 Welcome to lecture 39 on measure and integration. Today we will be looking at a special topic which is called various modes of convergence for measurable functions. So, topic for today's discussion is modes of convergence. Let us recall some of the ways of convergence that we might have already looked into earlier courses. So, let us take a sequence of functions fn is a sequence of functions on measurable space xs. Now, saying that fn converges to f point wise means that fnx is same as saying that fnx converges to f of x. These are numbers for every x belonging to x. So, this is converging point wise and you might have already come across something called fn converges to f uniformly. So, what is the difference between this point wise convergence and uniform convergence? Let us just write down in terms of our definitions of epsilon delta. So, point wise means for every x, for every epsilon bigger than 0, there is a n naught which will depend upon the point x and epsilon, such that fnx minus f of x is less than epsilon for every n bigger than n naught. So, that essentially means that the numbers fnx converges to the number f of x as n goes to infinity. So, given epsilon bigger than 0, there is a stage after which fnx is close to f of x. Of course, this stage may depend upon the point x and the number epsilon. Let us look at what is point wise convergence. So, saying that fn converges to f uniformly means for every x, for every epsilon bigger than 0, there is a stage n naught which does not depend on x, which depends only on epsilon such that fnx converges to f of x. That means that is mod fnx minus f of x is less than epsilon for every n bigger than n naught and the same stage works for every x. You must have already seen that the uniform convergence implies point wise convergence and the converse need not be true. So, uniform convergence implies point wise convergence, the converse need not be true. That you must have seen in your earlier courses in analysis. But we are going to look at today's functions which are defined not only on measure spaces. Actually, they are defined on measure spaces, not only on measurable spaces. So, we are going to look at a sequence fn of measurable functions on a measure space xs mu and we had already looked at some concept called convergence almost everywhere. So, let us define formally again that fn converges to f almost everywhere if everything is defined on a measure space xs mu. So, these are functions defined on the measure space xs mu and we say fn converges to f almost everywhere. If the set n, all x belonging to x is that fnx does not converge to f of x. If that is the set n, then we want n should belong to the sigma algebra s and mu of n should be equal to 0. So, that is convergence almost everywhere. So, obviously, point wise convergence implies convergence almost everywhere. Obvious examples one can construct to show that this otherwise implication need not hold. Convergence almost everywhere need not apply point wise convergence. Now, let us try to look at a relation between convergence almost everywhere. Let us just recall once again what we are saying. So, we are saying fn converges to f point wise if the sequence of numbers fnx converges to f of x. So, that is same as saying that for every epsilon bigger than 0, there is a n naught such that which may depend upon the point x and the number epsilon such that absolute value of fnx minus f of x is less than epsilon for every n bigger than n naught. So, the numbers fnx are converging. The sequence fnx is converging to f of x. Saying fn converges almost everywhere to f means the set of points where fnx does not converge to f of x that set is a set of measure 0. So, that is convergence almost everywhere or saying that fn converges uniformly to f meaning for every epsilon there is a stage n naught which depends only on epsilon n naught on the point x such that for every x the distance between fnx and f of x is less than epsilon for every n bigger than or equal to n naught and that stage works for every x. That is important thing for uniform convergence. So, instead of writing English all the time whenever fn converges to a point wise we will write fn arrow with a p above indicating it is a point wise convergence fn converges to f point wise. Convergence almost everywhere will be indicated by fn arrow pointing right side to f with a symbol a dot e dot saying almost everywhere to f. This is also written as fn converges to f fn right arrow f a e indicating this convergence is almost everywhere. Similarly, for uniform convergence we will denote it by symbols fn arrow to pointing towards the right and u above the arrow indicating it is uniform convergence. So, as pointed out earlier that fn converges to f uniformly implies fn converges to f point wise and that implies fn converges to f almost everywhere and none of the backward implications is true. So, one can easily construct counter examples to show that convergence almost everywhere and n naught imply convergence point wise and point wise convergence of n naught imply convergence which is uniform. However, we will like to still analyze whether some conclusions can be drawn from point wise convergence or almost everywhere convergence in terms of a uniform convergence. So, to analyze that let us assume that fn converges to f almost everywhere and let us take a set e which is in the sigma algebra S such that the measure of this sigma set e is finite and let us be given two numbers epsilon and delta arbitrary. Then we would like to show that there exists a set e lower epsilon a measurable set e lower epsilon of and a positive integer n naught which will depend on epsilon and delta such that with the following properties namely that e epsilon is a subset of e and the difference e minus e epsilon is small and for every x belonging to e epsilon on e epsilon fn x minus f of x is less than delta for every n bigger than or equal to n naught and that meaning that this for every x the same stage n naught works. So, essentially saying that the difference between fn and f stays less than delta for all n bigger than n naught the end of the stage does not depend on epsilon. So, let us prove this result. So, to prove this so we are given that fn converges to f almost everywhere. So, let us that means if we define the set n to be the set of all x belonging to x such that fn x does not converge to f of x then this set then mu of the set n is equal to 0. So, that is what is given to us. Now, let us look at the set that means on n complement on the complement of this set fn x converges to f of x for every x belonging to n complement. So, let us define. So, let us write the set e say m. So, we are also given delta. So, let us write e m delta to be the set of all points x belonging to x such that fn x such that the difference fn x minus f of x fn x minus the difference fn of x is less than delta for every n bigger than or equal to m. So, let us look at this set e m delta. So, this set depends on m and on delta. So, the difference between fn and f is less than delta for every n bigger than or equal to m. So, let us note that these sets e m deltas they belong to the sigma algebra and if the difference between fn and fx is less than delta for all n bigger than m there is also going to be also true for all n bigger than n plus 1. So, that means that e m delta is an increasing sequence. So, e m delta is a subset of e m plus 1 delta and further not only it is increasing because we are given that fn x converges to f of x for every x. So, that means the sequence is increasing and on n complement whether know the difference is. So, the sequence is increasing sequence and we can say that n complement is contained in union of e m delta m equal to 1 to infinity because on n complement it converges. So, for every x we can find some stage after which it will be less than for a given delta. So, the fact that on n complement fn x converges to f of x that implies that the set n complement is a subset of the union of e m deltas. So, thus if you look at the sequence e intersection e m complement delta then this is a decreasing sequence. It is a decreasing sequence of sets and mu of e being finite implies that limit n going to infinity mu of e intersection e m complement delta must be equal to this is a because e m was increasing sequence. So, the complements will be decreasing sequence. So, it must converge to intersection of all of them and that is contained in the complement e intersection n complement. So, that is equal to 0 because this sequence of sets decreases to a intersection n complement. So, that is equal to 0. So, what does that imply? So, that implies saying that this converges to 0. So, that implies that so given there is a stage. So, let us find some. So, let us find e m m naught such that mu of e intersection e complement m is m naught is less than epsilon is less than n and on e intersection e m naught complement we know that f n x minus f of x is less than delta for every n greater than or equal to m naught. So, given epsilon delta and mu of e finite we have found a stage m naught say that and a set e m naught say that e intersection. So, this is nothing but e minus the set that is less than epsilon and on the set this is less than epsilon and on the set e intersection this is less than the difference between f n and f is less than delta. So, that proves that required claim namely. So, this proves the required claim that namely if f n converges to f almost everywhere and we are given a set e of positive finite measure mu of e is finite then for every epsilon and delta we can find a subset e epsilon inside e says that the measure of on e epsilon f n x minus f x less than delta for every n greater than the stage n naught and the measure of the difference e minus e epsilon is small. So, this is consequence of the fact that f n converges to f almost everywhere and mu of e has got finite measure. As a consequence of this we prove theorem called Ligorov's theorem it says that if f n converges to f almost everywhere and e is a set of finite measure then given epsilon there is a set e epsilon such that measure of the set e minus e epsilon is less than epsilon and f n converges to f uniformly on e epsilon. So, let us prove so we are given that f n converges to f almost everywhere and mu of e is given to be finite. So, by the result just now proved for every delta equal to 1 over n let us apply the previous result for epsilon equal to delta equal to 1 over n there exists a set e n contained in e such that mu of e minus e n is less than 1 over n and on e n on the set let us write e m here just for the sake of clarity because we are going to apply it for every n and for every e m we also such that there is a set and a stage n m such that f n of x minus f of x is less than 1 over n m for every n greater than that stage n m. So, this is by the result just now we have proved that whenever f n converges to f almost everywhere and e is a set of finite measure then with delta equal to 1 over n sorry this is less than there is a set e m say that this is less than epsilon for every epsilon. So, let us do it with epsilon to the power 2 to the power m. So, we are applying the previous result with delta equal to 1 over m replacing epsilon by epsilon divided by 2 to the power m. So, we will get a set so there is a difference between e and e m is less than epsilon by 2 to the power m and on the set e m the difference f m f n minus f x is less than 1 over m for every n bigger than or equal to e m. So, now, let us define so define the set e epsilon equal to intersection of e m m equal to 1 to infinity. So, with that let us compute that this is the required set. So, mu of e minus e epsilon is equal to mu of union of e minus e m m equal to 1 to infinity because e epsilon is intersection so minus that will make it union. So, which is less than or equal to sigma m equal to 1 to infinity mu of e minus measure of e minus e m which is less than epsilon by 2 m. So, which is less than or equal to m equal to 1 to infinity of epsilon by 2 to the power m which is less than or equal to epsilon. So, we get that measure of the set e minus e epsilon is less than epsilon for is less than epsilon. Also, if we look at x belongs to e epsilon so that is equal to intersection of e m m equal to 1 to infinity. So, what does that imply? So, that means because it belongs to intersection so implies that x belongs to e m for every m. Mod of f n x minus f of x will be because it belongs to e m for every m so it is less than 1 over m for every n bigger than or equal to n m. So, that means for every m we can find a stage so that after which the difference f n x minus f x less than 1 over m for every x so that implies that f n converges to f uniformly on e epsilon that converges to e epsilon uniformly. So, that proves Igorov's theorem namely so this proves Igorov's theorem. So, which says that if f n converges to f almost everywhere then given any set e of finite measure we can find a part of it so that on that part of that set e f n converges to f uniformly and the measure of the remaining that is e minus e epsilon is small is less than epsilon. So, for every epsilon we can find a set e epsilon contained in e such that mu of e minus e epsilon is less than epsilon and on e epsilon f n converges to f uniformly. So, this is what is called Igorov's theorem. So, as a particular case of Igorov's theorem in case so let us define make a new definition for measurable functions f n and f and a set e in the sigma algebra. One says f n converges almost uniformly to f on e if for every epsilon there is a set e epsilon belonging to the sigma algebra such that the measure of the difference e minus e epsilon is small and on e epsilon f converges uniformly. So, such convergence is called almost uniform convergence. So, it is essentially saying it is uniform except on a set of measure small so that is what almost uniform is to be understood. So, Igorov's theorem can be restated as that if f n converges to f almost everywhere on a set e of finite measure then f n converges to f almost uniformly on e. So, almost everywhere convergence implies almost uniform convergence on every set of finite positive measure. So, in particular case when mu of x is finite this will imply f n converges to f almost everywhere on the set x. So, will imply that f n converges to f almost uniformly on x. So, on finite measure space is almost everywhere convergence implies convergence which is almost uniform. And in fact, converse of this proposition is also true when we have got the measure space finite. So, the converse says that if mu of x is finite and f n converges to f almost uniformly on x then f n converges to f almost everywhere. So, the proof of that is almost obvious because by the given hypothesis for every integer n we can find a set f n in s in the sigma algebra say that the measure of the set f n is small and f n converges to f uniformly on the complement of it. So, let us define the set f which is intersection of all these f n's then measure of the set f is less than or equal to measure of each f n which is less than 1 over n. So, the set f is a set of measure 0 and outside this if f if a point x belongs to f complement then that will mean it belongs to some f n complement for some n and on that convergence is uniform. So, f n converges to f in particular. So, that will imply that on f complement f n x converges to f of x. So, that proves the converse part of the Igorov theorem for finite measure spaces. So, let us draw a picture and try to see what these implications mean. So, we have got convergence is one is which is convergence point wise. So, this is the point wise convergence and we have got convergence almost everywhere. Convergence which is uniform and convergence which is almost uniform. So, we know that the point wise convergence implies convergence which is almost everywhere and convergence uniform implies which is the point wise convergence. So, convergence uniform implies convergence point wise and point wise implies almost everywhere. The other way around inequality, this is not true and this implication is also not true in general. We just now showed that convergence almost everywhere obviously does not imply this convergence. Convergence uniform obviously implies convergence almost uniform because almost uniform means outside a set of measures small. So, this is uniform implies convergence almost uniform also. So, almost everywhere convergence or point wise convergence cannot imply this because it does not imply this. So, and this implication convergence almost everywhere obviously does not imply in general this. But what we can say is that when it is finite, so almost this implies this when mu of x is finite and of course this also implies the other way around when this is finite. So, this is what one way almost everywhere implies convergence almost uniform that is Egorov's theorem and the converse always hold. So, these two are same if underlying measure space is a finite measure space. These are the implications that we can say are true. . So, next let us point out the fact that Egorov's theorem says that almost everywhere convergence on finite measure space is gives rise to almost uniform convergence. This can be used to prove a important theorem called Lugens theorem which says that if f is a real valued function on reals which is measurable and epsilon greater than 0 is arbitrary. Then there exists a continuous function g say that where f differs from g is a set of measure small. So, what it says that every measurable function is almost continuous you can look at it this way. So, we will not prove this result. So, basically the idea is on every interval one tries to apply this apply Egorov's theorem and then try to patch it up. So, those interested should look up the text book referred for this result saying that Egorov's theorem has an application namely every measurable real valued measurable function on reals is almost equal to a continuous function. That means the set of we can find a given any epsilon one can find a continuous function such that the difference f x not equal to g x is the measure of that set is small. So, these are the relationships between point wise convergence, convergence almost everywhere, uniform convergence and almost uniform convergence. But keep in mind almost uniform convergence is not uniform almost everywhere. So, these two are different things almost uniform means that except outside a set of measure small the convergence is uniform. But if you say uniform almost everywhere that means outside a set of measure 0 the convergence is uniform. So, almost uniform is not same as uniform almost everywhere. So, keep that in mind. Next we will like to discover another important modes of convergence called convergence in measure. This mode of convergence is very useful in the theory of probability analyzing what are called convergence of random variables. So, we will not go into the probabilistic aspect of this. We will just look at the measure theoretic aspect of what is convergence in measure. So, a sequence of measurable functions f n is set to converge in measure to a measurable function f on a measure space x s mu. If for every epsilon look at the set where f n minus f x is bigger than or equal to epsilon. So, this is the kind of set where f n is not going to the difference remains bigger than epsilon. So, collect all such points and if look at the measure of this set. If the measure of this set goes to 0 becomes smaller and smaller as n goes to infinity then we say f n converges to f in measure. So, saying f n converges to f in measure is for every epsilon. This is important for every epsilon bigger than 0. Look at the measure of the set where f n minus f is bigger than or equal to epsilon. Measure of that set limit n going to infinity should be equal to 0. So, this is called convergence in measure for function and this will write as f n with an arrow f and above the arrow will put the symbol m indicating that f n converges to f in measure. Let us look at some examples to understand convergence in measure vis-à-vis almost everywhere convergence. So, let us look at the sequence of functions f n which are defined on the Lebesgue measure space or Lebesgue measure space that is real line Lebesgue measurable sets and the length function. Let us take the function f n to be equal to the indicator function of n to n plus 1. So, what we are doing is where f n is the indicator function of the interval n to n plus 1. So, if this is the line, this is 1 2 3 n n plus 1. So, what is f 1? So, f 1 is the indicator function of 1 to 2. So, f 1 is nothing but the function. So, this is f 1. What is f 2? So, this is f 1. What is f 2? f 2 is the indicator function from 2 to 3. So, this is f 2. So, this is f 3 and so on and this is f n. So, what is f n is the function f n is nothing but the constant function 1 on the interval n to n plus 1. So, this graph is shifting as you move. Clearly, f n x converges to f x for every x. And that is obvious because if we are given a point x somewhere. So, here is a point x. Then after some stage, we can find a n which is bigger than this. So, you can find something some n naught which is bigger than this. Then f n naught of x will be equal to 0, which is. So, that means for every x, we can find a stage n naught, so that this is equal to 0. That will happen for every n bigger than that. That means the f n x converges to f of x, which is identically equal to 0. So, the sequence of functions f n x, this sequence of functions point wise, it converges to the function f of x, which is identically 0. So, f n converges to f identically 0 point wise. But let us look at the measure of the set, where f n minus f is bigger than or equal to say 1. So, because f n gives the value 1 or 0 only, so the set where f n x differs from f of x is precisely the interval n to n plus 1. So, Lebesgue measure of the set where f n x minus f of x is bigger than or equal to epsilon equal to 1 is equal to 1 and that never goes to 0. So, this is the sequence of measurable functions, which converges point wise, but it does not converge in measure to the function f identically equal to 0. So, point wise convergence need not imply convergence in measure. Point wise convergence, convergence almost everywhere in particular also does not imply convergence in measure. Let us look at another example, which is easy to describe pictorially. So, what we are going to look at is, look at the interval 0 1 divided into two equal parts. So, that is two equal parts 0, half and 1. At the next stage, divided into four equal parts 1, 2, so that is 0, 1 by 4, half, 3 by 4 and 1 and so on. So, define f 1 to be the indicator function of the whole interval 0 to 1. Define f 2 to be the indicator function of the interval 0 to half. So, this is f 2, which is the indicator function of 0 to half and this is f 3, which is equal to indicator function of half to 1. So, that is f 3. Similarly, this is f 4, this is f 5, this is f 6, f 7 and so on. So, each f n is nothing but the indicator function of the interval with n points, which are obtained by dividing the interval into equal parts at every stage. So, it is an indicator function of let us write an interval i k n, where the length of i k n is equal to 1 over 2 to the, it is going to be, depending on which stage it is something like 1 over 2 to the power, interval of length 1 over 2 to the power n or something like this, because it is going to be shrinking. So, if this is how the sequence f n is obtained, then it is clear that the set on which f n is 1 is becoming smaller and smaller. So, guess is this f n converges to f identically 0 in measure, but this sequence f n does not converge to f point wise. And for that, the reason is given any point x, given any point x, we can always find some f n like this, where the value of the function is f n is going to be equal to 1, because what is happening is f 1 is the indicator function of the whole interval, f 2 is this. So, the set over which the function is non-zero is taking the value 1 is fluctuating, but it is moving at no stage, it becomes 0. And actually at every point x, there will be some f n, which will be giving the value 1. So, this will not converge almost everywhere or point wise. So, this will be example of a sequence, which converges in measure, but not point wise. One can formally write this as follows. To write this more rigorously, let us look at for every n. First of all, choose a unique integer m, say that n lies between 2 to the power m and 2 to the power m plus 1. So, that is obvious that for any n, you can find a integer with this property. And now, next look at integer k between, for any integer k between 0 and 2 to the power m, this number n can be written as 2 to the power m plus k. So, because n to n plus 1 can be divided into intervals of length 1 over 2 to the power m. And once you do that, so you can find a k, say that this n is equal to 2 to the power m plus k. So, what we are saying is every natural number n can be expressed uniquely as in the form to every n, you can associate a pair, namely small m comma k, where m is such that n lies between 2 to the power m less than or equal to and less than 2 to the power m plus 1 and k is between 0 and 2 to the power m. So, this correspondence n with this pairs is unique. And obviously, as n goes to infinity, this m will also go to infinity and conversely as m goes to infinity n goes to infinity. So, how is this function f n defined? So, define f n for any n represent it uniquely as k plus 2 to the power m and take f n to be the indicator function of i k m. So, this is the interval starting at k by 2 to the power m and going to k plus 1 2 to the power m. That means, it is a interval of length 1 over 2 to the power m. So, this is, so f n is 1 on this interval and you can see that this interval is shifting but never becoming, is becoming smaller and smaller but never vanishing. So, f n is a sequence of measurable functions because of the indicator functions of intervals and for every x, you can find a stage. If n is for any n, you can find a, if n is equal to this, we can find a stage. So, that x will also belong to a interval of starting with l 0 with the base as 2 to the power m plus 1. So, that will mean that for every n, there is a stage n dash says that f n, n dash of f n dash at the point x is also equal to 1. So, that will prove that the sequence f n does not converge to the function identically 0 point wise. On the other hand, it is quite obvious that the set of points where this is bigger than epsilon for every epsilon bigger than 1, it is empty set and less than or equal to 1, that is the interval which is becoming smaller and smaller. So, that says that this is a sequence of functions which, so that will go to 0. So, this is a sequence of functions which converges in measure but not point wise. So, what we have shown is that the convergence in measure is neither implied by point wise convergence or convergence almost everywhere and neither implies convergence in measure. So, convergence in measure and point wise convergence are neither implied or implied by each other. So, these are, this is the concept of convergence in measure which is quite different from point wise convergence. However, when the underlying space is with finite measure, one can draw some conclusions and because in the theory of probability, the underlying measure space has got total mass 1. So, we want to look at this concept of convergence in measure when the underlying measure space has got finite measure. So, we want to prove that if mu of x is finite and f n converges to f almost everywhere, then f n converges also in measure to the set f. So, what we want to show is that convergence for finite measure space is convergence almost everywhere implies convergence in measure. So, let us look at a proof of this fact. So, recall that saying that f n converges in measure to f is same as saying, so this is what we want to show, we want to prove f n converges to f in measure. So, what we have to show? We have to show that for every epsilon, if we look at the set of those points where f n minus f is bigger than or equal to epsilon, then the measure of this set goes to 0 as n goes to infinity for every epsilon. So, let us call this set x belonging to x, f n x minus f of x bigger than or equal to epsilon, a set. So, it depends on epsilon and it depends on n. So, let us call this set as a n epsilon. So, a n epsilon is equal to the set of points where f n minus f is bigger than or equal to epsilon and we want to show that the measure of this set goes to 0 as n goes to infinity. Now, let us observe that a n epsilon is obviously a subset of the union of the sets a m epsilon from m equal to n, because this is one of those sets in the union. So, this is showing that mu of a n epsilon goes to 0 is same as is enough, if we can show that measure of this union goes to 0. But look at this union a m epsilon m bigger than or equal to n, this is a sequence of sets as n increases this union is going to become smaller and smaller. So, this is a sequence of sets which is decreasing. So, it is a decreasing sequence of sets and it decreases to, so where it will decrease, it will decrease to the intersection of the sets n equal to 1 to infinity of these sets and mu of the set x being finite. So, this is a sequence of sets which is decreasing to this set and mu of x being finite implies mu of the limit is equal to limit of the mu of the sets. So, limit n going to infinity mu of this is equal to this and let us observe that mu of the set which is intersection n equal to 1 to infinity union m equal to 1 to infinity a m epsilon. What is this set? This if x belongs to this set that means that for every n there is a m such that x belongs to a m for some m bigger than or equal to n and that is same as saying, which is same as saying. So, x does not converge to f of x because a m epsilon is the set where is bigger than x. What it says is that this is a set where f n does not converge to f of x and but we are given f n converges to f almost everywhere. So, this set has got measure 0. So, this limit is equal to measure 0. So, that proves the fact that this is of measure 0. So, that proves that limit of a n epsilon is 0 that means f n converges to f n measure. So, we started looking at the convergence in measure and we have proved given examples to illustrate that neither point wise convergence implies convergence in measure nor convergence in measure implies point wise convergence. However, when the underlying space is finite convergence almost everywhere does imply convergence in measure. So, we will continue this study of convergence in measure and also look at its relations with what is called convergence in mean or convergence in L P spaces. So, we will do that in the next lecture. Thank you.