 Welcome to lecture number 18 on measure and integration. If you recall in the previous lecture, we have defined the notion of integral for non-negative simple measurable functions. And we are started looking at the properties of this integral. So, we will continue this study of the properties of the integral for non-negative simple measurable functions. Then later on we will extend it to this integral to non-negative measurable functions. So, topics for today would be properties of integral for non-negative simple measurable functions. Continue the study of that and then define integral for non-negative simple measurable functions. If you recall, we had the last property that we have proved in the previous lecture was that if S n is any increasing sequence of non-negative measurable functions such that they converge to a simple non-negative simple function S of x, then integral of S is equal to limit of integral of S and d mu. So, that means under increasing limits, if the limit is again a non-negative simple measurable function, then you can interchange the order of integration and the notion of limit. So, integral of S d mu, which is S is the limit of S n d mu, S n is same as limit of n going to infinity of integral S n d mu. So, integrals of S n converge to integral of S. Let us observe one more simple property of this integral for any non-negative simple measurable function S. The integral S d mu can also be represented as the supremum of the integrals of S prime d mu, where S primes are non-negative simple measurable functions less than S. This property is obvious because S is less than or equal to S. So, the supremum has to be at least integral S d mu and it cannot be more because S prime less than S implies that integral of S prime is less than or equal to integral S. So, the supremum cannot be bigger than or equal to S d mu also. So, this is obvious property, but we will see later on an extension of this property later on. Next, let us observe another important property about the integral of non-negative simple measurable functions and that is the following. Suppose S n and S n dash are two increasing sequences of non-negative simple measurable functions both converging to the same limit. So, limit of S n x is same as limit of S n dash x. Then the claim is that the limit of integral S n d mu has to be equal to the limit integral S n prime d mu. That means, if two sequences of non-negative simple measurable functions have the same limit, then their integrals also converge to the same values. So, let us prove this result. So, we have got S n is a sequence of non-negative simple measurable functions. S n are increasing and also S n dash is another sequence which is also increasing and the limit n going to infinity of S n is equal to limit n going to infinity of S n dash. So, that is given to us. Now, let us fix any positive integer m bigger than or equal to 1 and consider the sequence S n minimum S m dash. Look at this sequence n bigger than or equal to 1. So, look at these functions. So, these are the functions which have the property that S n veg S m dash is always, this is the minimum. So, it is less than or equal to S n for every n. Also, as n goes to infinity, look at the sequence S n veg S m dash. So, the minimum of S n and S m dash as n goes to infinity, S n is going to increase to a limit. So, it will take over S m dash at some stage. So, that means this is going to converge to S m dash. So, this is obvious because S n and S m dash both the sequences have the same limit. So, at some stage S n will cross over S m dash for every m fixed. So, we have fixed n integer m. So, that implies that integral of S n veg S m dash d mu will converge to integral of S m dash d mu. So, once n, this is bigger than or equal to, so this is less than or equal to integral of S n d mu and that is because of this. So, we have got integral S n d mu is bigger than or equal to integral S m dash d mu for every n. So, that means, so this implies for all n large enough. So, that implies that limit of n going to infinity of integral S n d mu is bigger than or equal to integral S m dash d mu for every m fixed and hence because this is true for every m fixed. So, this implies that this is true for every m fixed. So, that implies that limit n going to infinity integral S n d mu is bigger than or equal to limit m going to infinity of integral S m dash d mu. So, we have shown that limit of S n, integral S n is bigger than or equal to limit of integral S m and because now we can interchange the two, so that implies similarly limit n going to infinity integral S n dash d mu is bigger than or equal to limit n going to infinity of integral S n d mu. So, that proves that the two are equal. So, this will prove that limit m going to infinity of integral S m dash d mu is equal to limit n going to infinity of integral S n d mu. So, that proves the result, namely if S n and S n dash are two increasing sequences of non-negative simple functions having the same limit, then their integral also converts to the same limit. This will be used soon. Let us look at an observation that in general, the class of non-negative simple measurable functions need not be closed under limiting operations. We showed that sum of non-negative simple measurable functions is a non-negative simple measurable function and we also just now observed that if a sequence S n is increasing to a sequence in l plus 0, that means if a sequence of non-negative simple measurable functions converges to a non-negative simple measurable function, then the integrals converge. In general, for decreasing sequences, for example, this need not hold or for even the limit of non-negative simple measurable functions may not be a non-negative simple measurable function. So, to give an example of that, let us consider the Lebesgue measurable measure space R, L and lambda. R is the real line, L is the space of the sigma algebra of Lebesgue measurable sets and lambda is the Lebesgue measure. So, let us define S n of x for every n to be equal to the indicator, sum of the indicator functions of k minus 1 to k, where k goes from 1 to n. So, this is just, it takes the constant value k on the interval k minus 1 to k. So, as is quite clear that as this is a non-negative simple measurable function on the real line and as n increases, this is going to be an increasing sequence that also is clear and actually it increases to a function which is equal to k on every interval k minus 1 to k. So, that the limit function is not going to be a non-negative simple measurable function. Of course, it will be a non-negative measurable function. So, what we are saying is that the class L plus 0 is not closed under limiting operations. So, that says that we should go over to a bigger class of functions namely the class of non-negative measurable functions and define integral there also. So, let us define for, let us now we will denote by L plus the class of all non-negative functions x to r star which are s measurable. Keep in mind we have got a fixed measure space which is complete that is x, s and mu. So, look at all non-negative s measurable functions on the set x and let us denote that this class of functions by L plus and if you recall, we had proved a theorem that for a non-negative measurable function, there is a sequence of non-negative simple measurable functions which is increasing to f. So, for every f belonging to which is a non-negative measurable function, so the function in the class L plus, we know that there exists a sequence s n of non-negative simple measurable functions s n which increases to f. So, f is a limit of a increasing sequence of non-negative simple measurable functions and we know, we are now just, we have defined the concept of integral for non-negative simple measurable functions s n. So, it is natural to define integral of f to be nothing but the limit of integrals of s n's. So, we define, so for a function f in L plus, we define its integral with respect to mu. So, denoted by integral f x d mu x or simply by f d mu to be limit n going to infinity of s n x d mu x, where s n is any sequence in L 0 plus increasing to f. And the first obvious claim is that this integral is well defined. It does not depend upon the choice of the sequence s n that we take which increases to the function f. That is because just now we have proved, just now we have proved the result that if there are two different sequences s n and s n prime non-negative simple measurable both increasing to f, then their limits are same. So, whichever sequence we take s n which increases to f, its limit is going to be the same extended real number and that extended real number is called the integral of f d mu. So, integral f d mu is well defined, so well defined. So, that means whatever sequence s n increasing to f we choose, it does not matter that limit of that sequence is same, limit of integrals of s n is same, so that is called integral of f d mu. And because it is limit of integrals of s n which are non-negative simple measurable function, so each integral s n is a non-negative simple measurable function. So, as a result the limit also is non-negative. So, integral of a non-negative measurable function f by this process is a well defined number and it is bigger than or equal to 0. So, now let us look at the properties of this that the class L 0 plus is a subset of L plus. So, that is obvious and we want to claim that integral of s d mu as an element of s is same as an element of L plus. So, that also is obvious because of the following fact. So, we have got, so let us take s belonging to L plus 0 and then we have its integral, integral s d mu as a integral of a non-negative simple function and L plus 0, we are now treating it as a subset of L plus. So, if you treat s as an element in L plus, then we can take the constant sequence s n is equal to s for every s and that will imply that integral of s d mu as an element in L plus is same as limit of integral s n which is same as integral s d mu as an element in L plus 0. So, as if you take a non-negative simple measurable function as an element of L plus and look at the integrals as a element of L plus, then that integral is same as an element of the non-negative simple measure. That means that the new integral that we have defined is in fact an extension of the notion of integral from non-negative simple measurable functions to non-negative measurable functions. Now, let us next look at the property that if f is a function in L plus and s is a function in L 0 plus, say that 0 is less than or equal to s less than or equal to f, then integral of s d mu is less than or equal to integral f d mu. So, let us prove this property. So, we have got s a non-negative simple measurable function and f is a non-negative measurable function and we are given that s is less than or equal to f. Now, since f belongs to L plus implies there is a sequence s n of non-negative simple measurable functions such that s n increases to f. Now, let us look at s n increases to f and the integral of s n d mu converges to integral f d mu. Next, let us look at, so consider, so observe here is s and s of x for any point and here will be some f of x and s n is going to increase to f. So, s n x is going to cross over s of x for some n. So, let us define b n to be the set of all those points x belonging to x such that s of x is less than or equal to s n of x. So, observations that this set b n is in the sigma algebra s and because s n is increasing this sequence b n of sets is also increasing to the whole space x because s n is converging to f of x. So, b n is going to increase to s of x. So, these are obvious properties because if s n is less than or equal to s of x then s n plus 1 is also bigger than. So, that means b n is inside b n plus x and as we observed that for every x there will be some n such that s n of x will cross over s of x. So, every x belonging to x belongs to some b n. So, b n is going to increase to x. So, now we observe the property that look at integral of the non-negative simple measurable function s d mu. So, that we can write it as limit integral n going to infinity integral over b n of s d mu. So, this is because s n is an increasing sequence s n increases to b n is an increasing sequence of sets b n increases to x and the integral over a set is a measure. So, keep in mind that the integral of a non-negative simple measurable function over a set E gives you a measure. So, that measure mu of that measure at b n will go to that value at x. So, that is same as saying that integral s d mu is limit of integral s over b n d mu. Now, on b n we know that on b n s n is bigger than s. So, let us use that fact. So, this is less than or equal to limit n going to infinity integral over b n of s n d mu. So, that is the non-negative simple function one non-negative simple measurable function is less than another than the integral of one will be less than the other. Now, this is integral s n over b n. So, if you replace that set b n by the whole space this will still be less than or equal to limit n going to infinity integral over the whole space x of s n d mu. So, that is equal to integral f d mu. So, that proves that integral of s d mu is less than or equal to integral of. So, implies that integral s d mu is less than or equal to integral f d mu whenever s is less than f and s is non-negative simple measurable function. So, that proves this property that if f is a non-negative simple measurable function f is a non-negative measurable function and s is a non-negative simple measurable function such that s is less than or equal to f then the integral of s is less than or equal to integral of f. And now as a consequence of this let us observe the property that integral of f d mu which we defined as the limit of integrals of s n d mu for any sequence s n can also be represented as supremum over of integrals s d mu where s is less than or equal to f and f is s is a non-negative simple measurable function. So, let us prove this property that. So, let us call let us define beta to be the supremum of integral of non-negative simple measurable function s d mu where 0 is less than or equal to non-negative simple measurable function s less than or equal to f. So, let us call this. So, we want to show that integral f d mu is equal to beta. So, let us prove that integral of f d mu can also be written as supremum of integral s d mu where s is less than or equal to f. So, to prove this property let us define. So, let beta be equal to supremum of integral s d mu where 0 is less than or equal to s is less than f and s is a non-negative simple measurable function. So, we want to show that beta is equal to integral f d mu. So, now one possibility is in case integral f d mu is equal to plus infinity then we know that f belongs to L plus. So, there is a sequence s n in L plus 0 non-negative simple measurable functions s n increasing to f and integral s n d mu converging to integral f d mu. But now in this case we know this is equal to plus infinity that means for every positive integer n there exist some n naught such that integral s n naught d mu will be bigger than or equal to n because this number is going to converge to infinity. So, this must exceed every n. So, there is n naught and this s n naught is less than or equal to f. So, we have found a non-negative simple measurable function s n naught less than or equal to f such that its integral is bigger than or equal to n. So, that implies that the supremum beta must also be bigger than or equal to n. So, this implies that the supremum beta is bigger than or equal to n for every n and hence that implies beta is equal to plus infinity. So, in the case integral f d mu is equal to plus infinity that implies beta is also equal to plus infinity. So, that is beta is equal to plus infinity is equal to integral f d mu. Now, let us look at the case when this integral is finite. So, in case integral f d mu is finite that means and we know that integral s n d mu converges to integral f d mu. So, for every epsilon bigger than 0, there is some n naught such that integral f d mu minus integral s n naught d mu is less than epsilon and that means integral f d mu is less than or equal to integral s n naught d mu plus epsilon and s n naught is one function which is less than or equal to f. So, this is less than or equal to beta plus epsilon. So, integral f d mu is less than or equal to beta plus epsilon and this holds for every epsilon. So, implying integral f d mu is less than or equal to beta. So, once again we have proved that integral f d mu is less than or equal to beta and clearly beta is less than or equal to integral f d mu. That is obvious because beta is the supremum overall non-negative simple functions s d mu less than or equal to integral of beta is the supremum. So, definition beta is the supremum of integral s d mu where s is less than or equal to f. So, integral s d mu is less than or equal to integral f. So, beta is always less than or equal to integral of f d mu. So, hence this implies beta is equal to integral of f d mu. So, that proves another way of defining the integral of a non-negative simple measurable function that if f is a non-negative simple measurable function then its integral can also be defined as the supremum over all integral s d mu where s is a non-negative simple measurable function. So, using these two definitions let us prove that various properties of the integral. So, for every function f in l plus we are defined integral f d mu. Now, we are going to look at the properties. So, the first property that we are saying is if f 1 is bigger than f 2 then integral f 1 is bigger than integral f 2. That follows from the above definition itself because integral f 1 d mu integral f 1 d mu is going to be the supremum over integrals of all non-negative simple measurable functions s. So, that s is less than or equal to f 1 and similarly for f 2 it is going to be the supremum over all non-negative simple measurable functions s less than or equal to f 2, but if s is less than or equal to f 2 and f 2 is less than or equal to f 1. So, s is going to be less than or equal to f 1. So, this supremum for f 1 is taken over a larger class and then that of f 2. So, that supremum is going to be for integral f 1 the supremum is going to be bigger than or equal to integral over f 2. So, that follows directly from here. From the above definition that this integral is supremum over integral of non-negative simple measurable functions below f. So, as a consequence of this we immediately have this theorem that if f 1 is bigger than f 2 then integral f 1 d mu is bigger than or equal to integral f 2. Next, let us look at the linearity property of this integral namely if alpha and beta are non-negative real numbers extended real numbers then and f 1 and f 2 are in l plus then alpha times f 1 plus beta times f 2 belongs to l plus and integral of alpha f 1 plus beta f 2 is equal to alpha times integral f 1 plus beta times integral f 2. So, to prove that f belongs to l 1 l plus implies there is a sequence S n of non-negative simple measurable functions S n increasing to f and limit n going to infinity integral S n d mu giving us the integral of f 1 d mu. And similarly g belongs to l plus implies that there is a sequence let us call it S n prime of non-negative simple measurable functions S n prime increasing to g and its limit n going to infinity integrals of S n prime d mu giving us the integral of g d mu. So, now from these two let us just simply observe that if S n increases to f then alpha S n will increase to alpha f and similarly beta S n prime will increase to beta of g and integral of alpha S n for non-negative simple measurable functions is equal to alpha times integral S n. So, combining all these properties we will have the required result. So, let us just write it out. So, implies that alpha S n increases to alpha of f and limit alpha S n n going to infinity integral d mu will be equal to alpha times limit n going to infinity of integral S n d mu, because for non-negative simple measurable functions alpha times S n is same as alpha times the integral and that is equal to alpha integral f d mu. And similarly limit of n going to infinity of beta S n prime integral of S n prime integral d mu will be equal to beta times integral of g d mu. On the other hand if we look at the sequence alpha S n plus beta g n if we look at the sequence alpha S n plus beta S n dash then this is a sequence of non-negative simple measurable functions and that increases to alpha f plus beta g by the properties of sequences. So, as a result we will have that the integral S alpha S n d mu. So, let us write it. So, because of this we have the property that the integral alpha S n plus beta S n dash d mu limit n going to infinity will be equal to integral of alpha f plus beta g d mu. But integration is linear for non-negative simple measurable functions. So, this side there is nothing but limit n going to infinity of alpha integral S n d mu plus beta integral g n d mu. And now by the properties of limits of sequences this is equal to alpha times limit integral of S n d mu plus beta times this is S n dash. So, beta times limit n going to infinity of integral S n dash d mu and this we know is alpha times integral f d mu plus beta times integral g d mu. So, integral of alpha f plus beta g is equal to alpha times integral f plus beta times integral of g. So, that proves the property that when alpha and beta are non-negative extended real numbers then alpha f 1 plus beta f 2 belongs to l plus that we had already shown also. And now we are claiming that the integral of alpha f 1 plus beta f 2 is equal to alpha times integral of f 1 plus beta times integral of f 2. I have written for g. So, that is same as for this. Next we prove an important property and an extension of the earlier version for non-negative simple functions namely for every measurable set E. If we look at the function the indicator function of E times f then that is also a non-negative measurable function. And if its integral is denoted as nu of E. So, nu of E is the integral chi E f d mu where E belongs to S then this is a measure this nu is a measure on the sigma algebra S and has the property that nu of a set E is 0 whenever mu of the set E is equal to 0. So, let us prove this property also and this property again we are going to use the fact that the integral of a non-negative simple measurable of a measurable function is given by as a limit of the integrals of a sequence increasing sequence of non-negative simple measurable functions. So, f belonging to l plus implies we have a sequence S n belonging to l plus 0 such that S n increases to f and integral f d mu is written as limit n going to infinity of integral S n d mu. So, that is by the fact that f belongs to l plus and integral of f is defined as limit of integral S n d mu for any sequence S n which increases to f. Now, for E a set in the sigma algebra S because S n is increasing to f. So, clearly indicator function of E times S n will increase to indicator function of E times f and observe we have done it earlier also that this is a non-negative simple measurable function it is increasing to this function. So, that implies that indicator function of E times f. So, this implies that the indicator function of E times f is a non-negative measurable function. And because we have got this sequence increasing to this non-negative measurable function so integral of the indicator function of E times f d mu is nothing but limit n going to infinity of integral indicator function of E times S n d mu times integral of S n d mu. So, that is this is how the integral is defined and now we want to claim that if we call this number nu of E as integral over E f d mu then nu is a claim is nu is a measure. So, to prove this let us so what we have to prove is the following. So, to prove this is what we have to show that if a set E is a countable disjoint union of sets E i E i is in the sigma algebra S then we want to show that nu of E is equal to sigma nu of E i i equal to 1 to infinity. So, this is what we have to show. So, let us start looking at nu of E by definition nu of E just now we saw that nu of E is nothing but integral of the indicator function of E times f d mu and that so integral of E f d mu is nothing but limit of integral indicator function of E S n d mu by the fact that S n is increasing to f. So, just now observed that. So, this can be written as limit n going to infinity of integral indicator function of E S n d mu. So, this is just from the fact that S n is increasing to f. So, indicator function of E times S n will increase to indicator function E times f hence integral of indicator function of E times f is nothing but the limit of the integrals of the indicator function E S n d mu and now this E is a disjoint union of sets E i. So, that implies. So, let us write this limit n going to infinity of. So, this I can write as summation chi E i of S n d mu i equal to 1 to infinity. So, here we are using the fact that E is a disjoint union of sets E i and for a non-negative measurable function S n if we integrate it over a set E then that is a measure. So, that is the property. So, the corresponding property for non-negative simple measurable functions which we had already proved is true. So, we are using that fact to bring it here. And now this is limit of a series i equal to 1 to infinity of sorry this is a integral here integral of chi E. E is union. So, this is integral of the union. So, now I am going to interchange this summation and or limit and summation y equal to 1 to infinity and that is allowed because all the quantities involved are non-negative. So, this interchange is possible. So, I can write it as summation i equal to 1 to infinity limit n going to infinity of integral chi E i S n d mu. And now we simply observe that this last quantity is nothing but summation i equal to 1 to infinity limit n going to infinity and the last quantity is nothing but nu limit of n going to infinity. So, that limit is nothing but integral of chi E i F d mu because S n is increasing to F. So, chi E i times S n increases to chi E i times F. So, this limit of n going to infinity integral of chi E i S n d mu is nothing but integral of chi E i times F. So, that value you put. So, and that is nothing but our definition of nu of E i. So, that proves that nu is a measure on E i. And once again observe that here we have used basically what we have done is we have used the fact that F is a limit of increasing sequence of non-negative simple measurable functions. So, integrals of non-negative simple measurable functions that sequence gives you integral of F. And then so go to that sequence use the property for non-negative simple measurable functions that property is true. So, and come back. And finally, to prove that nu of E equal to 0 implies mu of E equal to 0. So, that is so let us suppose mu of E equal to 0. Then what is nu of E? So, nu of E which was defined as integral chi E of F d mu which was nothing but integral of chi E times S n d mu limit of that. So, let us write limit n going to infinity of this, but this mu of E being 0 this integral is 0. So, this is equal to 0. So, once again for a non-negative simple measurable function its integral over E is 0 if mu of E is 0. So, that property is being used once again here. So, this proves the fact that this measure nu of E which is constructed as integral of chi of E F d mu is a special measure which has the property that it is null sets nu of E is 0 whenever mu of E is 0. So, till now what we have done is we have defined the integral of a non-negative simple measurable function as a limit of integral of non-negative simple measurable functions. Because if F is non-negative measurable it is a limit of non-negative simple measurable functions which increase to this function F. So, integrals of those non-negative simple measurable functions is defined. Take their limit and define integral of F to be limit of the integrals of non-negative simple measurable functions and using this we have proved that this integration is linear. So, the next property we want to analyze is how does this class of non-negative measurable functions and the operation of integral behave for sequences in the class L plus. So, here is the first important theorem that we are going to prove that is before that let us just prove just a simple observation that if F 1 and F 2 are non-negative simple and F 1 is equal to F 2 almost everywhere then integral of F 1 is equal to integral of F 2. So, that property is quite obvious because let us write the set n to be the set all x belonging to x where F 1 x is not equal to F 2 of x. Then the whole space can be written as n union and complement and we are given so we are given F 1 is equal to F 2. So, F 1 of n complement we are given F 1 is equal to F 2 almost everywhere. So, where they are not equal so this set has got F 1 sorry we are given that sorry we are given that this set n F 1 is equal to F 2 almost everywhere. So, where they are not equal that is a set of measure 0. So, mu of n is equal to 0. So, now integral of F 1 d mu can be written as integral over n F 1 d mu plus integral over n complement of F 1 d mu. This is by the fact just now we proved that integral over a set is a measure. So, this is integral over n and integral over n complement that gives you integral over the whole space and mu of n being equal to 0. So, this first term is 0 plus integral over n complement F 1 d mu. But this is also same as integral over n of F 2 d mu because measure of n is 0 and on n complement F 1 is equal to F 2. So, I can write as n complement F 2 d mu and that once again is equal to integral F 2 d mu. So, integral F 1 d mu is equal to integral of F 2 d mu. So, that essentially says that the integral of a function does not change if it changes its values on a set of measure 0. So, let us now come back to the property that I was trying to state earlier namely if we have a sequence F n of non-negative simple measurable functions and F n increase to F that is F of x is equal to limit of F 1 x then the claim is F belongs to L plus and integral F d mu is equal to limit n going to infinity integral of F n d mu. So, this is one of the important theorems in our subject it is called monotone convergence theorem. Monotone because we are looking at sequence F n which is a increasing sequence is a sequence of functions which is increasing. So, it is a monotonically increasing sequence of non-negative measurable functions increasing to a function F. We have already seen that F will be a measurable function and it is non-negative, but the important thing is integral of F. So, F is the limit integral of the limit is equal to limit of the integrals. So, that is the important property we want to prove for integral for non-negative simple non-negative measurable function. The proof of this theorem requires some construction and we do not have time today to complete the proof. So, we will do the proof of this theorem next time. So, we stop here today by having stated the monotone convergence theorem and look at the proof of this in the next lecture. Thank you.