 Welcome to lecture 40 on measure and integration. In the past two lectures, we had been looking at the various modes of convergence for measurable functions. We will continue this study in today's lecture also. We will recall some of the properties of convergence in measure that we had proved last time and then we will go on to what is called convergence in the pth mean. Let us just recall what we have been doing. f and f n are measurable functions on a measure space. Then we said that f n converges to f in measure f for every epsilon bigger than 0. The measure of the set all x, where f n x minus f x is bigger than or equal to epsilon, that measure of that set goes to 0 as n goes to infinity for every f. In some sense, the measure of the set, where f n is away from f by a distance epsilon, that goes to 0 for every epsilon bigger than 0. This was called convergence in measure. We denoted this by saying that f n with a arrow and the symbol m above it f. f n converges to f in measure. It is noted by this symbol. In the previous lecture, we showed that the convergence in measure neither implies nor is implied by convergence point almost everywhere. That means, if f n converges to f in measure, then it need not imply that f n converges point wise or almost everywhere. Conversely, if a sequence converges point wise or almost everywhere, then need not imply that it converges in measure. However, we proved that if the underlying measure space is finite and f n converges to f almost everywhere, then it also converges in measure. That means, convergence in measure is implied by convergence almost everywhere if the underlying measure space is finite. After that, we looked at what is called almost uniform convergence for functions. We said a sequence f n converges almost uniformly to f on a set E, if we can find for every epsilon a subset E epsilon such that the measure of the set E intersection E epsilon complement is finite and f n converges uniformly to f on E. That means, except for a small set of small measure, f n converges to f uniformly. This is called almost uniform convergence. Note, this is different from saying that the convergence is almost, convergence is uniform almost everywhere. Saying that f n converges to f uniformly almost everywhere will mean that except for an L set f n converges to f. And almost uniform convergence is saying that outside a set of measure 0 for every epsilon, there is a set E epsilon of such that outside that set, the convergence is uniform. In general, we showed that convergence in measure does not imply convergence almost everywhere, because if convergence almost everywhere, almost uniform implies convergence almost everywhere. We want to conclude that convergence in measure does not imply convergence almost uniform, because almost uniform convergence implies almost everywhere convergence. So, if convergence in measure implies almost uniform convergence, then it will also imply a convergence almost everywhere, which is not true in general. So, convergence in measure does not imply convergence almost uniform. However, if f n converges almost uniformly to f, then it also converges in measure. So, the convergence is always true. Namely, if f n converges almost uniformly to f, then f n converges to f in measure. So, that the proof of this is quite simple. So, because f n converges to f almost uniformly, so that means for every delta bigger than 0, we can select a set E delta such that outside the measure of the set outside E delta is small, f n converges uniformly to f on E delta. So, that is why the property that f n converges almost uniformly to f. So, for every delta, we can find a subset. So, let us say we are analyzing this on the set E, then measure of the set outside E delta is small and f n converges uniformly to E delta. But, what does that mean? That means, because it is converging uniformly to f on E delta, so that means for every n, integer n bigger than 0, f n must come close to f for every x in E delta. So, that means for every epsilon bigger than 0, if you look at, there exists a set n, this is a natural number n, say that the distance f n minus f x is less than epsilon for all n bigger than n and for every x belonging to E delta complement. But, then that means, if we look at the set of points where this is bigger than or equal to epsilon, that will be contained in E delta and measure of E delta is less than epsilon. So, that will prove that for every delta, we have got a stage n bigger than n naught, say that x, say that f n x minus f x bigger than epsilon is less than delta. That means, f n converges to f in measure. So, we have shown that almost uniform convergence implies convergence in measure. So, these are the basic properties of relations between convergence in measure with almost uniform convergence, convergence point wise and so on. Now, we just now said that convergence in measure need not imply convergence almost everywhere in general. However, one can prove a partial result in this direction namely, if f n converges to f in measure, then there is a subsequence which converges almost everywhere. And this result is quite useful sometimes when you are analyzing sequences of measurable functions which converge in measure. So, this theorem is called Rie's theorem. So, let us prove Rie's theorem. So, it says that f n be a sequence of measurable functions converging in measure to a measurable function f. Then there exists a subsequence f n k, such that f n k converges to f everywhere or almost to almost everywhere actually, we should be saying to f n almost. So, every sequence which converges in measure has a subsequence which converges almost everywhere. So, let us see a proof of this. So, to prove this, we have whatever looking at, we are looking at how to construct a subsequence f n k which converges almost everywhere to f. So, that means we have to find f n k such that, so we reformulate the problem as follows that look at the set of points. So, f n k we want with the property that whenever f n k x minus f of x is bigger than 1 over k, the set of these points for k equal to union of such sets for k equal to j to infinity, intersection over j equal to 1 into infinity, that must be equal to 0. So, we want a sequence f n k such that this measure of this set is equal to 0. Why is that enough? So, let us just look at it. So, we are saying that f n k will converge to f almost everywhere. If we look at the set intersection j equal to 1 to infinity, union over k equal to j to infinity, the set of points x belonging to x such that mod of f n k x minus f of x bigger than 1 over k, this set has got measure 0. Why is that? Because if we take an element x belonging to this, what will that imply? So, that is if and only if x belongs to, this is belonging to intersection, so x belongs to union k over j to infinity, x belonging to x such that f n k x minus f of x is bigger than or equal to 1 over k for every j. That must happen for every j, because it belongs to intersection. But what does that mean? That is if and only if this belongs to j, that means for every j, there exists k equal to j, k equal to j to infinity. So, that means for every j, there exists k bigger than or equal to j such that saying that x belongs to this union means x will belong to at least one of them. That means x belongs to for this set for some k bigger than or equal to j. So, there is a k bigger than or equal to j such that x belongs to this such that means f n k x minus f of x is bigger than or equal to k. For such an x, this must hold is bigger than or equal to k. So, that means it does not converge. So, if we show that this set has got measure 0, that means if x does not belong, then it must converge. So, what is the meaning of saying? So, if we want to say that x does not, if x does not belong to this set, that will mean what? It does not belong to the intersection. That means it does not, that means there exists. So, x belongs to this, does not belong to intersection. That means at least it does not belong to one of them. So, there exists some j not such that x does not belong to this. And if x does not belong to this for some j not, it does not belong to union. So, there exists. So, that means there exists a j not such that from j not onwards x does not belong to this. That means x cannot belong to any one of them. That means for every k bigger than or equal to j, there is a j not such that for every k bigger than or equal to j, bigger than or equal to does not hold equal to j. That means this must happen for less than 1 over k. So, saying x does not belong to this set will mean there is a j not. So, that for every k bigger than or equal to j not, the distance is less than k. That will imply that f and k x converges to f of x. So, what we are saying is essentially that this is the set of points where f and k does not converge to f and we want that set to have measure 0. So, once again let me revise saying that f and k does not converge to f x is same as saying x does not belong to this set. So, because if x does not belong, so for every does not belong to intersection over j of some sets. That means there is a set, there is a j not such that x does not belong to j not. So, it belongs to this for some it should be union of k. So, that is a union here. So, that means for every k bigger than j not this is less than k. So, we have to only find a subsequence. So, we have to complete the proof. We have to construct a subsequence. So, that measure of this set is equal to 0. So, let us write this set as A. So, A is the set of points where intersection over j equal to 1 to infinity union over k equal to j to infinity of where f n k x minus f x is bigger than or equal to 1 over k. And let the inner portion which is a union let us call that set as A j. So, what we want to prove is that mu of intersection A j is equal to mu of. So, we want to prove that mu of A is 0, but this set A is contained in A j because this is the intersection is smaller. So, A is subset of A j. So, showing that the mu of the set A is 0, A is contained in A j that means mu of A is less than or equal to mu of A j. So, we will be through if we can prove that mu of A j is converged to 0 as j goes to infinity. So, we have to construct a subsequence f n k with this property that mu of A j is of these sets must go to 0 as j goes to infinity. So, this will be true. So, we want a subsequence that mu of A j is go to 0. So, that will be true If we could find say for example, a sequence f n k says that the measure of the set where f n k minus f x is bigger than or equal to k, say it is less than 1 over 2 to the power of k plus 1. Suppose, we can choose our subsequence in that way then what will happen? Then mu of A j which is nothing but union of from k equal to j to infinity of these sets. So, that is by countable sub additive property mu of A j will be less than or equal to summation k equal to j to infinity mu of these sets and each mu of this set is less than 1 over 2 to the power k plus 1. So, that is summation j equal to infinity. So, that will be this is a geometric series with common ratio less than half. So, that will give you 1 over 2 to the power j. So, we will get mu of A j less than 1 over 2 to the power j. So, as j goes to infinity we will get mu of A j equal to 0 and hence we will be through. So, we have to only find a subsequence f n k with this property and that is done by using the fact that f n converges to f in measure. So, because it converges in measure, so by the property of convergence in measure, let us start with epsilon equal to 1 then convergence in measure says that mu of the set of those points where f n minus f x is bigger than 1 will be less than say half for every that goes to 0. So, that means after some stage the difference of the measure. So, measure of the set where f n minus f x is bigger than or equal to 1 will be small after some stage. So, let us that stage be called as n 1. So, using the fact the convergence in measure we can choose n 1 such that measure of the set where f n 1 minus f x bigger than 1 is less than half. Now, we proceed inductively supposing we have selected n 1 less than n 2 up to n k minus 1 have been selected with those required property. Then by the fact that the sequence f n is converging in measure for epsilon equal to 1 over 2 to the power k plus 1, select n k stage bigger than n k minus 1 such that the measure of the set f n k minus f x bigger than 1 over k is less than 1 over 2 to the power k plus 1. So, by induction this existence is a sequence f n k with this property is complete and hence we will have that this up sequence converges almost everywhere to f. So, this is an important result which is used sometimes to analyze sequences which converge in measure. So, this is called Rie's theorem. So, we have looked at various properties of. So, till now we looked at various properties of convergence, convergence almost everywhere, convergence point wise, convergence almost uniform, convergence uniform and convergence in measure and relations between such modes of convergence. There is another mode of convergence which arises when the functions f n are in LP spaces. Recall we have defined LP spaces the p th power integrable functions. So, one would like to know that when sequences converge in LP, does this convergence have any relation with convergence in measure, point wise convergence or convergence almost uniformly. So, we will analyze these things next. So, let us recall what is the meaning of saying that have a sequence converges in the p th mean or LP. So, saying that a sequence f n where p is bigger than 1 less than infinity and f belongs to LP. So, let us take functions f n and f in LP. So, saying that f n converges to f in LP or sometimes one also writes this as f n converges in the p th mean to f if the p th norm of f n minus f goes to 0 as n goes to infinity. So, this is called convergence in the p th mean. So, we would like to find out the relations between the p th means and other modes of convergence. So, the first property is that in general, convergence in the p th mean does not imply uniform convergence and it need not imply almost uniform convergence or convergence almost everywhere. So, convergence in p th mean need not imply any one of uniform convergence almost uniform convergence or convergence almost everywhere. So, to give an example of measure space and a sequence of functions in LP say that f n converges to f in LP, but f n does not converge in uniformly or almost uniformly or almost everywhere. So, we will give an example of a sequence which does not which converges in LP, but does not converge almost everywhere. So, that example itself will imply that p th mean cannot imply uniform convergence because uniform implies convergence almost everywhere. And similarly, same example will be sufficient to say that p th mean does not imply almost uniform convergence because uniform convergence implies almost uniform convergence. So, we want to construct a sequence of functions in LP such that the sequence converges in LP, but does not converge almost everywhere. So, for that, let us look at the measure space, the interval 0 1, Lebesgue measurable sets in 0 1 and lambda the length function. And let us write, if you recall, we had constructed a sequence of measurable functions on this measure space by looking at dividing the interval into equal parts by using the binary points. And that example showed that, that was the example which showed that convergence in measure does not imply convergence almost everywhere. So, that is the sequence we are looking at again. So, let us recall that sequence. So, f n for every n bigger than or equal to 1, let n be represented as k plus 2 to the power m, where this k is between 1 and 2 to the power m and m is. So, this we showed that for every n can be written uniquely in the form k plus 2 to the power m, where m is a positive integer and k is between 1 and 2 to the power m. So, for whenever m n is represented this way, we define f n to be the indicator function of the interval k by 2 to the power m and k plus 1 to k plus 1 by 2 to the power m. So, look at the k th interval of length 1 over 2 to the power m and define f n to be the indicator function of this interval. So, we showed earlier that this sequence of measurable functions does not converge almost everywhere and this converges in measure to the function identically equal to 0. So, this sequence of measurable functions converges in measure to f identically 0, but it does not converge to f at any point in 0, 1. So, this does not converge almost everywhere or at any point actually. Let us just prove that this sequence of measurable functions is in L p because this is the indicator function of an interval. So, the function takes the value 1 on this interval and interval is finite. So, obviously it implies this is a L p integrable function. So, indicator function of a sub-interval, so f n belongs to L p. So, we have got a sequence of functions in L p such that and let us look at what is the L p norm of f n because it is the indicator function of a interval of length 1 over 2 to the power m. So, it is precisely equal to 1 over 2 to the power m. The function takes the value 1 on that interval of length 1 over 2 to the power m. So, its L p norm is 1 over 2 to the power m raise to power 1 over p. So, this implies that the norm of f n minus f which is identically 0 goes to 0 as n goes to infinity because as n goes to infinity m also goes to infinity. So, this is a sequence of functions in L p which converges to the function identically 0, but the sequence does not converge almost everywhere. So, that proves our claim namely convergence in p, convergence in the pth norm does not imply convergence uniform or convergence almost uniform or convergence point wise. So, convergence in the pth mean does not imply this and so as we indicated this cannot imply uniform or almost uniform because both of them imply convergence almost everywhere. Next, let us look at fact that convergence in the pth mean always implies convergence in a measure. So, let us take a sequence f n which converges to f in L p and let epsilon greater than 0 be arbitrary. So, what we have to show? Let us just look at what we have to show. So, f n converges to f in L p. So, we are given that f n converges to f in pth mean to show f n converges to f in measure. That means, so what does it mean? That is for every epsilon bigger than 0, we have to look at the measure of the set x belonging to x such that f n x minus f of x bigger than or equal to epsilon limit n going to infinity must be equal to 0. So, this is what we have to show. So, let us call e to be the set x belonging to x where f n x minus f of x is bigger than or equal to epsilon. Now, we know that f n goes to f in L p. So, f n goes to f in L p that is equivalent to saying that f n minus f norm of that pth norm goes to 0. Let us take the power p also. So, what will that mean is the following. So, saying that in L p, let us look at the norm f n minus f p, this is equal to integral mod f n minus f to the power p d mu. So, that is what and this goes to 0. But this is equal to integral over e of the same thing f n minus f to the power p d mu plus integral over e complement of the same thing. So, mod f n minus f to the power p d mu. Now, on the set e, on the set e, on the set e f n is bigger than or equal to epsilon. So, on the set e, this difference is bigger than epsilon. So, it is bigger than or equal to epsilon to the power p into mu of the set e plus integral over e complement. But this is a non-negative function. So, even if I drop this term, it will still remain bigger than or equal to mu of e. So, that implies that mu of e is less than or equal to mod mu of e is norm of f n minus f to p divided by epsilon to the power p. So, this is the inequality that we get for, this is actually an important inequality in the theory of probability and this is called Chebyshev's inequality, a very simple inequality, but you see it gives a powerful outcome. So, that is following, because f n converges to f in L p. So, this goes to 0. So, that means that mu of e equal to 0. So, the set where f n is bigger than f by distance epsilon, mu of that depends on epsilon and so that goes to 0 for every epsilon. So, that will prove that if f n, so that proves that if f n converges to f in the p th mean, then it converges in measure. So, as we looked at just now, the proof is simple. Look at the set where f n minus f x is bigger than or equal to epsilon, then the integral mod of f n minus f n, norm of f n minus f to the power p can be split into two parts. So, and one part where the function f n minus f is bigger than epsilon is this and the remaining part will drop. So, inequality still stays, so mu of e is less than or equal to this, which goes to 0. So, that proves that convergence in the p th mean implies convergence in measure. However, in general, the convergence almost everywhere does not imply convergence in the p th mean. So, for that let us look at a simple example. Look at the Lebesgue measure space or Lebesgue measurable sets lambda. So, that is the Lebesgue measure space and look at the function f n which is defined as n raise to the power minus 1 by p into the indicator function of the interval 0 to n. So, this function first of all let us observe that this function belongs to L p because integral of f n raise to the power p is just n to the power minus 1 to the power p. So, that is 1 over n and into the measure of the interval 0 n that is n. So, all the each f n is a L p function, its integral norm is equal to L p norm is 1 and this converges uniformly to f. That is obvious because the values of the function that is taking on a larger and larger interval is 1 over n. So, becoming smaller and smaller. So, we can always find for every x, we can find a stage such that after which the distance will be less than 0. So, that means f n converge. So, it is easy to check. Let us verify. This is easy to verify that f n converges to f uniformly. However, we have now said that the L p norm of each f p is 1. So, that does not converge to f in the p th norm. So, what we are saying is convergence almost everywhere does not imply convergence in the p th mean. This is convergence almost everywhere. Here is another observation that none of uniform convergence or almost uniform convergence or convergence in measure imply convergence in the p th mean. So, that means the none of this each one. So, that means in general uniform convergence does not imply convergence in the p th mean or convergence almost uniform does not imply and similarly convergence in measure need not imply convergence in the p th mean. Because obviously, because both uniform convergence and almost uniform convergence apply convergence almost everywhere. So, if this were true, then we will have a contradiction. So, this is true. Next we want to say that convergence in the measure need not imply convergence in the p th mean. So, for that let us look at the example of the measure space 0, 1, Lebesgue measure space and let define g n of x equal to n to the power 1 over p. It is something similar to the previous one. Instead of minus 1 over p, it is n to the power 1 over p and into the indicator function of 0 to 1 over n. So, here the interval where function is non-zero is shrinking, but the value is increasing. In the previous one in the previous one, the value was decreasing, the length was increasing. So, it is the other way round of this. So, look at this function g n. All of these functions g n are in L p because integral mod g n will be equal to the power p will be n into indicator function. So, integral is equal to 1. So, each one of them has got is in L p with integral equal to 1. And it is obvious that the sequence g n converges in measure. It converges in measure that is obvious because the set where it is going to be together or equal to epsilon is going to be shrinking, it is 0 to 1 over n. So, that will go to 0. So, converges in measure and to identically equal to 0 and its integrals L p norms are equal to 1. So, hence g n does not converge to f identically 0 in the pth mean. So, that means we have produced a sequence of functions which is convergence in measure, but does not imply convergence in L p spaces. However, if the underlying measure space is finite, then the uniform convergence does imply convergence in the pth mean. So, that is quite obvious to verify. Let us just look at mu of a measure space so that mu of x is finite and let f n be a sequence of functions in L p converging uniformly to a measurable function f. We want to show that f n converges to f in L p also, but what is uniform convergence? Uniform convergence means for every epsilon bigger than 0, there is a stage such that after which, given every epsilon bigger than 0, you can find a stage n naught such that the distance between f n and f is less than epsilon for every n bigger than or equal to n 0 and for every x. So, uniform means for every x, the same stage satisfies the required property. So, for every x, there is a single stage n naught such that f n x minus f of x is less than epsilon. .. Now, let us look at the absolute value of the function mod f to the power p. See, we have just given that f is converging uniformly and each f n is in L p, but we do not know whether f is in L p or not. So, we have to first prove that this is in L p, but look at the absolute value of f to the power p. So, by triangle inequality, I can write it as mod of f n minus f plus. So, this should be also less than or equal to by triangle inequality add and subtract f n to the power p. Now, absolute value of a plus b is always less than or equal to 2 times the maximum of the two values. So, this is less than or equal to 2 times the maximum value of f n minus f and absolute value of f n. Of course, everything is to power p, but that is same as less than or equal to 2 to the power p into the maximum of this to the power p. Now, 2 to the power p and the maximum will always be less than or equal to the sum. So, you can write this is less than or equal to 2 to the power p mod of f n minus f to the power p plus mod of f n to the power p. So, what does that imply? So, we can integrate both sides with respect to mu. So, integral will be less than or equal to 2 to the power p into f n minus f n naught minus f to the power p d mu plus 2 to the power p of f n naught and this we just now showed is less than epsilon. So, this will be less than 2 to the power p epsilon to the power p mu of x plus 2 to the power p norm of f n to the power p and each of them being finite that says that the function f is in L p. So, what we have shown is if the underlying space is having finite measure and f n belongs to L p and converges uniformly to f, then the limit function is also in L p. So, that is what we have shown and now look at the difference. So, the difference of f n minus f to the p th norm is by definition integral of mod f n minus f to the power p raise to the power 1 over p, but for n bigger than n naught this difference is less than epsilon to the epsilon to the power p raise to the power 1 over p. So, that is epsilon into mu x raise to the power 1 over p. So, as epsilon goes to 0, this will go to 0. So, that proves that f n converges to f in the p th mean. Let us also analyze what happens the relations between L p spaces and other modes of convergence when the underlying space is of finite measure. So, the result says even if the underlying measure space is finite, then none of uniform convergence or convergence almost everywhere need imply convergence in the p th mean. So, this condition is not good enough to ensure that almost uniform convergence that will imply. That we have just now looked at the function g n x is equal to n to the power 1 over p. So, these are all L p functions and they with norm equal to L p norm equal to 1. So, they cannot converge in L p, but we know that this converges in measure and hence almost uniform. So, also the convergence in the p th mean need not imply almost uniform convergence or convergence almost everywhere. The other way around inequality, even when this is finite the convergence in p th mean need not imply almost uniform or convergence. So, that earlier example of the measure space 0 1 0 1, Lebesgue measurable sets and f n to be the indicator function of the interval i k m, where i k m is the interval of length to the power m. Just now we consider this example. So, this is the example of functions that converges to f uniformly. So, it converges to f in the p th, f identically 0 in the p th mean, but we know that it does not converge almost everywhere and hence it cannot also converge almost uniformly, because almost uniform convergence implies convergence almost everywhere. So, these are the various ways of looking at modes of convergence and analyzing them. Let me just look at this. All the implications put together in the form of a diagram. So, here is the diagram going to look at this. So, here we have got the notion of uniform convergence, uniform almost everywhere in the p th mean. Here is point wise, point wise almost everywhere in measure and almost uniform. As we all know that uniform convergence is the strongest one. So, uniform we have already seen earlier that uniform implies point wise and of course, point wise implies point wise almost everywhere and this other way around implications need not hold. Simple examples. In uniform, we will imply uniform almost everywhere. Actually, uniform is uniform almost everywhere with the set to be empty set. So, uniform implies uniform almost everywhere and just now we proved when mu of x is finite. So, this green arrow indicates that implication under the condition that mu of the whole space is finite. So, uniform convergence almost everywhere and underlying space finite implies convergence in the p th mean that we saw just now. And this way, we have already seen uniform implies point wise, apply point wise almost everywhere. And we also saw that point wise convergence, point wise almost everywhere. In general, linear implies convergence in measure, but when underlying measure space is finite, point wise will imply convergence in measure. Of course, point wise almost everywhere need not imply almost uniform, but we just showed today that if the underlying space is finite measure, then the point wise implies almost uniform. And obviously, almost uniform we have shown implies point wise almost everywhere. And finally, we also showed today that almost uniform implies in measure and when the underlying space is finite, in measure implies convergence in the p th mean, we use Chebyshev's inequality. And just now we said that convergence always in the p th mean always implies convergence in measure. So, this is the overall picture of various modes of convergence. So, with this we come to the end of the basic concepts of measure theory. Whatever we have not done is essentially looking at, one can also look at various ways of necessary and sufficient conditions under which a sequence f n converges to f in L p. That is one thing we have not done, but in the limited scope of lectures, we wanted to cover almost in 40 lectures the basic concepts of measure theory and that we have covered. Another thing that we have not proved is looking at the change of variables formula for R n. That again requires a bit of work and technical things. So, we have not covered that in our discussions. So, with that we come to an end of this course video lectures on measure theory. In the next one lecture, I will just try to give an overall view of the things that we have done in our course. Thank you.