind of akin to, in Riemann integration, 00:06:37.070 --> 00:06:39.520 align:middle line:84% the integral over a point being 0. 00:06:39.520 --> 00:06:41.590 align:middle line:84% But now we have more interesting sets 00:06:41.590 --> 00:06:43.760 align:middle line:84% of measure 0 other than just a point. 00:06:43.760 --> 00:06:44.930 align:middle line:90% So what's the proof of this? 00:06:44.930 --> 00:06:46.513 align:middle line:84% Well, we don't have much to go off of. 00:06:46.513 --> 00:06:47.900 align:middle line:90% We just have the definition. 00:06:47.900 --> 00:06:49.360 align:middle line:90% So let's use the definition. 00:06:49.360 --> 00:07:16.390 align:middle line:84% Let phi be simple with this sort of canonical representation 00:07:16.390 --> 00:07:18.460 align:middle line:84% with the less than or equal to f. 00:07:18.460 --> 00:07:21.160 align:middle line:84% So what I'd like to show is that the integral of phi is 0. 00:07:21.160 --> 00:07:23.590 align:middle line:84% But this is-- and therefore, the integral 00:07:23.590 --> 00:07:26.440 align:middle line:84% of f, which is the sup over all of these integrals, 00:07:26.440 --> 00:07:27.790 align:middle line:90% is therefore 0. 00:07:27.790 --> 00:07:36.980 align:middle line:84% But this is clear because since all of these Aj's are subsets 00:07:36.980 --> 00:07:44.520 align:middle line:84% of E, which is a set of measure 0, 00:07:44.520 --> 00:07:49.380 align:middle line:84% this implies the measure of Aj equals 0 for all j. 00:07:49.380 --> 00:07:56.890 align:middle line:84% And therefore, the integral of phi over E, 00:07:56.890 --> 00:08:02.680 align:middle line:84% which is equal to Aj measure of Aj equals 0. 00:08:02.680 --> 00:08:05.440 align:middle line:84% And thus, the integral of f, which 00:08:05.440 --> 00:08:08.065 align:middle line:84% is the sup over all of these, is just a sup of 0 then. 00:08:08.065 --> 00:08:13.590 align:middle line:90% 00:08:13.590 --> 00:08:15.990 align:middle line:84% So I guess I should have started this proof off with let 00:08:15.990 --> 00:08:20.550 align:middle line:84% f be an L plus of E. So I left that off, 00:08:20.550 --> 00:08:27.210 align:middle line:84% but you know what I was doing, hopefully. 00:08:27.210 --> 00:08:32.760 align:middle line:84% OK, so it's only interesting to take the integral 00:08:32.760 --> 00:08:34.290 align:middle line:90% over sets of positive measure. 00:08:34.290 --> 00:08:37.289 align:middle line:90% 00:08:37.289 --> 00:08:41.309 align:middle line:84% Now, we have a few facts which carry over. 00:08:41.309 --> 00:08:47.722 align:middle line:90% 00:08:47.722 --> 00:08:48.680 align:middle line:90% Well, not really carry. 00:08:48.680 --> 00:08:50.270 align:middle line:84% Well, one of them carries over from what 00:08:50.270 --> 00:08:52.853 align:middle line:84% we did for simple functions and the others just kind of follow 00:08:52.853 --> 00:08:54.920 align:middle line:90% from the definition. 00:08:54.920 --> 00:08:57.060 align:middle line:84% And it'll be an exercise in the assignment, 00:08:57.060 --> 00:08:58.850 align:middle line:90% so it's the following. 00:08:58.850 --> 00:09:20.060 align:middle line:84% If phi is-- then the two definitions 00:09:20.060 --> 00:09:24.225 align:middle line:90% of the integral of phi agree. 00:09:24.225 --> 00:09:26.750 align:middle line:90% 00:09:26.750 --> 00:09:31.730 align:middle line:84% And so I'll say exactly what I mean in just a second. 00:09:31.730 --> 00:09:50.370 align:middle line:84% Two, f and g are in L plus of E. c is a non-negative real number 00:09:50.370 --> 00:10:01.770 align:middle line:84% and f is less than or equal to g on E. Then a couple of things. 00:10:01.770 --> 00:10:04.920 align:middle line:84% The integral of c times f is equal to c 00:10:04.920 --> 00:10:10.690 align:middle line:84% times the integral of f and the integral 00:10:10.690 --> 00:10:17.500 align:middle line:84% over E of f is less than or equal to the integral of g 00:10:17.500 --> 00:10:20.290 align:middle line:90% over E. 00:10:20.290 --> 00:10:23.080 align:middle line:84% And one final property is the following. 00:10:23.080 --> 00:10:25.710 align:middle line:90% 00:10:25.710 --> 00:10:35.120 align:middle line:84% If f is a measurable function on E 00:10:35.120 --> 00:10:43.300 align:middle line:84% and f is a measurable subset of E, 00:10:43.300 --> 00:10:47.510 align:middle line:84% then I can integrate f also on capital F. So this is in the, 00:10:47.510 --> 00:10:52.210 align:middle line:84% I think, maybe assignment 5 that if I have a function which 00:10:52.210 --> 00:10:54.490 align:middle line:84% is measurable on a set and I take 00:10:54.490 --> 00:10:57.950 align:middle line:84% any subset of that measurable set, which is measurable, 00:10:57.950 --> 00:11:01.120 align:middle line:84% then f restricted to this set is also measurable. 00:11:01.120 --> 00:11:03.160 align:middle line:84% So what I'm about to say makes sense. 00:11:03.160 --> 00:11:10.090 align:middle line:84% Then the integral of little f over capital F 00:11:10.090 --> 00:11:15.820 align:middle line:84% is equal to the integral of E of little f times the indicator 00:11:15.820 --> 00:11:24.810 align:middle line:84% function of f, which is less than or equal to the integral 00:11:24.810 --> 00:11:27.720 align:middle line:90% over E of little f. 00:11:27.720 --> 00:11:31.255 align:middle line:90% 00:11:31.255 --> 00:11:36.740 align:middle line:84% So I think the rest of the statements, statements 2 and 3 00:11:36.740 --> 00:11:39.530 align:middle line:90% are completely unambiguous. 00:11:39.530 --> 00:11:43.670 align:middle line:84% Maybe you were wondering what exactly did I mean by 1. 00:11:43.670 --> 00:11:47.750 align:middle line:84% So we had kind of two definitions of the integral 00:11:47.750 --> 00:11:49.010 align:middle line:90% of a simple function, right? 00:11:49.010 --> 00:11:55.070 align:middle line:84% We defined it first as the sum of the coefficients 00:11:55.070 --> 00:11:57.860 align:middle line:90% times the measure of these sets. 00:11:57.860 --> 00:12:01.460 align:middle line:84% And then we also have a second definition 00:12:01.460 --> 00:12:04.250 align:middle line:90% because a simple function-- 00:12:04.250 --> 00:12:08.160 align:middle line:84% a measurable non-negative simple function is also in this set. 00:12:08.160 --> 00:12:11.750 align:middle line:84% So I should have that what's here should be 00:12:11.750 --> 00:12:16.460 align:middle line:90% equal to what's on the right. 00:12:16.460 --> 00:12:24.600 align:middle line:84% So statement 1 is the statement that what's 00:12:24.600 --> 00:12:27.570 align:middle line:84% underlined three times is equal to what 00:12:27.570 --> 00:12:31.213 align:middle line:90% was underlined four times. 00:12:31.213 --> 00:12:32.130 align:middle line:90% That's a lot of lines. 00:12:32.130 --> 00:12:36.520 align:middle line:90% 00:12:36.520 --> 00:12:43.330 align:middle line:84% All right, so this will be a fairly straightforward exercise 00:12:43.330 --> 00:12:44.530 align:middle line:90% just using the definitions. 00:12:44.530 --> 00:12:47.230 align:middle line:84% And also, what we did for simple functions. 00:12:47.230 --> 00:12:49.830 align:middle line:90% 00:12:49.830 --> 00:12:55.870 align:middle line:84% One consequence of this theorem and the one before it 00:12:55.870 --> 00:12:56.830 align:middle line:90% is the following. 00:12:56.830 --> 00:13:00.700 align:middle line:84% Is that I can relax this condition here 00:13:00.700 --> 00:13:03.670 align:middle line:84% in 2 to an almost everywhere statement. 00:13:03.670 --> 00:13:06.280 align:middle line:90% 00:13:06.280 --> 00:13:16.126 align:middle line:84% So if f and g are non-negative measurable functions, 00:13:16.126 --> 00:13:24.090 align:middle line:84% and f is less than or equal to g almost everywhere on E, 00:13:24.090 --> 00:13:30.810 align:middle line:84% then the integral of E of f is less than 00:13:30.810 --> 00:13:35.580 align:middle line:90% or equal to the integral of g. 00:13:35.580 --> 00:13:42.760 align:middle line:90% 00:13:42.760 --> 00:13:45.400 align:middle line:84% So let's write the proof of this. 00:13:45.400 --> 00:13:47.920 align:middle line:90% 00:13:47.920 --> 00:13:56.520 align:middle line:84% So let f be the set of all x's and E such that 00:13:56.520 --> 00:13:59.325 align:middle line:84% f of x is less than or equal to g of x. 00:13:59.325 --> 00:14:03.810 align:middle line:90% 00:14:03.810 --> 00:14:09.390 align:middle line:84% So it's not difficult to realize that this is a measurable set. 00:14:09.390 --> 00:14:15.990 align:middle line:84% This is, if you'd like, f minus g. 00:14:15.990 --> 00:14:19.340 align:middle line:90% The inverse image of-- 00:14:19.340 --> 00:14:20.435 align:middle line:90% or let's do it this way. 00:14:20.435 --> 00:14:23.730 align:middle line:90% 00:14:23.730 --> 00:14:32.070 align:middle line:84% Write this as g minus f inverse image of 0 infinity. 00:14:32.070 --> 00:14:35.970 align:middle line:90% 00:14:35.970 --> 00:14:38.040 align:middle line:84% And since g and f are measurable functions, 00:14:38.040 --> 00:14:44.730 align:middle line:84% their difference is measurable, non-negative. 00:14:44.730 --> 00:14:48.296 align:middle line:90% So this is always non-negative. 00:14:48.296 --> 00:14:52.278 align:middle line:84% OK, so maybe there's a small issue 00:14:52.278 --> 00:14:54.320 align:middle line:84% with what happens at infinity, but you're dealing 00:14:54.320 --> 00:14:55.350 align:middle line:90% with that in the assignment. 00:14:55.350 --> 00:14:57.267 align:middle line:84% So I'm just going to erase this from the board 00:14:57.267 --> 00:14:59.060 align:middle line:84% and you'll just have to accept that this 00:14:59.060 --> 00:15:02.510 align:middle line:84% is measurable under the wisdom that I gave you 00:15:02.510 --> 00:15:05.210 align:middle line:84% that if you can write it down, typically it's measurable. 00:15:05.210 --> 00:15:08.090 align:middle line:90% 00:15:08.090 --> 00:15:13.440 align:middle line:84% And what is the measure of the complement 00:15:13.440 --> 00:15:17.575 align:middle line:84% is 0 because this is supposed to hold almost everywhere. 00:15:17.575 --> 00:15:48.240 align:middle line:90% 00:15:48.240 --> 00:15:51.930 align:middle line:84% OK, so I left off the fact, which strictly speaking, 00:15:51.930 --> 00:15:54.765 align:middle line:90% we might need for this, but. 00:15:54.765 --> 00:16:06.080 align:middle line:90% 00:16:06.080 --> 00:16:10.250 align:middle line:90% OK, so this is also an exercise. 00:16:10.250 --> 00:16:16.730 align:middle line:90% 00:16:16.730 --> 00:16:23.870 align:middle line:84% So then the integral of little f over E, 00:16:23.870 --> 00:16:28.760 align:middle line:84% which is equal to the integral over E of f union, f 00:16:28.760 --> 00:16:30.020 align:middle line:90% complement here-- 00:16:30.020 --> 00:16:33.110 align:middle line:90% 00:16:33.110 --> 00:16:38.180 align:middle line:84% this is equal to the integral over f, little f, 00:16:38.180 --> 00:16:43.670 align:middle line:84% plus the integral over f complement of little f 00:16:43.670 --> 00:16:49.680 align:middle line:84% So strictly speaking I didn't write why this is true down. 00:16:49.680 --> 00:16:53.070 align:middle line:84% But let's think about it just for a moment. 00:16:53.070 --> 00:16:54.620 align:middle line:84% So these are two disjoint subsets 00:16:54.620 --> 00:17:00.990 align:middle line:84% that make up E. Why is this going to be equal to this? 00:17:00.990 --> 00:17:04.109 align:middle line:84% Well, it's true for simple functions. 00:17:04.109 --> 00:17:06.980 align:middle line:84% So I mean, if I make this statement and assume phi-- 00:17:06.980 --> 00:17:11.270 align:middle line:84% I mean, f is simple, then it's not 00:17:11.270 --> 00:17:14.130 align:middle line:84% hard to convince yourself that this is true. 00:17:14.130 --> 00:17:16.040 align:middle line:84% And if it's true for simple functions, 00:17:16.040 --> 00:17:20.810 align:middle line:84% then from how we've defined the integral as to be this sup, 00:17:20.810 --> 00:17:24.380 align:middle line:84% it will carry over to general, non-negative measurable 00:17:24.380 --> 00:17:25.349 align:middle line:90% functions. 00:17:25.349 --> 00:17:30.100 align:middle line:84% So let's just accept this and you can prove it on your own. 00:17:30.100 --> 00:17:32.540 align:middle line:90% It's not difficult. 00:17:32.540 --> 00:17:38.680 align:middle line:84% But since we have this and f complement 00:17:38.680 --> 00:17:45.495 align:middle line:84% is a set of measure 0, this is the integral of little f. 00:17:45.495 --> 00:17:50.460 align:middle line:90% 00:17:50.460 --> 00:17:52.370 align:middle line:90% All right, since this is 0. 00:17:52.370 --> 00:17:55.970 align:middle line:84% Now, on capital F, little f is less than or equal to g. 00:17:55.970 --> 00:17:58.190 align:middle line:84% So this is less than or equal to the integral 00:17:58.190 --> 00:18:01.400 align:middle line:90% of g over capital F. 00:18:01.400 --> 00:18:03.530 align:middle line:84% And just going backwards, this is 00:18:03.530 --> 00:18:10.490 align:middle line:84% equal to the integral of fg plus f complement, which 00:18:10.490 --> 00:18:20.070 align:middle line:84% is equal to the integral of g, G. So 00:18:20.070 --> 00:18:28.560 align:middle line:84% modulo this equality, which I leave to you to fill in, 00:18:28.560 --> 00:18:29.370 align:middle line:90% proves the theorem. 00:18:29.370 --> 00:18:39.408 align:middle line:90% 00:18:39.408 --> 00:18:42.230 align:middle line:84% OK, so now we have the definition 00:18:42.230 --> 00:18:46.560 align:middle line:84% of Lebesgue integral for non-negative measurable 00:18:46.560 --> 00:18:47.060 align:middle line:90% functions. 00:18:47.060 --> 00:18:48.920 align:middle line:90% We have some properties of it. 00:18:48.920 --> 00:18:50.960 align:middle line:84% What's kind of missing from this list 00:18:50.960 --> 00:18:56.390 align:middle line:84% that I've given so far is that linearity, I guess, right? 00:18:56.390 --> 00:18:59.000 align:middle line:84% The integral of the sum of two non-negative measurable 00:18:59.000 --> 00:19:01.190 align:middle line:84% functions is equal to the sum of the integrals. 00:19:01.190 --> 00:19:04.280 align:middle line:84% We had that for simple functions. 00:19:04.280 --> 00:19:13.520 align:middle line:84% How do we get that for general non-negative measurable 00:19:13.520 --> 00:19:14.430 align:middle line:90% functions? 00:19:14.430 --> 00:19:21.020 align:middle line:90% 00:19:21.020 --> 00:19:23.310 align:middle line:84% OK, so what I'm about to prove is not 00:19:23.310 --> 00:19:27.690 align:middle line:84% just a tool for proving that but is one of the big three 00:19:27.690 --> 00:19:29.310 align:middle line:84% convergence theorems that you find 00:19:29.310 --> 00:19:31.410 align:middle line:84% in Lebesgue measure in integration or Lebesgue 00:19:31.410 --> 00:19:34.830 align:middle line:84% integration, which is the following-- 00:19:34.830 --> 00:19:40.090 align:middle line:90% 00:19:40.090 --> 00:19:51.290 align:middle line:84% monotone convergence theorem, which 00:19:51.290 --> 00:20:03.600 align:middle line:84% is the following that if fn is a sequence 00:20:03.600 --> 00:20:07.110 align:middle line:84% of non-negative measurable functions 00:20:07.110 --> 00:20:11.280 align:middle line:84% such that f1 is less than or equal to f2 00:20:11.280 --> 00:20:18.050 align:middle line:84% is less than or equal to f3 on E. 00:20:18.050 --> 00:20:21.742 align:middle line:84% So pointwise they are increasing. 00:20:21.742 --> 00:20:24.160 align:middle line:90% The sequence is increasing. 00:20:24.160 --> 00:20:30.340 align:middle line:84% And there exists a function f so that fn 00:20:30.340 --> 00:20:37.380 align:middle line:84% goes to f pointwise on E. So let me just recall, 00:20:37.380 --> 00:20:44.070 align:middle line:84% this means for all x and E limit as n goes to infinity 00:20:44.070 --> 00:20:47.205 align:middle line:90% of fn of x equals f of x. 00:20:47.205 --> 00:20:51.320 align:middle line:90% 00:20:51.320 --> 00:20:55.010 align:middle line:84% So in particular, f is going to be a non-negative measurable 00:20:55.010 --> 00:20:57.140 align:middle line:84% function because remember, the pointwise limit 00:20:57.140 --> 00:21:00.580 align:middle line:84% of measurable functions is measurable, 00:21:00.580 --> 00:21:03.910 align:middle line:84% so what I'm about to say makes sense. 00:21:03.910 --> 00:21:16.700 align:middle line:84% Then the integrals converge to the integrals of the limit. 00:21:16.700 --> 00:21:21.880 align:middle line:84% So the limit of the integrals is the integral of the limits. 00:21:21.880 --> 00:21:25.410 align:middle line:84% So this is a much stronger statement than anything 00:21:25.410 --> 00:21:28.650 align:middle line:84% you come across in Riemann integration. 00:21:28.650 --> 00:21:32.130 align:middle line:84% Riemann integration usually requires uniform convergence 00:21:32.130 --> 00:21:37.350 align:middle line:84% while here, at least for monotone sequences, 00:21:37.350 --> 00:21:39.030 align:middle line:84% we just need pointwise convergence. 00:21:39.030 --> 00:21:43.950 align:middle line:84% So I think there is a version of this theorem 00:21:43.950 --> 00:21:48.960 align:middle line:84% that one could state for Riemann integration. 00:21:48.960 --> 00:21:51.600 align:middle line:84% But still, just on the face of it, 00:21:51.600 --> 00:21:56.310 align:middle line:84% you have a pointwise statement implying 00:21:56.310 --> 00:21:57.670 align:middle line:90% convergence of integrals. 00:21:57.670 --> 00:22:03.150 align:middle line:84% So that should immediately kind of suggest to you 00:22:03.150 --> 00:22:05.760 align:middle line:84% that what we've built up, this Lebesgue integration, 00:22:05.760 --> 00:22:07.860 align:middle line:84% is much more powerful than Riemann integration. 00:22:07.860 --> 00:22:11.625 align:middle line:90% So let's prove the theorem. 00:22:11.625 --> 00:22:15.300 align:middle line:90% 00:22:15.300 --> 00:22:21.330 align:middle line:84% So since f1 is less than or equal to f2 00:22:21.330 --> 00:22:25.290 align:middle line:84% is less than or equal to f3 and so on, 00:22:25.290 --> 00:22:28.410 align:middle line:84% this implies the integral of E of f1 00:22:28.410 --> 00:22:32.880 align:middle line:84% is less than or equal to the integral of f2 and so on. 00:22:32.880 --> 00:22:35.680 align:middle line:90% 00:22:35.680 --> 00:22:38.040 align:middle line:90% And what else? 00:22:38.040 --> 00:22:49.930 align:middle line:84% So-- which implies that the limit as n 00:22:49.930 --> 00:22:58.820 align:middle line:84% goes to infinity of the integrals of fn 00:22:58.820 --> 00:23:03.710 align:middle line:90% exists in 0 infinity. 00:23:03.710 --> 00:23:07.570 align:middle line:84% So this is a non-negative increasing sequence 00:23:07.570 --> 00:23:09.220 align:middle line:90% of real numbers now. 00:23:09.220 --> 00:23:12.230 align:middle line:84% The integral over E of f1, that's a real number. 00:23:12.230 --> 00:23:13.180 align:middle line:90% This is a real number. 00:23:13.180 --> 00:23:14.800 align:middle line:84% All of these are non-negative numbers 00:23:14.800 --> 00:23:17.690 align:middle line:84% because this is a sup over non-negative numbers. 00:23:17.690 --> 00:23:20.870 align:middle line:84% So I have an increasing sequence of non-negative numbers. 00:23:20.870 --> 00:23:23.560 align:middle line:84% So that either has a limit, a finite limit, 00:23:23.560 --> 00:23:25.610 align:middle line:90% or it must go to infinity. 00:23:25.610 --> 00:23:28.090 align:middle line:84% That's not difficult to prove knowing what 00:23:28.090 --> 00:23:30.790 align:middle line:90% from basic analysis. 00:23:30.790 --> 00:23:34.010 align:middle line:90% 00:23:34.010 --> 00:23:53.580 align:middle line:84% Moreover, since their pointwise increasing and converging 00:23:53.580 --> 00:24:00.640 align:middle line:84% for all x, this implies that we still 00:24:00.640 --> 00:24:04.040 align:middle line:84% have f1 is less than or equal to f3 and so on. 00:24:04.040 --> 00:24:07.420 align:middle line:90% But they all sit below f. 00:24:07.420 --> 00:24:11.170 align:middle line:84% So for each x-- this is a f1 of x, f2 of x, 00:24:11.170 --> 00:24:14.230 align:middle line:84% f3 of x and so on this is an increasing sequence of real 00:24:14.230 --> 00:24:18.700 align:middle line:84% numbers converging to f of x, which is either a finite number 00:24:18.700 --> 00:24:20.880 align:middle line:90% or infinity. . 00:24:20.880 --> 00:24:25.790 align:middle line:84% So these numbers are increasing to this limit, 00:24:25.790 --> 00:24:29.285 align:middle line:84% and therefore, they must always sit below the limit. 00:24:29.285 --> 00:24:35.150 align:middle line:90% 00:24:35.150 --> 00:24:40.490 align:middle line:84% And since all of these functions sit below f, 00:24:40.490 --> 00:24:49.680 align:middle line:84% this implies that for all n, the integral of fn over E 00:24:49.680 --> 00:24:52.890 align:middle line:84% is less than or equal to the integral of f. 00:24:52.890 --> 00:24:55.800 align:middle line:84% And therefore, the limit, which we 00:24:55.800 --> 00:25:02.260 align:middle line:84% know exists as either a finite number or infinity, 00:25:02.260 --> 00:25:07.420 align:middle line:84% is less than or equal to the integral of f. 00:25:07.420 --> 00:25:10.900 align:middle line:84% So just based on the assumptions, 00:25:10.900 --> 00:25:15.660 align:middle line:84% we immediately get that one of these quantities that we want 00:25:15.660 --> 00:25:17.410 align:middle line:84% to show is equal to the other is less than 00:25:17.410 --> 00:25:21.890 align:middle line:90% or equal to the other quantity. 00:25:21.890 --> 00:25:24.700 align:middle line:90% So standard trick of analysis. 00:25:24.700 --> 00:25:26.587 align:middle line:84% If you get kind of for free one quantity is 00:25:26.587 --> 00:25:29.170 align:middle line:84% less than or equal to the other quantity that you want to show 00:25:29.170 --> 00:25:32.270 align:middle line:84% are equal, let's try and go the reverse direction. 00:25:32.270 --> 00:25:44.740 align:middle line:84% So now we show that the integral of f over E 00:25:44.740 --> 00:25:49.660 align:middle line:84% is less than or equal to the integral of the limit 00:25:49.660 --> 00:25:55.940 align:middle line:84% as n goes to infinity of the integral over E of fn. 00:25:55.940 --> 00:25:57.315 align:middle line:84% And therefore, the two are equal. 00:25:57.315 --> 00:26:01.620 align:middle line:90% 00:26:01.620 --> 00:26:03.128 align:middle line:90% All right. 00:26:03.128 --> 00:26:04.670 align:middle line:84% Now to show this, we're going to show 00:26:04.670 --> 00:26:07.400 align:middle line:84% that for every simple function less than 00:26:07.400 --> 00:26:09.320 align:middle line:84% or equal to f, the integral of that 00:26:09.320 --> 00:26:10.895 align:middle line:90% simple function sits below this. 00:26:10.895 --> 00:26:13.620 align:middle line:90% 00:26:13.620 --> 00:26:15.830 align:middle line:84% Now, we know these fn's so here's the game plan. 00:26:15.830 --> 00:26:18.230 align:middle line:84% We know the fn's are increasing to f. 00:26:18.230 --> 00:26:22.460 align:middle line:84% So if I take a simple function less than or equal to f-- 00:26:22.460 --> 00:26:26.030 align:middle line:84% if the simple function is less than f, then eventually, 00:26:26.030 --> 00:26:28.250 align:middle line:84% fn is going to pass it up, right? 00:26:28.250 --> 00:26:30.950 align:middle line:84% Because the fn's are increasing to f and the simple function 00:26:30.950 --> 00:26:32.600 align:middle line:90% sits below f. 00:26:32.600 --> 00:26:35.538 align:middle line:84% And therefore, eventually, we should 00:26:35.538 --> 00:26:37.580 align:middle line:84% have this is bigger than or equal to the integral 00:26:37.580 --> 00:26:38.580 align:middle line:90% of that simple function. 00:26:38.580 --> 00:26:41.300 align:middle line:84% Now, we're only requiring the simple function 00:26:41.300 --> 00:26:44.180 align:middle line:84% to be less than or equal to f so we'll give ourselves 00:26:44.180 --> 00:26:47.180 align:middle line:84% a little bit of room and then send that bit of room to 0. 00:26:47.180 --> 00:26:57.225 align:middle line:84% So let phi be a non-negative simple function. 00:26:57.225 --> 00:27:02.890 align:middle line:90% 00:27:02.890 --> 00:27:17.160 align:middle line:84% Phi equals sum j equals 1 to n Aj chi Aj with phi less than 00:27:17.160 --> 00:27:17.850 align:middle line:90% or equal to f. 00:27:17.850 --> 00:27:19.890 align:middle line:84% And our goal is to show that the integral of phi 00:27:19.890 --> 00:27:21.330 align:middle line:84% is less than or equal to the limit 00:27:21.330 --> 00:27:22.860 align:middle line:90% as n goes to infinity of fn. 00:27:22.860 --> 00:27:27.520 align:middle line:90% 00:27:27.520 --> 00:27:31.500 align:middle line:84% So here's that little bit of room I was referring to. 00:27:31.500 --> 00:27:35.160 align:middle line:84% Let epsilon be a small number between 0 and 1. 00:27:35.160 --> 00:27:44.490 align:middle line:84% And let En be the set of all x's and E 00:27:44.490 --> 00:27:50.400 align:middle line:84% such that fn of x is greater than or equal to 1 00:27:50.400 --> 00:27:53.770 align:middle line:90% minus epsilon f of x-- 00:27:53.770 --> 00:27:54.610 align:middle line:90% I mean, phi of x. 00:27:54.610 --> 00:28:02.130 align:middle line:90% 00:28:02.130 --> 00:28:14.200 align:middle line:84% So note for all x in E, 1 minus epsilon times phi of x, this 00:28:14.200 --> 00:28:16.660 align:middle line:84% is strictly less than now f of x. 00:28:16.660 --> 00:28:19.900 align:middle line:84% So we had phi of x is less than or equal to f of x. 00:28:19.900 --> 00:28:23.890 align:middle line:84% But now, if I multiply this by a small number that's close to-- 00:28:23.890 --> 00:28:27.160 align:middle line:84% or a number that's slightly less than 1, 00:28:27.160 --> 00:28:29.170 align:middle line:84% then I will have strict inequality. 00:28:29.170 --> 00:28:37.870 align:middle line:90% 00:28:37.870 --> 00:28:51.650 align:middle line:84% So in particular, then since for all x and E I have a-- 00:28:51.650 --> 00:28:59.640 align:middle line:90% 00:28:59.640 --> 00:29:01.950 align:middle line:90% equals f of x. 00:29:01.950 --> 00:29:05.340 align:middle line:84% This implies that every x must eventually 00:29:05.340 --> 00:29:07.650 align:middle line:84% lie in one of these E sub fn's right? 00:29:07.650 --> 00:29:11.220 align:middle line:84% Because 1 minus epsilon phi of x is less than f of x. 00:29:11.220 --> 00:29:14.610 align:middle line:84% The fn's are approaching f of x so it must pass up 00:29:14.610 --> 00:29:17.130 align:middle line:84% this value at some point in its quest 00:29:17.130 --> 00:29:20.910 align:middle line:84% to get to f of x, or at least close to f of x. 00:29:20.910 --> 00:29:27.950 align:middle line:84% So simply from this fact, this implies that the union over n 00:29:27.950 --> 00:29:33.693 align:middle line:84% equals 1 to infinity of the En's gives me E. 00:29:33.693 --> 00:29:34.610 align:middle line:90% Let me highlight this. 00:29:34.610 --> 00:29:43.640 align:middle line:90% 00:29:43.640 --> 00:29:48.210 align:middle line:84% Now, since these functions are increasing-- 00:29:48.210 --> 00:29:51.020 align:middle line:84% I should say, they're pointwise increasing. 00:29:51.020 --> 00:29:52.820 align:middle line:84% Not that they are increasing functions, 00:29:52.820 --> 00:29:54.620 align:middle line:84% but they are pointwise increasing. 00:29:54.620 --> 00:29:58.550 align:middle line:84% So f1 is less than or equal to f2 is less than or equal to f3 00:29:58.550 --> 00:29:59.840 align:middle line:90% and so on. 00:29:59.840 --> 00:30:04.640 align:middle line:84% This implies that E1 is contained in E2 00:30:04.640 --> 00:30:08.000 align:middle line:90% is contained in E3 and so on. 00:30:08.000 --> 00:30:11.790 align:middle line:84% If I have some n so that x is in this set-- 00:30:11.790 --> 00:30:13.760 align:middle line:84% so fn of x is bigger than or equal to 1 00:30:13.760 --> 00:30:15.980 align:middle line:90% minus epsilon phi of x-- 00:30:15.980 --> 00:30:19.053 align:middle line:84% then f of n plus 1 of x is bigger than or equal to fn 00:30:19.053 --> 00:30:20.720 align:middle line:84% of x, which is bigger than or equal to 1 00:30:20.720 --> 00:30:22.280 align:middle line:90% minus epsilon phi of x. 00:30:22.280 --> 00:30:27.480 align:middle line:84% And therefore, that x is then En plus 1. 00:30:27.480 --> 00:30:33.630 align:middle line:84% So this is not only a sequence of sets whose union gives me E, 00:30:33.630 --> 00:30:37.270 align:middle line:84% they're an increasing sequence of sets. 00:30:37.270 --> 00:30:39.250 align:middle line:84% Increasing in the sense of inclusion. 00:30:39.250 --> 00:30:42.830 align:middle line:90% 00:30:42.830 --> 00:30:43.330 align:middle line:90% OK? 00:30:43.330 --> 00:30:53.450 align:middle line:90% 00:30:53.450 --> 00:30:58.250 align:middle line:84% Now, we're going to use these two highlighted things in just 00:30:58.250 --> 00:31:03.360 align:middle line:84% a minute, along with continuity of Lebesgue measure 00:31:03.360 --> 00:31:05.500 align:middle line:90% to get what we want. 00:31:05.500 --> 00:31:16.610 align:middle line:90% So we have the integral of E fn. 00:31:16.610 --> 00:31:20.120 align:middle line:84% This is less than or equal to the integral of fn 00:31:20.120 --> 00:31:24.080 align:middle line:90% over a smaller set, so E sub n. 00:31:24.080 --> 00:31:25.940 align:middle line:90% Now, on E sub n-- 00:31:25.940 --> 00:31:27.500 align:middle line:84% remember, the E sub n's are defined 00:31:27.500 --> 00:31:29.990 align:middle line:84% as where fn is bigger than or equal to 1 00:31:29.990 --> 00:31:32.000 align:middle line:90% minus epsilon phi of x. 00:31:32.000 --> 00:31:37.175 align:middle line:84% So this is bigger than or equal to E sub n 1 minus phi-- 00:31:37.175 --> 00:31:40.730 align:middle line:90% 00:31:40.730 --> 00:31:42.470 align:middle line:90% sorry, what am I doing? 00:31:42.470 --> 00:31:50.080 align:middle line:84% Minus epsilon phi, which is equal to 1 minus 00:31:50.080 --> 00:31:54.490 align:middle line:84% epsilon times the integral of E sub n over phi. 00:31:54.490 --> 00:31:59.860 align:middle line:84% And this is, by definition, equal to a sum 00:31:59.860 --> 00:32:02.110 align:middle line:90% from j equals 1 to n. 00:32:02.110 --> 00:32:05.380 align:middle line:84% Over the set E sub n, I get integral Aj measure 00:32:05.380 --> 00:32:09.460 align:middle line:90% of Aj intersect E sub n. 00:32:09.460 --> 00:32:15.940 align:middle line:84% And I made a small notational error. 00:32:15.940 --> 00:32:22.650 align:middle line:84% Let's change this into an m since we 00:32:22.650 --> 00:32:24.570 align:middle line:90% have n already denoting the-- 00:32:24.570 --> 00:32:28.690 align:middle line:90% 00:32:28.690 --> 00:32:31.610 align:middle line:84% indexing the functions, we do not want this n right here. 00:32:31.610 --> 00:32:32.590 align:middle line:90% So that should be an m. 00:32:32.590 --> 00:32:34.360 align:middle line:84% It's just a fixed finite number depending 00:32:34.360 --> 00:32:35.318 align:middle line:90% on the simple function. 00:32:35.318 --> 00:32:38.100 align:middle line:90% 00:32:38.100 --> 00:32:41.520 align:middle line:90% So I have this for all n. 00:32:41.520 --> 00:32:46.640 align:middle line:84% And therefore, the limit as n goes 00:32:46.640 --> 00:32:49.610 align:middle line:84% to infinity of the integral over E 00:32:49.610 --> 00:32:55.070 align:middle line:84% of f sub n is less than or equal to the limit as n goes 00:32:55.070 --> 00:32:59.540 align:middle line:84% to infinity of 1 minus epsilon times sum from j 00:32:59.540 --> 00:33:09.130 align:middle line:84% equals 1 to m Aj measure of Aj E sub n. 00:33:09.130 --> 00:33:16.120 align:middle line:84% Now, the E sub n's are increasing to E, and therefore, 00:33:16.120 --> 00:33:18.430 align:middle line:90% for each fixed j-- 00:33:18.430 --> 00:33:22.390 align:middle line:84% so in fact, let's pause on this real 00:33:22.390 --> 00:33:28.470 align:middle line:84% quick and come back to this thing. 00:33:28.470 --> 00:33:30.600 align:middle line:84% We're going to eventually take the limit as n goes 00:33:30.600 --> 00:33:32.140 align:middle line:90% to infinity of this quantity. 00:33:32.140 --> 00:33:34.170 align:middle line:84% So let's look at what this does as n 00:33:34.170 --> 00:33:39.480 align:middle line:84% goes to infinity since by those two things that I highlighted, 00:33:39.480 --> 00:33:45.900 align:middle line:84% that En's are increasing subsets of E whose union gives me E, 00:33:45.900 --> 00:33:47.010 align:middle line:90% I get-- 00:33:47.010 --> 00:34:00.730 align:middle line:84% since E1 Aj is a subset of E2 Aj subset of E3 and so on. 00:34:00.730 --> 00:34:07.690 align:middle line:84% And the union n equals 1 to infinity of E sub n intersect 00:34:07.690 --> 00:34:16.380 align:middle line:84% Aj equals Aj, because then this is just 00:34:16.380 --> 00:34:19.409 align:middle line:84% going to be equal to the union of the E sub n's intersect Aj. 00:34:19.409 --> 00:34:22.770 align:middle line:84% That just gives me E intersect Aj, which is just Aj. 00:34:22.770 --> 00:34:29.204 align:middle line:84% We get by the continuity of Lebesgue measure-- 00:34:29.204 --> 00:34:31.110 align:middle line:90% this implies for all j-- 00:34:31.110 --> 00:34:39.210 align:middle line:84% that the limit as n goes to infinity of the measure of Aj 00:34:39.210 --> 00:34:49.719 align:middle line:84% intersect En is equal to the measure of Aj, 00:34:49.719 --> 00:35:02.500 align:middle line:84% which, as we just said, this En intersect Aj, which-- 00:35:02.500 --> 00:35:07.240 align:middle line:84% remember, this set is equal to Aj, so this is measure of Aj. 00:35:07.240 --> 00:35:12.160 align:middle line:84% So from the two yellow boxes we had before, 00:35:12.160 --> 00:35:14.470 align:middle line:90% we get this useful one. 00:35:14.470 --> 00:35:16.120 align:middle line:84% For all j, we have the limit as n 00:35:16.120 --> 00:35:19.720 align:middle line:84% goes to infinity of the measure of Aj intersect En is 00:35:19.720 --> 00:35:22.120 align:middle line:90% equal to the measure of Aj. 00:35:22.120 --> 00:35:27.250 align:middle line:84% So now, we'll stick this into this inequality 00:35:27.250 --> 00:35:29.390 align:middle line:90% after we take the limit. 00:35:29.390 --> 00:35:32.680 align:middle line:84% So I got ahead of myself a minute ago. 00:35:32.680 --> 00:35:35.770 align:middle line:84% Thus, limit as n goes to infinity 00:35:35.770 --> 00:35:39.700 align:middle line:84% of the integral over E of fn, this 00:35:39.700 --> 00:35:45.370 align:middle line:84% is bigger than or equal to limit as n 00:35:45.370 --> 00:35:51.800 align:middle line:84% goes to infinity of 1 minus epsilon m 00:35:51.800 --> 00:35:59.600 align:middle line:84% now Aj measure of Aj intersect En. 00:35:59.600 --> 00:36:03.440 align:middle line:84% Now, these numbers here all converge to-- 00:36:03.440 --> 00:36:05.930 align:middle line:84% each of these numbers here for each j 00:36:05.930 --> 00:36:09.950 align:middle line:84% converges as n goes to infinity to the measure of A sub j. 00:36:09.950 --> 00:36:14.420 align:middle line:84% So the limit is then equal to 1 minus epsilon 00:36:14.420 --> 00:36:18.080 align:middle line:84% times the integral j equals 1 to m times-- 00:36:18.080 --> 00:36:22.166 align:middle line:90% 00:36:22.166 --> 00:36:24.400 align:middle line:84% what am I writing the integral for? 00:36:24.400 --> 00:36:28.000 align:middle line:90% Aj measure of Aj. 00:36:28.000 --> 00:36:32.200 align:middle line:84% And this is just equal to 1 minus integral, 00:36:32.200 --> 00:36:36.290 align:middle line:84% by definition, the integral of phi. 00:36:36.290 --> 00:36:41.780 align:middle line:84% So I've shown that for all epsilon between 0 and 1, 00:36:41.780 --> 00:36:54.680 align:middle line:84% this 1 minus epsilon times the integral of phi is less than 00:36:54.680 --> 00:37:01.370 align:middle line:84% or equal to this number over here, which may be infinite, 00:37:01.370 --> 00:37:02.090 align:middle line:90% may be finite. 00:37:02.090 --> 00:37:05.590 align:middle line:90% 00:37:05.590 --> 00:37:07.180 align:middle line:84% And since this hold for all epsilon, 00:37:07.180 --> 00:37:08.485 align:middle line:90% I can send epsilon to 0. 00:37:08.485 --> 00:37:19.550 align:middle line:90% 00:37:19.550 --> 00:37:23.750 align:middle line:84% So I have this inequality between now fixed things 00:37:23.750 --> 00:37:27.470 align:middle line:84% along with an epsilon here, so I can send epsilon to 0 00:37:27.470 --> 00:37:31.400 align:middle line:84% and I get the integral of phi is less than or equal to the limit 00:37:31.400 --> 00:37:35.775 align:middle line:90% as n goes to infinity of fn. 00:37:35.775 --> 00:37:40.200 align:middle line:90% 00:37:40.200 --> 00:37:43.320 align:middle line:84% And since phi is an arbitrary simple function that's less 00:37:43.320 --> 00:37:46.200 align:middle line:84% than or equal to f, the sup over all of these-- which is, 00:37:46.200 --> 00:37:49.850 align:middle line:84% by definition, the integral of f-- 00:37:49.850 --> 00:37:59.340 align:middle line:84% is less than or equal to the limit of the integrals. 00:37:59.340 --> 00:38:07.310 align:middle line:90% 00:38:07.310 --> 00:38:07.823 align:middle line:90% All right. 00:38:07.823 --> 00:38:08.990 align:middle line:90% That's the end of the proof. 00:38:08.990 --> 00:38:14.300 align:middle line:90% 00:38:14.300 --> 00:38:28.570 align:middle line:84% So monotone convergence theorem, a very useful theorem, 00:38:28.570 --> 00:38:31.180 align:middle line:84% important theorem throughout all of this. 00:38:31.180 --> 00:38:37.340 align:middle line:90% 00:38:37.340 --> 00:38:43.990 align:middle line:84% OK, so let's get a few applications from this. 00:38:43.990 --> 00:38:46.870 align:middle line:84% So this first one is kind of a way 00:38:46.870 --> 00:38:48.667 align:middle line:84% how would you evaluate now this integral? 00:38:48.667 --> 00:38:50.500 align:middle line:84% Remember, the integral, which I just erased, 00:38:50.500 --> 00:38:53.620 align:middle line:84% is defined as the sup over all integrals of simple functions. 00:38:53.620 --> 00:38:56.740 align:middle line:84% So in order to actually compute the integral 00:38:56.740 --> 00:38:59.740 align:middle line:84% of a non-negative measurable function, 00:38:59.740 --> 00:39:02.320 align:middle line:84% I would have to find out the integral 00:39:02.320 --> 00:39:04.810 align:middle line:84% of every simple function less than or equal to it 00:39:04.810 --> 00:39:06.610 align:middle line:84% and take the sup over that whole set, which 00:39:06.610 --> 00:39:12.010 align:middle line:84% is kind of a useless or impossible way of computing 00:39:12.010 --> 00:39:14.290 align:middle line:90% the integral. 00:39:14.290 --> 00:39:18.520 align:middle line:84% It's similar to when you come across Riemann integration 00:39:18.520 --> 00:39:20.290 align:middle line:84% and the Riemann integral is defined 00:39:20.290 --> 00:39:24.520 align:middle line:90% as the limit of Riemann sums. 00:39:24.520 --> 00:39:28.480 align:middle line:84% Nobody-- you can compute maybe three integrals just 00:39:28.480 --> 00:39:29.350 align:middle line:90% from Riemann sum. 00:39:29.350 --> 00:39:32.377 align:middle line:84% So we need a more efficient way of being 00:39:32.377 --> 00:39:33.960 align:middle line:84% able to compute the Lebesgue integral, 00:39:33.960 --> 00:39:35.640 align:middle line:84% and the monotone convergence theorem 00:39:35.640 --> 00:39:38.350 align:middle line:90% gives us that kind of for free. 00:39:38.350 --> 00:39:45.060 align:middle line:84% So we have the following, if f is 00:39:45.060 --> 00:39:51.445 align:middle line:84% a non-negative measurable function and phi 00:39:51.445 --> 00:40:07.540 align:middle line:84% n is a sequence of simple functions, which are all 00:40:07.540 --> 00:40:21.610 align:middle line:84% non-negative and pointwise increasing and converging 00:40:21.610 --> 00:40:33.240 align:middle line:84% pointwise to f then the integral over E of f 00:40:33.240 --> 00:40:36.990 align:middle line:84% is equal to the limit as n goes to infinity 00:40:36.990 --> 00:40:42.570 align:middle line:84% of the integral of the simple functions. 00:40:42.570 --> 00:40:45.760 align:middle line:84% So back when we discussed measurable functions, 00:40:45.760 --> 00:40:48.510 align:middle line:84% we actually constructed such a sequence of simple functions 00:40:48.510 --> 00:40:51.130 align:middle line:84% that satisfies the hypotheses of this theorem. 00:40:51.130 --> 00:40:52.650 align:middle line:84% So this is not a vacuous theorem. 00:40:52.650 --> 00:40:54.480 align:middle line:84% But this theorem tells you that if you 00:40:54.480 --> 00:40:56.130 align:middle line:84% want to compute the integral of f, 00:40:56.130 --> 00:40:59.970 align:middle line:84% just take any sequence of simple functions increasing up 00:40:59.970 --> 00:41:02.340 align:middle line:84% to f and compute the limit of the integral. 00:41:02.340 --> 00:41:04.965 align:middle line:84% And that'll give you the integral of f. 00:41:04.965 --> 00:41:07.590 align:middle line:90% 00:41:07.590 --> 00:41:09.660 align:middle line:84% Now, there's just this-- this follows immediately 00:41:09.660 --> 00:41:11.243 align:middle line:84% from the monotone convergence theorem. 00:41:11.243 --> 00:41:15.600 align:middle line:84% I have the taking fn is equal to the phi n's. 00:41:15.600 --> 00:41:18.000 align:middle line:84% So there's no proof to go with that. 00:41:18.000 --> 00:41:22.320 align:middle line:90% 00:41:22.320 --> 00:41:26.260 align:middle line:84% The next theorem is linearity of the integral. 00:41:26.260 --> 00:41:32.220 align:middle line:84% So if f and g are two non-negative measurable 00:41:32.220 --> 00:41:39.020 align:middle line:84% functions, then the integral of f plus g 00:41:39.020 --> 00:41:45.515 align:middle line:84% is equal to the integral of f plus the integral of g. 00:41:45.515 --> 00:41:48.770 align:middle line:90% 00:41:48.770 --> 00:41:53.000 align:middle line:84% Now, note there's no ambiguity with how to define, 00:41:53.000 --> 00:41:54.400 align:middle line:90% so we kind of had a-- 00:41:54.400 --> 00:41:58.550 align:middle line:84% there's some touchy business about adding and subtracting 00:41:58.550 --> 00:42:00.740 align:middle line:84% two extended real valued measurable functions, 00:42:00.740 --> 00:42:02.448 align:middle line:84% but there's none of that here since these 00:42:02.448 --> 00:42:06.480 align:middle line:84% are both non-negative measurable extended real valued functions. 00:42:06.480 --> 00:42:08.240 align:middle line:84% So this will always only be of the form 00:42:08.240 --> 00:42:11.160 align:middle line:84% infinity plus infinity, which we define to be infinity. 00:42:11.160 --> 00:42:14.300 align:middle line:84% So just let me make that small note. 00:42:14.300 --> 00:42:17.975 align:middle line:84% So the integral is linear, so what's the proof? 00:42:17.975 --> 00:42:22.370 align:middle line:90% 00:42:22.370 --> 00:42:42.400 align:middle line:84% Let phi n and psi n be two sequences of simple functions 00:42:42.400 --> 00:42:46.070 align:middle line:84% such that they're increasing to f and g, respectively. 00:42:46.070 --> 00:42:50.740 align:middle line:84% So 0 is less than or equal to phi 1 00:42:50.740 --> 00:42:54.790 align:middle line:84% is less than or equal to phi 2 and so on. 00:42:54.790 --> 00:43:02.050 align:middle line:84% And phi n converges to f pointwise on E. 00:43:02.050 --> 00:43:04.120 align:middle line:90% OK, so I should have-- 00:43:04.120 --> 00:43:07.030 align:middle line:84% everything's happening on this set E. 00:43:07.030 --> 00:43:08.425 align:middle line:90% And the same for the psis. 00:43:08.425 --> 00:43:12.510 align:middle line:90% 00:43:12.510 --> 00:43:18.195 align:middle line:84% And psi n converging to g pointwise. 00:43:18.195 --> 00:43:25.270 align:middle line:90% 00:43:25.270 --> 00:43:29.740 align:middle line:84% Then, if I take the sum of these two simple functions 00:43:29.740 --> 00:43:31.210 align:middle line:84% or sequences of simple functions, 00:43:31.210 --> 00:43:38.580 align:middle line:84% I get an increasing sequence of simple functions. 00:43:38.580 --> 00:43:41.740 align:middle line:90% 00:43:41.740 --> 00:43:52.430 align:middle line:84% And phi n plus psi n converges to f plus g pointwise. 00:43:52.430 --> 00:43:56.820 align:middle line:90% 00:43:56.820 --> 00:43:58.860 align:middle line:84% And by this theorem that followed immediately 00:43:58.860 --> 00:44:05.730 align:middle line:84% from the monotone convergence theorem, 00:44:05.730 --> 00:44:12.960 align:middle line:84% I get that the integral of f plus g over E-- this 00:44:12.960 --> 00:44:17.790 align:middle line:84% is equal to the limit as n goes to infinity of integral 00:44:17.790 --> 00:44:20.850 align:middle line:90% of E of phi n plus psi n. 00:44:20.850 --> 00:44:23.910 align:middle line:84% And now, we've proved linearity of the integral 00:44:23.910 --> 00:44:26.470 align:middle line:90% for simple functions. 00:44:26.470 --> 00:44:29.850 align:middle line:84% So this is equal to the limit as n 00:44:29.850 --> 00:44:33.225 align:middle line:84% goes to infinity of the integral of E 00:44:33.225 --> 00:44:39.950 align:middle line:84% of phi n plus the integral of E of psi n. 00:44:39.950 --> 00:44:44.630 align:middle line:84% And again, by the theorem that I just stated a minute ago, 00:44:44.630 --> 00:44:47.030 align:middle line:84% by the monotone convergence theorem, this converges 00:44:47.030 --> 00:44:49.310 align:middle line:84% through the integral of f, this converges 00:44:49.310 --> 00:44:50.510 align:middle line:90% through the integral of g. 00:44:50.510 --> 00:44:59.950 align:middle line:84% So the limit of the sum is the sum of the limits, and I get g. 00:44:59.950 --> 00:45:08.505 align:middle line:90% 00:45:08.505 --> 00:45:10.970 align:middle line:90% OK. 00:45:10.970 --> 00:45:13.670 align:middle line:84% Using the same kind of argument, if you like-- 00:45:13.670 --> 00:45:15.920 align:middle line:84% except now not for two functions, 00:45:15.920 --> 00:45:17.760 align:middle line:90% but for one function-- 00:45:17.760 --> 00:45:21.730 align:middle line:84% you can show that the integral of a function 00:45:21.730 --> 00:45:24.793 align:middle line:84% over a union of two disjoint sets 00:45:24.793 --> 00:45:25.960 align:middle line:90% is the sum of the integrals. 00:45:25.960 --> 00:45:29.000 align:middle line:84% This is something that I pointed to 00:45:29.000 --> 00:45:31.840 align:middle line:84% but didn't prove at the very beginning of this lecture. 00:45:31.840 --> 00:45:36.040 align:middle line:84% But using that that is true for simple functions and this 00:45:36.040 --> 00:45:38.240 align:middle line:84% argument using the monotone convergence theorem-- 00:45:38.240 --> 00:45:41.440 align:middle line:84% which didn't require what I had proved earlier so this is not 00:45:41.440 --> 00:45:44.340 align:middle line:90% a circular argument-- 00:45:44.340 --> 00:45:46.950 align:middle line:84% you can prove that the integral of non-negative measurable 00:45:46.950 --> 00:45:49.650 align:middle line:84% function over a union of two disjoint sets 00:45:49.650 --> 00:45:53.340 align:middle line:84% is the sum of the integrals, one over the first set, 00:45:53.340 --> 00:45:54.600 align:middle line:90% one over the second set. 00:45:54.600 --> 00:45:57.312 align:middle line:90% 00:45:57.312 --> 00:46:02.150 align:middle line:90% All right, so that's cool. 00:46:02.150 --> 00:46:06.530 align:middle line:90% This integral is linear over-- 00:46:06.530 --> 00:46:10.940 align:middle line:84% so the integral of the sum of two measurable functions 00:46:10.940 --> 00:46:12.440 align:middle line:90% is the sum of the integrals. 00:46:12.440 --> 00:46:15.860 align:middle line:84% What's even better is that the integral over an infinite sum 00:46:15.860 --> 00:46:20.070 align:middle line:84% is equal to the infinite sum of the integrals. 00:46:20.070 --> 00:46:26.140 align:middle line:90% So actually, let me state-- 00:46:26.140 --> 00:46:27.550 align:middle line:90% or let's just do this. 00:46:27.550 --> 00:46:31.564 align:middle line:90% 00:46:31.564 --> 00:46:41.990 align:middle line:84% If fn is a sequence of non-negative measurable 00:46:41.990 --> 00:46:50.200 align:middle line:84% functions, then the integral of the sum of E 00:46:50.200 --> 00:46:56.600 align:middle line:84% is equal to the infinite sum of the integrals. 00:46:56.600 --> 00:47:01.550 align:middle line:90% 00:47:01.550 --> 00:47:05.090 align:middle line:84% Well, first off, this is a well-defined function 00:47:05.090 --> 00:47:07.190 align:middle line:84% because it's a sum of non-negative. 00:47:07.190 --> 00:47:09.860 align:middle line:84% So pointwise for each x, this is a sum 00:47:09.860 --> 00:47:11.820 align:middle line:90% of non-negative real numbers. 00:47:11.820 --> 00:47:13.730 align:middle line:84% So that's either going to be a finite number 00:47:13.730 --> 00:47:16.650 align:middle line:84% if this series converges or it's going to be infinite, 00:47:16.650 --> 00:47:17.150 align:middle line:90% all right? 00:47:17.150 --> 00:47:20.600 align:middle line:84% Remember, we're allowing extended real value 00:47:20.600 --> 00:47:24.200 align:middle line:84% non-negative measurable functions in our framework 00:47:24.200 --> 00:47:25.370 align:middle line:90% for now. 00:47:25.370 --> 00:47:27.450 align:middle line:90% So this is meaningful. 00:47:27.450 --> 00:47:30.140 align:middle line:84% And it's a measurable function by stuff 00:47:30.140 --> 00:47:34.420 align:middle line:84% we proved in the section on measurable functions. 00:47:34.420 --> 00:47:38.685 align:middle line:84% OK, so the proof is pretty short. 00:47:38.685 --> 00:47:48.190 align:middle line:90% 00:47:48.190 --> 00:47:55.330 align:middle line:84% By an induction argument and the previous theorem 00:47:55.330 --> 00:48:07.730 align:middle line:84% for the sum of two functions, we have the statement 00:48:07.730 --> 00:48:12.170 align:middle line:84% that for every fixed natural number of capital 00:48:12.170 --> 00:48:16.400 align:middle line:84% N, the integral of the sum, n equals 1 00:48:16.400 --> 00:48:26.620 align:middle line:84% to capital N fn E is equal to the finite sum 00:48:26.620 --> 00:48:27.610 align:middle line:90% of the integrals. 00:48:27.610 --> 00:48:34.180 align:middle line:90% 00:48:34.180 --> 00:48:37.360 align:middle line:84% So if you can do something for two, 00:48:37.360 --> 00:48:41.500 align:middle line:84% usually you can do something for n by an induction argument. 00:48:41.500 --> 00:48:44.260 align:middle line:84% So I'll leave the details of this to you 00:48:44.260 --> 00:48:47.298 align:middle line:84% or you can just believe it based on how many induction arguments 00:48:47.298 --> 00:48:48.340 align:middle line:90% you've done in your life. 00:48:48.340 --> 00:48:51.060 align:middle line:90% 00:48:51.060 --> 00:48:53.460 align:middle line:90% So we have this. 00:48:53.460 --> 00:49:04.060 align:middle line:84% And so since n equals 1, 1 fn is less than 00:49:04.060 --> 00:49:07.720 align:middle line:84% or equal to the sum from n equals 1 to 2 of fn 00:49:07.720 --> 00:49:11.650 align:middle line:84% is less than or equal to sum from n equals 1 to 3 fn. 00:49:11.650 --> 00:49:14.330 align:middle line:84% Because these are all non-negative functions, 00:49:14.330 --> 00:49:19.840 align:middle line:84% so adding non-negative functions to something only increases it. 00:49:19.840 --> 00:49:22.670 align:middle line:90% 00:49:22.670 --> 00:49:32.510 align:middle line:84% And sum from n equals 1 to n converges 00:49:32.510 --> 00:49:45.700 align:middle line:84% to pointwise simply by defining this to be the limit 00:49:45.700 --> 00:49:49.090 align:middle line:84% as capital N goes to infinity of fn of x pointwise. 00:49:49.090 --> 00:49:55.123 align:middle line:90% 00:49:55.123 --> 00:49:57.040 align:middle line:84% All right, since I have these two things, then 00:49:57.040 --> 00:50:06.130 align:middle line:84% by monotone convergence theorem, I get that the integral of n 00:50:06.130 --> 00:50:11.770 align:middle line:84% equals 1 to infinity of fn of E. this is equal to the limit 00:50:11.770 --> 00:50:18.340 align:middle line:84% as capital N goes to infinity of the integral of E sum from n 00:50:18.340 --> 00:50:27.740 align:middle line:84% equals 1 to capital N, which, by what we have right here, 00:50:27.740 --> 00:50:32.720 align:middle line:84% is equal to the limit as capital N goes to infinity of-- now 00:50:32.720 --> 00:50:37.550 align:middle line:90% the finite sum comes out. 00:50:37.550 --> 00:50:42.065 align:middle line:84% And this is by definition this infinite sum. 00:50:42.065 --> 00:50:45.490 align:middle line:90% 00:50:45.490 --> 00:50:53.830 align:middle line:84% So for non-negative measurable functions, 00:50:53.830 --> 00:50:57.940 align:middle line:84% the integral of the sum is equal to the sum of the integrals, 00:50:57.940 --> 00:50:59.440 align:middle line:90% even for an infinite sum. 00:50:59.440 --> 00:51:05.930 align:middle line:90% 00:51:05.930 --> 00:51:11.090 align:middle line:84% So again, this is simply false for if I replace everything 00:51:11.090 --> 00:51:12.140 align:middle line:90% by Riemann integration. 00:51:12.140 --> 00:51:15.710 align:middle line:84% Because, in fact, I can come up with a sequence of functions, 00:51:15.710 --> 00:51:18.110 align:middle line:84% fn, whose Riemann integral is always 0, 00:51:18.110 --> 00:51:19.880 align:middle line:84% but the sum is not Riemann integrable. 00:51:19.880 --> 00:51:23.390 align:middle line:84% Think of taking fn to be the function, which 00:51:23.390 --> 00:51:27.815 align:middle line:90% is 0 off of a rational number. 00:51:27.815 --> 00:51:31.130 align:middle line:90% 00:51:31.130 --> 00:51:36.870 align:middle line:84% And then so first, enumerate the rationals Q1, Q2, Q3, Q4, 00:51:36.870 --> 00:51:38.990 align:middle line:90% and so on. 00:51:38.990 --> 00:51:41.630 align:middle line:84% And take fn to be the function that 00:51:41.630 --> 00:51:46.940 align:middle line:84% is 0 when x is not equal to Qn and 1 when x is equal to Qn. 00:51:46.940 --> 00:51:48.890 align:middle line:84% Then the infinite sum is just going 00:51:48.890 --> 00:51:53.000 align:middle line:84% to be the indicator function of the rational, say, in 0, 1. 00:51:53.000 --> 00:51:56.270 align:middle line:84% That's not Riemann integrable, but the sum of these integrals 00:51:56.270 --> 00:51:57.440 align:middle line:90% is just 0. 00:51:57.440 --> 00:51:59.670 align:middle line:84% So this is not true for Riemann integrals again. 00:51:59.670 --> 00:52:03.125 align:middle line:84% So we're doing something much more powerful here. 00:52:03.125 --> 00:52:14.710 align:middle line:90% 00:52:14.710 --> 00:52:16.870 align:middle line:90% OK. 00:52:16.870 --> 00:52:26.960 align:middle line:84% Let's do some more properties of the integral. 00:52:26.960 --> 00:52:30.670 align:middle line:84% Now, back to properties of the integral. 00:52:30.670 --> 00:52:36.410 align:middle line:90% 00:52:36.410 --> 00:52:44.400 align:middle line:84% So if I have a non-negative measurable function, 00:52:44.400 --> 00:52:55.820 align:middle line:84% then the integral of f equals 0 if and only if x equals 0 00:52:55.820 --> 00:53:11.120 align:middle line:84% almost everywhere on E. So now what's-- 00:53:11.120 --> 00:53:14.300 align:middle line:84% this is a two-way street, so one direction. 00:53:14.300 --> 00:53:17.510 align:middle line:84% If f is equal to 0 almost everywhere, 00:53:17.510 --> 00:53:20.762 align:middle line:84% then it's less than or equal to 0 almost everywhere. 00:53:20.762 --> 00:53:22.220 align:middle line:84% And therefore, the integral of f is 00:53:22.220 --> 00:53:24.830 align:middle line:84% less than or equal to the integral of 0, 00:53:24.830 --> 00:53:26.390 align:middle line:90% and the integral of 0 is 0. 00:53:26.390 --> 00:53:36.530 align:middle line:84% So this direction follows from the fact 00:53:36.530 --> 00:53:40.400 align:middle line:84% that f is less than or equal to 0 almost everywhere, 00:53:40.400 --> 00:53:44.270 align:middle line:84% which implies that the integral of f over E 00:53:44.270 --> 00:53:48.020 align:middle line:84% is less than or equal to the integral of 0, which 00:53:48.020 --> 00:53:49.040 align:middle line:90% you can check is 0. 00:53:49.040 --> 00:53:52.090 align:middle line:90% 00:53:52.090 --> 00:53:54.760 align:middle line:84% And this is a non-negative quantity, so. 00:53:54.760 --> 00:53:57.360 align:middle line:90% 00:53:57.360 --> 00:54:00.060 align:middle line:84% So now, how about the other direction 00:54:00.060 --> 00:54:03.420 align:middle line:84% that the integral of f being 0 for a non-negative measurable 00:54:03.420 --> 00:54:09.090 align:middle line:84% function implies f is 0 almost everywhere. 00:54:09.090 --> 00:54:18.030 align:middle line:84% So let's let fn to be the set of all x's and E 00:54:18.030 --> 00:54:24.525 align:middle line:84% such that f of x is bigger than 1 over n. 00:54:24.525 --> 00:54:29.910 align:middle line:90% 00:54:29.910 --> 00:54:37.540 align:middle line:84% And let's let f be the set of all x's such that f of x 00:54:37.540 --> 00:54:40.030 align:middle line:90% is bigger than 0. 00:54:40.030 --> 00:54:46.460 align:middle line:84% Now, if x is-- so this is x and E, I should say. 00:54:46.460 --> 00:54:49.340 align:middle line:84% Now, if I have an x where f of x is bigger than 0, 00:54:49.340 --> 00:54:52.190 align:middle line:84% then at least for some large n, f 00:54:52.190 --> 00:55:05.330 align:middle line:84% of x will be bigger than 1 over n So then the union from n 00:55:05.330 --> 00:55:09.275 align:middle line:84% equals 1 to infinity of the fn's equals f. 00:55:09.275 --> 00:55:11.800 align:middle line:90% 00:55:11.800 --> 00:55:14.560 align:middle line:84% I mean, each of these is a subset of capital F, 00:55:14.560 --> 00:55:16.510 align:middle line:84% so their union is contained in capital F 00:55:16.510 --> 00:55:18.927 align:middle line:84% and I've just-- the argument I gave a minute ago shows you 00:55:18.927 --> 00:55:21.100 align:middle line:84% that capital F is contained in the union. 00:55:21.100 --> 00:55:25.390 align:middle line:90% So this union equals f. 00:55:25.390 --> 00:55:32.190 align:middle line:84% And just by how it's defined, 1 is bigger 00:55:32.190 --> 00:55:36.540 align:middle line:84% than or equal to a 1/2, which is bigger than the 1/3, 00:55:36.540 --> 00:55:42.930 align:middle line:84% f1 is contained in f2 is contained in f3 and so on. 00:55:42.930 --> 00:55:47.160 align:middle line:84% If f of x is bigger than 1/2, it's certainly bigger than 1/3. 00:55:47.160 --> 00:55:51.560 align:middle line:90% 00:55:51.560 --> 00:55:59.960 align:middle line:84% Right, so now we'll use again, continuity of Lebesgue measure. 00:55:59.960 --> 00:56:08.030 align:middle line:84% So then for all n, 0, which is less than or equal to 1 over n 00:56:08.030 --> 00:56:10.160 align:middle line:90% times the measure of fn-- 00:56:10.160 --> 00:56:19.020 align:middle line:84% this is equal to the integral over E of 1 00:56:19.020 --> 00:56:22.500 align:middle line:84% over n times the indicator function of fn. 00:56:22.500 --> 00:56:29.710 align:middle line:90% Now, on this set-- 00:56:29.710 --> 00:56:32.420 align:middle line:90% but what am I saying? 00:56:32.420 --> 00:56:33.670 align:middle line:90% Yeah, let's write it this way. 00:56:33.670 --> 00:56:36.720 align:middle line:90% 00:56:36.720 --> 00:56:40.740 align:middle line:90% This is equal to 1 over n. 00:56:40.740 --> 00:56:43.500 align:middle line:90% 00:56:43.500 --> 00:56:48.720 align:middle line:84% And now, on fn, 1 over n is less than or equal to f of x. 00:56:48.720 --> 00:56:51.210 align:middle line:84% So this is less than or equal to the integral 00:56:51.210 --> 00:56:56.400 align:middle line:84% of fn F. And capital F sub n is a subset of capital E, 00:56:56.400 --> 00:57:02.700 align:middle line:84% so this is less than or equal to the integral of E over E of f. 00:57:02.700 --> 00:57:06.690 align:middle line:84% But by assumption, this is 0, right? 00:57:06.690 --> 00:57:10.410 align:middle line:84% And sandwiched in between 0 and 0 is 1 over n times 00:57:10.410 --> 00:57:12.390 align:middle line:90% the measure of f sub n. 00:57:12.390 --> 00:57:16.800 align:middle line:84% And therefore, for all n, measure of f sub n 00:57:16.800 --> 00:57:22.260 align:middle line:84% equals 0, which tells me that the measure of f-- which 00:57:22.260 --> 00:57:32.010 align:middle line:84% is equal to that union, which is equal to this increasing 00:57:32.010 --> 00:57:35.820 align:middle line:84% union-- is by the continuity of Lebesgue measure, 00:57:35.820 --> 00:57:38.460 align:middle line:84% equal to the limit as n goes to infinity 00:57:38.460 --> 00:57:44.050 align:middle line:84% of the measure of f sub n, which equals 0. 00:57:44.050 --> 00:57:47.950 align:middle line:84% And therefore, the set of all x's where f of x is positive 00:57:47.950 --> 00:57:51.325 align:middle line:90% has Lebesgue measure 0. 00:57:51.325 --> 00:57:54.970 align:middle line:90% 00:57:54.970 --> 00:57:58.393 align:middle line:84% And f equals 0 almost everywhere. 00:57:58.393 --> 00:58:04.190 align:middle line:90% 00:58:04.190 --> 00:58:06.890 align:middle line:84% Now, using what we've done here and the monotone convergence 00:58:06.890 --> 00:58:12.140 align:middle line:84% theorem, we can slightly relax the assumptions 00:58:12.140 --> 00:58:13.640 align:middle line:84% in the monotone convergence theorem. 00:58:13.640 --> 00:58:17.470 align:middle line:90% 00:58:17.470 --> 00:58:18.940 align:middle line:84% So we have the following theorem. 00:58:18.940 --> 00:58:32.500 align:middle line:84% If fn is sequenced in non-negative measurable 00:58:32.500 --> 00:58:44.220 align:middle line:84% functions such that now for almost every x in E, 00:58:44.220 --> 00:58:49.950 align:middle line:84% we have f1 of x is less than or equal to f2 of x is less than 00:58:49.950 --> 00:58:53.970 align:middle line:90% or equal to f3 of x and so on. 00:58:53.970 --> 00:58:58.410 align:middle line:84% And limit as n goes to infinity of fn 00:58:58.410 --> 00:59:03.090 align:middle line:90% of x equals a function f of x. 00:59:03.090 --> 00:59:06.510 align:middle line:84% So remember, in the statement of the monotone convergence 00:59:06.510 --> 00:59:08.970 align:middle line:84% theorem, we assume these two things for every x. 00:59:08.970 --> 00:59:12.857 align:middle line:84% Now, we're just assuming them for almost every x in E. 00:59:12.857 --> 00:59:14.190 align:middle line:90% Then we get the same conclusion. 00:59:14.190 --> 00:59:20.310 align:middle line:84% Then the integral of E of f is equal to the limit 00:59:20.310 --> 00:59:23.790 align:middle line:84% as n goes to infinity of the integral of E 00:59:23.790 --> 00:59:26.070 align:middle line:90% over E of f sub n. 00:59:26.070 --> 00:59:40.160 align:middle line:90% 00:59:40.160 --> 00:59:40.880 align:middle line:90% OK. 00:59:40.880 --> 00:59:46.235 align:middle line:84% So we call these two conditions star. 00:59:46.235 --> 00:59:51.550 align:middle line:90% 00:59:51.550 --> 00:59:57.730 align:middle line:84% Let capital F be the set while x is in E such 00:59:57.730 --> 01:00:03.865 align:middle line:84% that fn's are increasing to F, so star holds. 01:00:03.865 --> 01:00:07.950 align:middle line:90% 01:00:07.950 --> 01:00:11.920 align:middle line:84% Then the measure of the complement 01:00:11.920 --> 01:00:14.900 align:middle line:90% is, by assumption, equal to 0. 01:00:14.900 --> 01:00:16.760 align:middle line:84% And I should say here, the complement in E. 01:00:16.760 --> 01:00:18.485 align:middle line:90% So I should say E take away f. 01:00:18.485 --> 01:00:21.630 align:middle line:90% 01:00:21.630 --> 01:00:28.590 align:middle line:84% So let me-- and that should have been in the-- 01:00:28.590 --> 01:00:31.120 align:middle line:90% 01:00:31.120 --> 01:00:32.620 align:middle line:84% so if I write complement, you should 01:00:32.620 --> 01:00:35.840 align:middle line:84% interpret that as the complement within E. So E take away f. 01:00:35.840 --> 01:00:40.340 align:middle line:90% 01:00:40.340 --> 01:00:48.050 align:middle line:90% Then f minus chi sub f-- 01:00:48.050 --> 01:00:50.540 align:middle line:84% so this is the indicator function over capital F. 01:00:50.540 --> 01:00:55.940 align:middle line:90% This equals 0 almost everywhere. 01:00:55.940 --> 01:01:07.550 align:middle line:84% And fn minus chi sub f fn equals 0 almost everywhere for all n. 01:01:07.550 --> 01:01:13.780 align:middle line:84% These equal 0 when x is an f which 01:01:13.780 --> 01:01:16.090 align:middle line:84% the complement is as measure 0, which 01:01:16.090 --> 01:01:17.440 align:middle line:90% I didn't finish writing down. 01:01:17.440 --> 01:01:27.400 align:middle line:90% 01:01:27.400 --> 01:01:33.550 align:middle line:84% Now, by monotone convergence theorem applied now 01:01:33.550 --> 01:01:48.120 align:middle line:84% to these parts, if you like, and the previous theorem, we 01:01:48.120 --> 01:01:58.930 align:middle line:84% have that the integral of f of E, 01:01:58.930 --> 01:02:05.620 align:middle line:90% this is equal to the chi sub f. 01:02:05.620 --> 01:02:10.960 align:middle line:90% 01:02:10.960 --> 01:02:16.550 align:middle line:90% And so this is equal to-- 01:02:16.550 --> 01:02:21.050 align:middle line:90% so since f minus f-- 01:02:21.050 --> 01:02:21.550 align:middle line:90% wait. 01:02:21.550 --> 01:02:26.410 align:middle line:90% 01:02:26.410 --> 01:02:27.700 align:middle line:90% Yeah, so OK. 01:02:27.700 --> 01:02:32.030 align:middle line:84% So since f equals f times chi sub f almost everywhere, 01:02:32.030 --> 01:02:34.460 align:middle line:90% the integrals equal. 01:02:34.460 --> 01:02:36.580 align:middle line:84% And so this is equal to the integral 01:02:36.580 --> 01:02:39.040 align:middle line:90% over capital F of little f. 01:02:39.040 --> 01:02:42.250 align:middle line:84% And by the monotone convergence theorem applied here, 01:02:42.250 --> 01:02:45.370 align:middle line:84% this is equal to the limit as n goes 01:02:45.370 --> 01:02:53.000 align:middle line:84% to infinity of the fn's, because they are pointwise 01:02:53.000 --> 01:02:55.370 align:middle line:84% increasing on capital F to little f, 01:02:55.370 --> 01:02:58.005 align:middle line:90% and this is equal to-- 01:02:58.005 --> 01:03:00.050 align:middle line:84% OK, so I really didn't need the previous theorem. 01:03:00.050 --> 01:03:01.550 align:middle line:84% I could have used what I had earlier 01:03:01.550 --> 01:03:07.580 align:middle line:84% that if I have two functions which equal each other almost 01:03:07.580 --> 01:03:09.110 align:middle line:90% everywhere. 01:03:09.110 --> 01:03:11.150 align:middle line:84% So this previous theorem should not 01:03:11.150 --> 01:03:14.060 align:middle line:84% be referring to what I just proved a minute ago, but really 01:03:14.060 --> 01:03:15.950 align:middle line:84% to the theorem at the beginning of lecture 01:03:15.950 --> 01:03:18.325 align:middle line:84% that if I have two functions that equal each other almost 01:03:18.325 --> 01:03:21.290 align:middle line:84% everywhere, then their integrals equal each other. 01:03:21.290 --> 01:03:24.440 align:middle line:90% 01:03:24.440 --> 01:03:25.940 align:middle line:84% Although, maybe I didn't state that. 01:03:25.940 --> 01:03:27.607 align:middle line:84% I just stated the less than or equal to. 01:03:27.607 --> 01:03:29.630 align:middle line:84% But if they're equal almost everywhere, 01:03:29.630 --> 01:03:32.300 align:middle line:84% they're less than or equal to each other almost everywhere. 01:03:32.300 --> 01:03:33.500 align:middle line:90% Anyways, back to this. 01:03:33.500 --> 01:03:40.850 align:middle line:90% This is equal to the limit fn. 01:03:40.850 --> 01:03:44.860 align:middle line:84% So the whole point is that sets of measure 0 01:03:44.860 --> 01:03:49.570 align:middle line:84% don't affect statements that involve integrals. 01:03:49.570 --> 01:03:57.140 align:middle line:84% That should be the take home, that if your conclusions are 01:03:57.140 --> 01:04:00.320 align:middle line:84% in terms of integrals, conditions 01:04:00.320 --> 01:04:04.940 align:middle line:84% holding almost everywhere suffice, typically. 01:04:04.940 --> 01:04:07.130 align:middle line:84% So for example, the simplest one we had earlier 01:04:07.130 --> 01:04:09.667 align:middle line:84% was that if f is less than or equal to g, 01:04:09.667 --> 01:04:11.750 align:middle line:84% then the integral of f is less than or equal to g. 01:04:11.750 --> 01:04:14.450 align:middle line:84% We can relax that to the integral 01:04:14.450 --> 01:04:16.790 align:middle line:84% that if f is less than or equal to g almost everywhere, 01:04:16.790 --> 01:04:19.440 align:middle line:84% then the conclusion, which is stated in terms of integrals, 01:04:19.440 --> 01:04:19.940 align:middle line:90% still holds. 01:04:19.940 --> 01:04:22.482 align:middle line:84% The integral of f is less than or equal to the integral of g. 01:04:22.482 --> 01:04:29.660 align:middle line:90% 01:04:29.660 --> 01:04:49.730 align:middle line:84% So now, we'll do the second big convergence of integrals-- 01:04:49.730 --> 01:04:56.390 align:middle line:84% or this one's actually an inequality between integrals, 01:04:56.390 --> 01:04:58.590 align:middle line:90% but it's still extremely useful. 01:04:58.590 --> 01:05:01.790 align:middle line:84% In fact, it's equivalent to the monotone convergence theorem, 01:05:01.790 --> 01:05:04.640 align:middle line:84% so it is neither stronger nor weaker. 01:05:04.640 --> 01:05:17.740 align:middle line:90% 01:05:17.740 --> 01:05:28.930 align:middle line:84% So we have Fatou's lemma, stated as a theorem, 01:05:28.930 --> 01:05:41.790 align:middle line:84% of course which states that if fn 01:05:41.790 --> 01:05:50.220 align:middle line:84% is a sequence in L plus of E, then the integral of E 01:05:50.220 --> 01:05:59.020 align:middle line:84% of the liminf is the integral that's infinity of fn of x. 01:05:59.020 --> 01:06:01.870 align:middle line:90% This is a function. 01:06:01.870 --> 01:06:06.940 align:middle line:84% For each x, I take the liminf as n goes to infinity of fn of x. 01:06:06.940 --> 01:06:12.190 align:middle line:84% This is less than or equal to the liminf 01:06:12.190 --> 01:06:17.650 align:middle line:84% as n goes to infinity of the integrals of fn. 01:06:17.650 --> 01:06:22.980 align:middle line:90% So let me state it this way. 01:06:22.980 --> 01:06:27.900 align:middle line:84% So the liminf of f sub n, let me just 01:06:27.900 --> 01:06:29.780 align:middle line:90% recall, what is the liminf? 01:06:29.780 --> 01:06:36.250 align:middle line:90% 01:06:36.250 --> 01:06:48.270 align:middle line:84% This is equal to the sup over n equals 1 inf over k 01:06:48.270 --> 01:06:54.460 align:middle line:84% bigger than or equal to n fk of x. 01:06:54.460 --> 01:06:59.470 align:middle line:84% So that is the definition of the limsup, if you like, if-- 01:06:59.470 --> 01:07:02.560 align:middle line:84% in fact, let me not just be specific to fn 01:07:02.560 --> 01:07:09.310 align:middle line:84% of x, just of a sequence of real numbers, the liminf of a sub n 01:07:09.310 --> 01:07:12.610 align:middle line:84% is equal to this thing on the right hand side. 01:07:12.610 --> 01:07:23.750 align:middle line:90% 01:07:23.750 --> 01:07:26.030 align:middle line:90% OK. 01:07:26.030 --> 01:07:31.760 align:middle line:84% So this follows pretty easily from the monotone convergence 01:07:31.760 --> 01:07:32.857 align:middle line:90% theorem. 01:07:32.857 --> 01:07:34.940 align:middle line:84% I said a minute ago that it's, in fact, equivalent 01:07:34.940 --> 01:07:36.440 align:middle line:84% to the monotone convergence theorem. 01:07:36.440 --> 01:07:39.500 align:middle line:84% You can prove if you-- so we're going 01:07:39.500 --> 01:07:41.840 align:middle line:84% to use the monotone convergence theorem to prove it. 01:07:41.840 --> 01:07:44.120 align:middle line:84% You can also assume Fatou's lemma holds 01:07:44.120 --> 01:07:50.180 align:middle line:84% and then prove the monotone convergence theorem from it. 01:07:50.180 --> 01:07:54.420 align:middle line:84% You can also prove it independently 01:07:54.420 --> 01:07:56.730 align:middle line:84% from the monotone convergence theorem. 01:07:56.730 --> 01:07:59.370 align:middle line:84% I mean, using essentially what's a similar argument to how 01:07:59.370 --> 01:08:03.370 align:middle line:84% you prove the monotone convergence theorem. 01:08:03.370 --> 01:08:20.983 align:middle line:84% OK, so first off, so we have liminf 01:08:20.983 --> 01:08:26.359 align:middle line:84% of fn of x, which is, again, by what I've written up here, 01:08:26.359 --> 01:08:32.319 align:middle line:84% sup n bigger than or equal to 1 inf k bigger than 01:08:32.319 --> 01:08:37.300 align:middle line:90% or equal to n fk of x. 01:08:37.300 --> 01:08:41.620 align:middle line:90% This is now for a fixed n-- 01:08:41.620 --> 01:08:48.720 align:middle line:84% or what happens to what's in the bracket as n is increasing? 01:08:48.720 --> 01:08:52.350 align:middle line:84% Well, this inf is being taken over a smaller set. 01:08:52.350 --> 01:08:54.810 align:middle line:84% And the inf of a smaller set is bigger than 01:08:54.810 --> 01:08:57.689 align:middle line:84% or equal to the inf of the larger set. 01:08:57.689 --> 01:09:04.649 align:middle line:84% So this inf here, this thing in brackets, is increasing in n. 01:09:04.649 --> 01:09:09.279 align:middle line:84% So this sup is, in fact, the limit 01:09:09.279 --> 01:09:20.310 align:middle line:84% as n goes to infinity of this increasing 01:09:20.310 --> 01:09:23.279 align:middle line:84% sequence of real numbers defined as the inf over k bigger 01:09:23.279 --> 01:09:27.470 align:middle line:90% than or equal to n of fk of x. 01:09:27.470 --> 01:09:30.620 align:middle line:84% And so what I just told you is not specific to fk of x. 01:09:30.620 --> 01:09:34.939 align:middle line:84% It's specific to ak, for sequence ak. 01:09:34.939 --> 01:09:39.521 align:middle line:90% 01:09:39.521 --> 01:09:48.408 align:middle line:84% OK, and basically, I'm going to write down 01:09:48.408 --> 01:09:49.450 align:middle line:90% what I said a minute ago. 01:09:49.450 --> 01:09:54.340 align:middle line:84% Since fk bigger than or equal to 1 fk of x 01:09:54.340 --> 01:09:58.884 align:middle line:84% is less than or equal to inf a bigger than or equal to 2. 01:09:58.884 --> 01:10:02.355 align:middle line:84% We have fk of x is less than or equal to-- now 01:10:02.355 --> 01:10:06.430 align:middle line:90% it changes to 3 and so on. 01:10:06.430 --> 01:10:12.960 align:middle line:84% This implies by the monotone convergence theorem 01:10:12.960 --> 01:10:23.860 align:middle line:84% that the integral of the liminf of fn is equal to the limit 01:10:23.860 --> 01:10:29.070 align:middle line:84% as n goes to infinity of the integral over E of the inf 01:10:29.070 --> 01:10:32.710 align:middle line:90% k bigger than or equal to n fk. 01:10:32.710 --> 01:10:35.870 align:middle line:84% So I have this function here, which is defined in this way. 01:10:35.870 --> 01:10:38.080 align:middle line:84% So for each n, I get a function here. 01:10:38.080 --> 01:10:49.750 align:middle line:90% 01:10:49.750 --> 01:10:50.470 align:middle line:90% All right. 01:10:50.470 --> 01:10:58.390 align:middle line:84% Now, for all j bigger than or equal to m, 01:10:58.390 --> 01:11:03.310 align:middle line:84% this function given by the inf over k 01:11:03.310 --> 01:11:06.070 align:middle line:90% bigger than or equal to-- 01:11:06.070 --> 01:11:09.710 align:middle line:84% let me add one more quantifier in here. 01:11:09.710 --> 01:11:14.380 align:middle line:84% So for all j bigger than or equal to n, for all x in E, 01:11:14.380 --> 01:11:19.450 align:middle line:84% I have that the inf over k bigger than or equal to 01:11:19.450 --> 01:11:22.270 align:middle line:90% n of fk of x-- 01:11:22.270 --> 01:11:26.435 align:middle line:84% this is certainly less than or equal to fj of x. 01:11:26.435 --> 01:11:29.340 align:middle line:90% 01:11:29.340 --> 01:11:32.400 align:middle line:84% This is the inf overall fk of x for k 01:11:32.400 --> 01:11:33.600 align:middle line:90% bigger than or equal to n. 01:11:33.600 --> 01:11:36.900 align:middle line:84% And for any fixed j bigger than or equal to n, that's certainly 01:11:36.900 --> 01:11:38.490 align:middle line:90% less than or equal to fj of x. 01:11:38.490 --> 01:11:40.290 align:middle line:84% Because this is a lower bound for all 01:11:40.290 --> 01:11:44.070 align:middle line:84% of these guys for all j bigger than or equal to n. 01:11:44.070 --> 01:11:47.040 align:middle line:84% And therefore, since this function here 01:11:47.040 --> 01:11:52.950 align:middle line:84% sits below this function, I have for all j bigger than or equal 01:11:52.950 --> 01:11:59.880 align:middle line:84% to n, the integral of E of inf of fk 01:11:59.880 --> 01:12:03.680 align:middle line:84% is less than or equal to the integral of fj. 01:12:03.680 --> 01:12:06.530 align:middle line:90% 01:12:06.530 --> 01:12:10.910 align:middle line:84% So I have this number here sits below this number 01:12:10.910 --> 01:12:12.830 align:middle line:90% here for all j. 01:12:12.830 --> 01:12:16.310 align:middle line:84% This is a fixed number depending on n, 01:12:16.310 --> 01:12:19.370 align:middle line:84% this is a fixed number depending on j. 01:12:19.370 --> 01:12:21.740 align:middle line:84% And this holds for all j bigger than or equal to n. 01:12:21.740 --> 01:12:24.470 align:middle line:84% So this thing has to be a lower bound 01:12:24.470 --> 01:12:27.080 align:middle line:84% for the set of all numbers of this form for j 01:12:27.080 --> 01:12:28.640 align:middle line:90% bigger than or equal to n. 01:12:28.640 --> 01:12:33.080 align:middle line:84% And therefore, the integral of E of inf k bigger 01:12:33.080 --> 01:12:37.190 align:middle line:84% than or equal to n of fk is less than or equal to the inf 01:12:37.190 --> 01:12:45.360 align:middle line:84% overall j bigger than or equal to n of the integral of f sub 01:12:45.360 --> 01:12:50.610 align:middle line:84% j over E. Now, we're going to take this and stick it 01:12:50.610 --> 01:12:53.445 align:middle line:90% into this inequality here. 01:12:53.445 --> 01:13:02.388 align:middle line:90% 01:13:02.388 --> 01:13:04.850 align:middle line:84% So that's what we had before, which 01:13:04.850 --> 01:13:13.820 align:middle line:84% was that the liminf of fn over E, which is equal to limit as n 01:13:13.820 --> 01:13:23.060 align:middle line:84% goes to infinity of the inf k bigger than or equal to n fk. 01:13:23.060 --> 01:13:25.190 align:middle line:84% This is, by what we've just shown, 01:13:25.190 --> 01:13:26.690 align:middle line:84% is less than or equal to the limit 01:13:26.690 --> 01:13:31.550 align:middle line:84% as n goes to infinity of the inf over j bigger 01:13:31.550 --> 01:13:35.850 align:middle line:90% than or equal to n of fj. 01:13:35.850 --> 01:13:38.680 align:middle line:90% 01:13:38.680 --> 01:13:46.120 align:middle line:84% But this is just, by definition, equal to the liminf 01:13:46.120 --> 01:13:50.850 align:middle line:84% of the integrals of the fn's, which is Fatou's lemma. 01:13:50.850 --> 01:14:05.380 align:middle line:90% 01:14:05.380 --> 01:14:13.900 align:middle line:84% OK, so one more theorem about the Lebesgue integral, 01:14:13.900 --> 01:14:17.010 align:middle line:90% which is a very useful one. 01:14:17.010 --> 01:14:19.200 align:middle line:84% Throughout all this, we have had functions 01:14:19.200 --> 01:14:20.980 align:middle line:90% that are extended real value. 01:14:20.980 --> 01:14:24.300 align:middle line:84% So we're dealing with non-negative functions, which 01:14:24.300 --> 01:14:27.750 align:middle line:90% can equal infinity at points. 01:14:27.750 --> 01:14:29.370 align:middle line:84% And maybe that makes you nervous, 01:14:29.370 --> 01:14:32.580 align:middle line:84% but I'm going to tell you that as long 01:14:32.580 --> 01:14:34.170 align:middle line:84% as the integral is finite, you don't 01:14:34.170 --> 01:14:36.960 align:middle line:90% have to be nervous too often. 01:14:36.960 --> 01:14:43.260 align:middle line:84% So if f is a non-negative measurable 01:14:43.260 --> 01:14:54.800 align:middle line:84% function over a measurable set E and the integral is finite, 01:14:54.800 --> 01:15:00.980 align:middle line:84% then the set of x is where f of x 01:15:00.980 --> 01:15:09.065 align:middle line:84% is infinite as the set of measure 0. 01:15:09.065 --> 01:15:11.720 align:middle line:90% 01:15:11.720 --> 01:15:15.170 align:middle line:84% So the measure of the set where it's infinite is 0. 01:15:15.170 --> 01:15:16.130 align:middle line:90% So what's the proof? 01:15:16.130 --> 01:15:23.540 align:middle line:90% 01:15:23.540 --> 01:15:31.190 align:middle line:84% So it's kind of how we did, in spirit, 01:15:31.190 --> 01:15:36.300 align:middle line:84% the proof that if the integral is 0, 01:15:36.300 --> 01:15:38.620 align:middle line:84% then the function is 0 almost everywhere. 01:15:38.620 --> 01:15:42.960 align:middle line:84% So let f be the set of all x in E 01:15:42.960 --> 01:15:51.600 align:middle line:84% such that f of x equals infinity and fn be 01:15:51.600 --> 01:15:57.270 align:middle line:84% the set of all x's in E such that f of x is bigger than n. 01:15:57.270 --> 01:16:08.090 align:middle line:90% 01:16:08.090 --> 01:16:11.087 align:middle line:84% Oh well, that's what I had in my mind, 01:16:11.087 --> 01:16:12.920 align:middle line:84% but what I wrote in my notes is a little bit 01:16:12.920 --> 01:16:15.770 align:middle line:84% different than the proof I had in my mind just now. 01:16:15.770 --> 01:16:18.830 align:middle line:84% So let's go with what's in my notes that's 01:16:18.830 --> 01:16:21.170 align:middle line:90% a little more cautious. 01:16:21.170 --> 01:16:21.670 align:middle line:90% OK. 01:16:21.670 --> 01:16:24.480 align:middle line:90% 01:16:24.480 --> 01:16:32.130 align:middle line:84% Then for all n, a natural number, 01:16:32.130 --> 01:16:38.430 align:middle line:84% n times chi f is less than or equal to f times chi f, where 01:16:38.430 --> 01:16:40.830 align:middle line:84% this is the indicator function of capital 01:16:40.830 --> 01:16:43.920 align:middle line:84% F. Because f on capital F is just infinite, 01:16:43.920 --> 01:16:46.710 align:middle line:90% so this always holds, right? 01:16:46.710 --> 01:16:53.470 align:middle line:84% And therefore, n times the measure of f-- 01:16:53.470 --> 01:16:57.960 align:middle line:84% so of all n, n times a measure of f 01:16:57.960 --> 01:17:05.010 align:middle line:84% is less than or equal to the integral over E times f chi f, 01:17:05.010 --> 01:17:06.840 align:middle line:84% which is less than or equal to the integral 01:17:06.840 --> 01:17:10.620 align:middle line:90% of E of f, which is finite. 01:17:10.620 --> 01:17:11.655 align:middle line:90% That's a fixed number. 01:17:11.655 --> 01:17:15.460 align:middle line:90% 01:17:15.460 --> 01:17:20.050 align:middle line:84% Then for all n, the measure of f is less than 01:17:20.050 --> 01:17:23.740 align:middle line:84% or equal to 1 over n times the integral 01:17:23.740 --> 01:17:28.510 align:middle line:84% of f over E. Again, this is a fixed finite number which goes 01:17:28.510 --> 01:17:30.500 align:middle line:90% to 0 as n goes to infinity. 01:17:30.500 --> 01:17:32.005 align:middle line:84% This is just a fixed number as well. 01:17:32.005 --> 01:17:42.340 align:middle line:90% 01:17:42.340 --> 01:17:46.480 align:middle line:90% Thus, measure of f equals 0. 01:17:46.480 --> 01:17:51.230 align:middle line:90% 01:17:51.230 --> 01:17:54.050 align:middle line:84% OK, so that seems like a good place to stop. 01:17:54.050 --> 01:17:58.220 align:middle line:84% Next time-- so we've defined the Lebesgue integral 01:17:58.220 --> 01:18:01.580 align:middle line:84% of a non-negative measurable function. 01:18:01.580 --> 01:18:06.410 align:middle line:84% We will then define the class of Lebesgue integrable functions 01:18:06.410 --> 01:18:08.930 align:middle line:84% and extend the definition of integral 01:18:08.930 --> 01:18:12.575 align:middle line:84% to those functions in a fairly straightforward way. 01:18:12.575 --> 01:18:15.910 align:middle line:90% 01:18:15.910 --> 01:18:19.895 align:middle line:84% Prove some simple properties of the Lebesgue integral. 01:18:19.895 --> 01:18:22.650 align:middle line:90% 01:18:22.650 --> 01:18:25.390 align:middle line:84% And also, the last big convergence theorem, 01:18:25.390 --> 01:18:27.690 align:middle line:84% which is the dominated convergence theorem. 01:18:27.690 --> 01:18:33.000 align:middle line:84% And we may or may not finish by the end of next lecture 01:18:33.000 --> 01:18:38.220 align:middle line:84% the proof that Lp spaces, which are based on the Lebesgue 01:18:38.220 --> 01:18:41.280 align:middle line:84% integral-- so we built the Lebesgue integral to have 01:18:41.280 --> 01:18:52.860 align:middle line:84% a space of functions for which kind of-- 01:18:52.860 --> 01:18:53.730 align:middle line:90% OK, so let me stop. 01:18:53.730 --> 01:18:57.020 align:middle line:90% That alarm kind of threw me off. 01:18:57.020 --> 01:19:01.670 align:middle line:84% So it's not too difficult to show proof. 01:19:01.670 --> 01:19:04.430 align:middle line:84% Or you can just accept for now-- and we'll actually see why this 01:19:04.430 --> 01:19:05.450 align:middle line:90% is the case soon-- 01:19:05.450 --> 01:19:09.800 align:middle line:84% that the space of continuous functions with norm 01:19:09.800 --> 01:19:11.160 align:middle line:90% being the integral. 01:19:11.160 --> 01:19:14.300 align:middle line:84% So the norm of f being, let's say, 01:19:14.300 --> 01:19:18.740 align:middle line:84% the integral of the absolute value of f is a norm space, 01:19:18.740 --> 01:19:22.070 align:middle line:90% but it's not a Banach space. 01:19:22.070 --> 01:19:25.790 align:middle line:84% Or you could change the integral of the absolute value 01:19:25.790 --> 01:19:30.350 align:middle line:84% of f to what's called a big Lp-norm, the integral of f 01:19:30.350 --> 01:19:32.870 align:middle line:84% raised to the p all raised to the 1 over p. 01:19:32.870 --> 01:19:34.970 align:middle line:84% So the analog of the little Lp norms, which 01:19:34.970 --> 01:19:38.900 align:middle line:90% we encountered a few weeks ago. 01:19:38.900 --> 01:19:41.000 align:middle line:84% None of those are Banach spaces when 01:19:41.000 --> 01:19:43.010 align:middle line:84% restricted to continuous functions, or even 01:19:43.010 --> 01:19:45.810 align:middle line:90% Riemann integrable functions. 01:19:45.810 --> 01:19:50.870 align:middle line:84% So our goal-- at least it was a while back 01:19:50.870 --> 01:19:54.530 align:middle line:84% when we started this section, this big section on Lebesgue 01:19:54.530 --> 01:20:01.460 align:middle line:84% integration-- was to build, or at least come upon a space 01:20:01.460 --> 01:20:06.330 align:middle line:84% where the resulting integrable functions form a Banach space. 01:20:06.330 --> 01:20:10.160 align:middle line:84% And so we may or may not, by the end of next lecture, 01:20:10.160 --> 01:20:11.430 align:middle line:90% introduce those. 01:20:11.430 --> 01:20:16.650 align:middle line:90% But that's where we're headed. 01:20:16.650 --> 01:20:18.200 align:middle line:90% That's where we're almost at. 01:20:18.200 --> 01:20:23.450 align:middle line:84% And these spaces arose because one 01:20:23.450 --> 01:20:27.800 align:middle line:84% wants to apply functional analysis facts, tools 01:20:27.800 --> 01:20:31.700 align:middle line:84% to concrete questions, such as questions about convergence 01:20:31.700 --> 01:20:38.780 align:middle line:84% of Fourier series, which arose immediately after Fourier said 01:20:38.780 --> 01:20:44.200 align:middle line:84% that any periodic function can be expanded as a Fourier 01:20:44.200 --> 01:20:44.700 align:middle line:90% series. 01:20:44.700 --> 01:20:46.640 align:middle line:84% So a lot of people went to a lot of trouble 01:20:46.640 --> 01:20:54.630 align:middle line:84% to fill in precisely what it means expanded as, expanded 01:20:54.630 --> 01:20:56.340 align:middle line:90% as pointwise. 01:20:56.340 --> 01:21:00.080 align:middle line:84% Does this Fourier series converge 01:21:00.080 --> 01:21:04.610 align:middle line:84% to the function that was kind of hard to do on average? 01:21:04.610 --> 01:21:06.890 align:middle line:84% Do you mean average as in measured with respect 01:21:06.890 --> 01:21:11.300 align:middle line:84% to some norm that's integrated that involves integration? 01:21:11.300 --> 01:21:15.320 align:middle line:84% So which is why we're coming here. 01:21:15.320 --> 01:21:18.380 align:middle line:90% But we'll see that next time. 01:21:18.380 --> 01:21:23.120 align:middle line:84% Or we'll see the applications of this integration theory, 01:21:23.120 --> 01:21:25.280 align:middle line:84% along with the functional analysis later 01:21:25.280 --> 01:21:29.240 align:middle line:84% in the course when we circle around to Fourier series. 01:21:29.240 --> 01:21:31.630 align:middle line:90% All right, so we'll stop there. 01:21:31.630 --> 01:21:38.000 align:middle line:90%