 Welcome to lecture 28 on measure and integration. If you recall, we had started looking at the computation of product measure of a set E in the product sigma algebra and we had shown this can be computed via the sections of the set E, integrating the sections and taking the measures. So, let us recall this result and then we will continue to generalize this result for functions which are non-negative and integrable functions. So, let us recall the result that we had proved last time, namely if x a mu and y b nu are 2 sigma finite measure spaces and the product sigma algebra, a product space is x cross y, a times b and mu cross nu is a product measure space, then we showed that for any set E in the product sigma algebra, the measure mu cross nu of E can be computed by either taking the section of the set E at a point x and that gives as a subset of the set y and we showed that this is in the sigma algebra, you can take the measure of this section. So, this becomes the function of the variable x and then you can integrate out this function with respect to mu to get the product measure or equivalently you can take the y section of the set E that gives a subset of x which is a measurable set and then you can take the measure of mu measure of that that gives a function of y and then integrate out that with respect to y to get the measure of the set E. This result we want to re-interpret as follows. See the measure of the set E x can be written as the integral of the indicator function of the set E with respect to the product measure. So, the mu cross nu of the set E is nothing but the integral of the indicator function of the set E on one hand. On the other hand, if you look at the x section or the y sections, they are nothing but the indicator functions of the set E again. So, when you take the integration with respect to y that means you are fixing x. So, you are looking at the section of E at x. So, nu of E x is nothing but the integral over y of the indicator function of E with respect to the variable y and similarly the other variable. As we had mentioned, the importance of this result lies in the fact that the indicator function of a set E is a function of two variables and to find its integral with respect to the product measure, what we can do is we can fix one of the variables say x. This becomes a function of one variable y and one shows that this is integrable with respect to nu. So, when you integrate out with respect to nu, the variable y is a function of x, which again can be integrated with respect to mu and this integration gives you the integral of the indicator function of E. So, the important thing is that here when you are integrating with respect to y, the variable x is fixed. So, this is only a function of the variable y. So, for every x fix, you treat it as a function of the variable y, integrate that out and then integrate out that integral with respect to the other variable and similarly we can interchange x and y and get the result. So, what we want to show is today is that this result is true for all non-negative measurable functions on the product space x cross y. So, the theorem we want to prove is the following. Namely, if f is a function on the product space x cross y and it is a non-negative function which is. So, f is a non-negative measurable function on x cross y is measurable with respect to the product sigma algebra. Then, the following statement is sold namely if we fix one of the variables say x. So, if x naught is fixed, then consider the function f of x comma y naught and similarly y goes to f of x naught y. So, for the function f of x y, either you fix the variable y at y 0 or you fix the variable x at x 0 and treat it as a function of one variables only, then these functions are measurable on x and y respectively. So, what we are saying is that for a function of two variables, if it is measurable with respect to the product sigma algebra, then fixing either of the variables gives you a function of the one variable which is measurable on the corresponding basis with respect to the corresponding sigma algebras. These are non-negative functions. So, you can integrate them out. So, if you integrate this function x going to f of x y naught with respect to x mu, then that gives you a function which depends on y. So, the function y going to integral over x of f x y d mu x. So, you integrate it out the variable x, this gives you a function of y and similarly you integrate out f x y with respect to the variable y, you get a function with respect to x. So, the claim is these two are well defined non-negative measurable functions on their respective spaces. Finally, these are non-negative measurable. So, you can integrate them out with respect to the corresponding variables. So, if you integrate out them, so first integrate with respect to y and then with respect to x that is same as integrating first with respect to x and then with respect to y and both are equal to the integral of the function f of x y with respect to the product sigma algebra. So, this gives an extension of the earlier result and so it says that for non-negative measurable functions, if you want to integrate with respect to the product sigma algebra, product measure, then you can do it one variable at a time. So, these integrals, these two integrals are called the iterated integrals. So, the claim is that for a non-negative measurable function, the integral with respect to the product measure is equal to the two iterated integrals. Once again the importance being you are integrating one variable at a time. So, let us prove this result. So, this proof is going to be built up step by step and this is what I call as the simple function technique. So, the idea is that when f is the indicator function of a set, this result is true by the earlier result on product measures and then by looking at because everything involves integrals. So, an integration being a linear operation will get that from that this result is true for non-negative simple measurable functions and once it is true for non-negative simple measurable functions, an application of monotone convergence theorem will give us that the result is true for all non-negative measurable functions on the product space. So, that is the approach basically we are going to follow and this is what I call as the simple function technique. When you want to prove something for non-negative measurable functions on a measure space, verify it for the indicator functions, verify it for the non-negative simple measurable functions and then verify it for the limits of non-negative simple measurable functions. So, let us prove this. So, the step one is that the required claim holds when f is the indicator function of a product indicator function of a set E in the product sigma algebra. So, let us look at that. That was what we have already shown. When E is so step one, when E is an element in the product sigma algebra, we had already shown that the product measure mu cross nu of E on one hand is equal to you take the indicator function of E and integrate out with respect to x d mu and then integrate out that with respect to the variable y, so d nu. That is same as you first integrate out the indicator function x y with respect to d nu. That means keeping the variable x fix, you are integrating with respect to y and then compute the integral of that with respect to the variable x. So, these two are equal and the middle thing, if you recall, we said it is equal to the indicator function, integral of the indicator function of E with respect to the product sigma algebra. So, this is besides this saying that the claim holds for f equal to the indicator function of a set E, E belonging to the product sigma algebra. So, this is step one and now from here we want to go to step two. So, step two is let us take a function. So, step two, the required claim holds for non-negative simple measurable functions. For s f equal to s, a non-negative simple measurable function on x cross y. So, let us take, so what does the function look like? So, a non-negative simple function s on the product space looks like sigma A i indicator function of some sets E i, where i is 1 to n and these E i's are sets in the product sigma algebra A times B and their union of E i's they are pairwise disjoint and their union is equal to x cross y. Now, by step one, what does step one say? Step one says for each E i the claim holds. So, step one says for every i equal to 1, 2 and so on n, the integral of the indicator function of E i with respect to d mu cross nu on one hand it is equal to the integral over x, integral over y of indicator function E x y d nu and d mu and also equal to the integral over y, integral over x of the indicator function of E d mu and then d nu. So, this is what we know. Now, let us just observe, because this is true for every i and integration is linear. So, that means if I multiply, I can multiply throughout by A i. So, if I multiply by A i, so I can multiply here by A i, so this is E i. So, if I multiply, so this is A i and I can multiply by A i and I can multiply here by A i and then take the summation, so some over i, then summation over i, summation over i and here will be summation over i. Now, this summation integration will be linear, I can take the summation inside. So, when I take this summation inside the integral and then again take it inside, I will get summation of A i times indicator function of E i, integration with respect to nu and then integral with respect to mu is equal to this I take it inside, that will be summation of A i. So, let us just write that, this is more of writing than understanding. So, when I do this summation and take the summations inside, I will get integral over x, integral over y of summation i equal to 1 to n A i, indicator function of E i d nu d mu is equal to, so that is equal to and this summation inside will give me integral over x cross y of summation A i, indicator function of E i d mu cross nu and the last one will give me integral over y, integral over x of summation i 1 to n A i, indicator function of E i d mu d nu. So, this is just using the property that for every indicator function of a set, the result is true and integration is linear. So, the result is true for finite linear combinations of these things also. So, this gives us and this is precisely my function s. So, this says integral over x, integral over y of s x y d nu d mu is equal to the integral over the product space. So, this is the function s, so s of x y d mu cross nu and that is also equal to the other iterated integral, integral over y, integral over x of s x y d mu of x d nu over, we have already done it. This is over y, so over y that should be nu actually and this should be mu, sorry, because this was over x, so that should be, no that was d mu and this is d nu. So, this is over x, so d mu and d nu, so that is, so this says that the result holds, so this proves the second step namely the claim holds for f a non-negative simple measurable function. So, now from here to go to general non-negative functions, step 3 says we should be able to prove the result when f is, so let us take f on x cross y to r star, f non-negative f a times measurable with respect to the product sigma algebra. Now, we look at the characterization of non-negative measurable functions, f being non-negative measurable implies that there exists a sequence S n of non-negative simple measurable functions S n of x increasing to f of, S n of x y increasing to f of x y for every x y belonging to the product z x cross y. So, that is by, because of the fact that this is by the, so this is by the fact that for every non-negative measurable function there is a sequence of non-negative simple measurable functions converging to it. Now, by step 2, so by step 2 we know that for every S n the corresponding result holds, so what does it mean? That means for every S n the integral of the non-negative function S n x y d mu cross nu x y is equal to the iterated integral, so let us write say for example, integral over x, integral over y S n of x y d nu d mu and similarly the other iterated integral. Now, what we are going to do is observe the fact that for every S n this result is true and S n is a non-negative sequence of non-negative simple measurable functions increasing to f. So, by the definition of the integral this one converges to the integral of f x y d mu cross nu of x y. So, this is by the fact of monotone convergence, this is not really monotone convergence theorem, this is actually the definition of the integral. So, if f is a non-negative measurable function then its integral is defined as the limit of any sequence of non-negative simple measurable functions increasing to it. So, that is by the definition. On the other hand we will compute this integral and so it is the corresponding iterated integral of the function f. So, let us look at this integral. So, integral over x, integral over y S n of x y d nu d mu. So, this we want to show claim that this converges to integral over x integral over y of f x y d nu d mu. So, this is what we want to show and once that is shown, so on one end this converges, so this converges to this iterated integral on the other end this converges to this integral of f. So, these two will be equal and we will be through. So, let us try to prove that this iterated integral converges to this iterated integral. So, for that let us observe first that S n, so note, so here are the steps for proving this. So, first of all let us try to prove that integral over y with respect to nu converges to corresponding integral. So, for that let us note that S n of x y is increasing to f of x y for every x and y. So, if we fix x, then we get a function y going to S n of x y for every y belonging to y and because S n itself is increasing, so this sequence of functions we already know these are measurable functions. These are measurable functions for every simple function. We had seen that if we fix one of the variables, the other one is a measurable function for simple functions that is true. First of all this is clear. Therefore, fix x, this is an increasing sequence of non-negative functions. So, this is a sequence of increasing sequence of non-negative functions and each one of them is a measurable function on y. Second observation is that each one of them, so the function y going to S n of x y is a measurable function. That is in a sense obvious because, so to see this we note that so for this, so observation is, so let us say S n, so let S n be equal to something. So, sigma, let us say it is a i n indicator function of e i n for some i equal to 1 to some indexings at 1 to m n. Then for every fixed x, so for x fixed, y going to the indicator function of e i n x y is measurable, so that we have already observed that that is a measurable function while computing the product measure we saw that this is. So, each one of them is measurable with respect to the product sigma algebra because it is an indicator function of a set and for every fixed x, this will be a measurable function on y. So, that will be the sections, so that will be the measurable, scalar multiple of a measurable function is measurable and the sum of measurable functions is measurable. So, this is obvious fact that for a simple measurable function S n, if we fix one of the variables in the other variable, it becomes a measurable function. So, this is a measurable function. So, that means for x fixed, so what we have got on is for x fixed, the sequence S n x y over n is a sequence of non-negative b measurable functions on y and also it is increasing. In fact, it increases to, so what does it function? It increases because S n increases to f. So, when we fix one of the variables x, this is going to increase to the function f of x y, x fixed as a function of y. So, increases to the function y going to f of x y. So, it is a perfect setting for application of monotone convergence theorem. So, by monotone convergence theorem, we get the following. So, by monotone convergence theorem, we get, so to this applied monotone convergence theorem. So, by monotone convergence theorem, the integrals S n x y d nu y over n y limit of that must be equal to integral of f x y with respect to d nu y. So, this is what we get. Of course, for every x fixed, so for every x fixed, we get that this limit must be equal to this. So, that means what? That means this function, so this is a function of x. So, that implies that if I look at x going to y f of x y d nu y, if we treat this as a function of x, then it is a limit of these functions, limit of non-negative simple integrals of non-negative simple functions. So, that means that this function is measurable. So, that implies, so this implies that this function is non-negative measurable. It is a non-negative measurable function and this is a non-negative measurable function and it is a limit of this sequence of non-negative measurable functions. So, by star, I can apply again another application of, again another application of monotone convergence theorem. So, this limit of, this on the left hand side, the limit of integrals of S n with respect to nu that also is a limit of measurable function. So, this itself is a measurable function with respect to x and because S n's are increasing, these integrals are also increasing. So, this is, so if we call, so this limit is equal to this. So, now what we are saying is another application of monotone convergence theorem to the fact that if we look at the sequence of measurable functions, this is a measurable function. So, integral of that, so let us write, so the function x going to this is a non-negative measurable function. So, what is left to be proved? We want to integrate this with respect to mu. Now, so that is what we are saying that to this we apply monotone convergence theorem. So, we will get the limit of these functions is this function. So, integral limit of the integrals, so this says limit n going to infinity, integrals of this function, so integral over x of these functions, these functions are S n x y d nu y d mu x. So, this limit of this must be equal to integral of this with respect to x again by monotone convergence theorem. So, let us write that, so this is equal to integral over x, integral over y of S n x y d nu y d mu x. So, this limit is equal to this, but on the other hand we had seen that on the other hand we had seen that this iterated integral for S n's that is equal to the double integral. So, this is one thing upon observation also. Let us look at the other fact, so what is the other fact? So, also we have that integral over x, integral over y S n x y d nu y d mu x. For simple non-negative simple measurable functions the claim holds. That means this is equal to the double integral, the integral over x cross y of S n x y with respect to the product measure mu cross nu. So, this is because we have already proved in step 2 that the result holds for non-negative simple measurable functions. And now if I look at, so this result is equal to this, so limit of this must be equal to limit of that. So, implies the limit n going to infinity of this left hand side must be equal to limit n going to infinity of the right hand side. So, this one and that is coming here, but limit of the left hand side we have already seen is equal to the limit of the left hand side. We have already seen it is equal to this and what is the limit of the right hand side? S n is a sequence, so we have already seen that S n is a sequence of non-negative simple functions increasing to f. So, this must be equal to integral x cross y of f x y d mu cross nu. So, that proves that this must be equal to this. So, that proves that the step 3 proves that the integral of f x y with respect to integral of, let us just see how we proved that. So, what we have shown is this limit must be equal to this and what was that limit of that quantity? So, what we have shown is that integral of d mu cross nu is equal to over x cross y is equal to limit n going to infinity of integral over x, integral over y of S n x y d nu y d mu x. So, this is what we have proved just now that this limit on one hand side was this, other hand side was this. So, limit of these two quantities must be equal. So, this is what we have proved, but this quantity, let us see what is this? So, note that S n for every y fix was increasing. So, let us look at the sequence for every x fix that is an increasing sequence of non-negative measurable functions increasing to the function f of x y. So, monotone convergence theorem says this inner integral converges to integral of y f of x y d nu y. That is what we had already observed and then again this is a sequence of non-negative simple non-negative measurable functions. The application of monotone convergence theorem gives us that integral of this limit of that must be equal to d mu of x. So, that says that the double integral of the non-negative simple function is equal to the iterated integral of the non-negative measurable function iterated first with respect to nu and then with respect to mu. We can interchange x and y. So, same arguments you will imply. Let us say that this is also equal to integral over y, integral over x of f of x y d nu y d mu x. So, basically let us just go through the ideas in the proof that basically this proof is an application of the fact that integral for a non-negative simple function is built from the limits of integrals of non-negative simple measurable functions. So, that fact is used very effectively because we know that the corresponding result is true for indicator functions and integration is linear. So, that allows us to say that from the indicator functions you can go to non-negative simple measurable functions by just taking scalar multiplications and additions of characteristic functions. So, that will give us that the result is true for non-negative simple measurable functions. And then just some suitable applications of monotone convergence theorem will give us that the integral of a non-negative measurable function on the product space can be computed via the iterated integrals. Let us just go through this proof through the slides once again so that we have a clear idea what we are doing. So, the step one that the required claim holds when f is the indicator function of e that is the previous theorem that we have proved. And step two the required claim holds when f is a non-negative simple measurable function. So, from step one to step two one goes via the fact that integrals are linear operations. And then one goes to step three that the required claim holds when f is a non-negative measurable function. So, that requires applications of basically applications of monotone convergence theorem. So, step three is the crucial one where lot of applications of monotone convergence theorem are used. So, let us just go through that again. So, let S n be a sequence of non-negative simple measurable functions such that S n increases to f. So, that is by the fact that f is a non-negative measurable function. So, now let us fix x. So, then the sequence S n x is fixed. So, in the variable y is a sequence of non-negative simple measurable functions on y and it increases to the function f of x y for x fix. So, point wise S n x dot as a S x fix S n x as a function of y increases to the function f x as a function of y. So, an application of monotone convergence theorem is not required here. So, this is a limit of increasing sequence of measurable functions. So, that says that the function y going to f of x y is a non-negative measurable function because this function is a limit of measurable function. So, first fact to being used is that limits of measurable functions is a measurable function. Now, we can also apply monotone convergence theorem to conclude that the iterated integral of S n must come converge to the iterated integral of f with respect to y. So, that is the first application of monotone convergence theorem that for the non-negative measurable function f for the variable x fix its integral with respect to the variable y is well defined because this is a non-negative measurable function and it is equal to limit with respect to n of the non-negative the iterated integral of the non-negative simple measurable functions S n with respect to y. Now, this result also says that this equality also says that this the right hand side treated as a function of x. So, that means that converges to this integral and by the fact that the required result holds for non-negative simple measurable functions this function integral of S n with respect to y is a measurable function of x. So, here we were using the step 2. So, this is a sequence of measurable functions converging to a function. So, that means this integral must be a measurable function. So, again limits of measurable functions are measurable. So, that gives you that x going to integral over y f x y d y. So, the iterated integral of f with respect to y is a measurable function with respect to x and it is non-negative and once again this is a non-negative function and is a limit of these integrals limits of this measurable functions. So, another monotone convergence theorem application gives that integral of the iterated integral of the integral of S n with respect to y its integral with respect to x must come to the corresponding integral of f with respect to x. So, here we are applying monotone convergence theorem that the integral of x integral of integral over x of the integral of f with respect to y must come must be limit of the corresponding integrals with respect to the non-negative simple functions. Now, come back and for non-negative simple functions we know the result is true. So, this iterated integral must be equal to the double integral. So, that says so this is equal to the double integral and now S n is a sequence of non-negative measurable functions on the product space. So, this again by either you can say application of monotone convergence theorem or just by the definition this limit must be. So, this is equal to this and the limit of that must be equal to the integral of the function f over x cross y. So, that says the corresponding result holds. So, this iterated integral of f is equal to the double integral of f with respect to mu cross mu and similarly the other thing can be proved you can interchange the variables x and y. So, this result is true. So, this is the result which is called Fubini's theorem 1 which says so this is the first Fubini's theorem which says that for a non-negative measurable function on the product space if you want to integrate find its integral with respect to the product measure you can do it by integrating one variable at a time. Either you can fix x integrate out with respect to y and then integrate with respect to x or interchange choice is yours you can first integrate with respect to x and then with respect to y. So, the two iterated integrals for a function of two variables is equal to the double integral for non-negative measurable functions. So, this is called first Fubini's theorem which helps one to integrate functions of two variables. So, next we want to show that this result also holds for functions which are integrable. So, we want to prove that for a integrable function the corresponding result holds. .. So, let us look at the proof of that. So, let us look at so let us take a function F which is L1 on x cross y it is integrable with respect to x cross y and we want to say that that the integral of F x y over x cross y d mu cross nu we want to say that this integral on one hand is equal to you can integrate first x y with respect to nu we want to claim this with respect to y over y and then integrate out with respect to x d mu x or one should be able to say that this is also equal to you take the function F x y integrate out the variable with respect to mu so x and then integrate out with respect to y d nu of y we want to say that these two this results hold. Now, for that on obvious if these equations are to hold where F is not necessarily non negative so that means what first of all the inner integral for example integral of F x y with respect to y must exist that means we should be able to say for a function of two variables which is integrable when I fix the variable x as a function of y that is integrable so that is integrable and then that gives us a function of x and then we should be able to say that is integrable with respect to x and finally these two are equal and similarly the other result must hold. So, the theorem which we want to prove is the following namely that is called Feminist theorem 2. So, namely if F is a integrable function F is a integrable so we want to prove the following that if F is a integrable function then the following statements are true namely one that for the function of two variables if I fix either of the variable then with respect to other variable it is integrable not for all but we are able to say that the function x going to F x y and y going to F x y for the other variables are integrable for almost all y and for almost all x. So, for almost all fixing of coordinate the other variable it becomes a function which is integrable with respect to the other one. So, that is one and then secondly once these are integrable you can integrate out. So, it says that the function y going to integral of F over x with respect to mu and similarly the integral of y F with respect to y these two are defined almost everywhere and of course they are defined almost everywhere and are integrable and hence the third step says they are integrable and indeed the integrals the iterative integrals are equal to the double integral. So, we would like to prove this theorem. So, to prove this let us proceed as follows. So, we are given that the function F belongs to L 1 of x cross y. Let us write the positive and the negative parts of the function. So, F is equal to F plus the positive part minus the negative part and the integral of F x y with respect to the product measure mu cross mu is equal to integral of the double integral of F plus with respect to the product measure minus the integral of the negative part d mu cross mu over x cross y. So, that is the definition of the integral. If F is integrable then the integral of the function is nothing but the integral of the positive part minus the integral of the negative part of the function. .. Now, let us look at them separately. So, F plus x y of course d mu cross mu over x cross y. So, F plus is a non-negative function is a non-negative measurable function. So, by the result fubinous theorem 1, I can write this as integral over x integral over y F plus x y integral over F x y d mu with respect to y and then d mu with respect to x. So, implies by Fubinous theorem 1 that is Fubinous theorem for non-negative measurable functions that integral of a non-negative measurable function can be computed by iterated integrals. So, this is n also equal to let us write the other one also. You can interchange integral over x F plus of x y d mu x and d mu of y. So, this is for F plus we have used the Fubinous theorem 1. .. Now, let us observe F being integrable this quantity is finite. So, all these integrals are finite quantities. So, that means, all these being finite implies. So, for example, the first one implies. So, implies because of integrability that integral over x integral over y F plus x y of d of mu y d mu x is finite. Now, here is an important observation that we have earlier proved that if the integral of a function is finite, then the function must be finite. So, here we are using integral finite implies function finite almost everywhere. So, this we had already proved. So, this fact we are going to use now. So, look at this inner this integral with respect to mu of this function is finite. So, that implies that the function. So, this as a function of x, x going to integral over y of F plus x y d mu y is finite almost everywhere, almost everywhere with respect to x. So, we have used the fact that integrable function implies that the function is finite almost everywhere. Now, once again for almost all x, this is finite. That means, so this also implies that the function. So, y going to F plus x y is finite almost everywhere and of course, integrable, because this integral is finite almost everywhere. So, it is a function which is integrable and finite almost everywhere. So, implies I can integrate it out and this is a non-negative function. It is integrable and non-negative integrable function. So, we have for this already seen this is equal to. So, that we have already seen that for non-negative integrable function, this is equal to this integral. Similarly, the function x going to, similarly the function for x going to F plus x y is finite almost everywhere and integrable is also. And the corresponding results also hold for, similar results holds for F minus. So, that means what? So, all those four functions, so all the four functions are finite and integrable. So, we can integrate them out and we have the results corresponding results for, so these are all integrable. So, that is the first part. So, that is the first part that these functions are integrable almost everywhere and correspondingly almost everywhere with respect to x and y and these functions are also defined and are integrable. So, that means we have the following. So, they are all integrable and finite and for F plus x y with respect to x cross y d mu cross nu, we have got this is equal to the iterated integral with respect to x with respect to y of F plus x y d nu d mu. Similarly for the negative part, we have x y d mu cross nu x cross y is equal to integral over y integral over x of F minus x y d nu d mu. So, now we can and all these are finite quantities because F plus and everything is integrable. So, these are all finite quantities. This is finite and this is finite. So, I can take the difference of the two. So, the difference of the left hand side, so subtract second from the first and use the fact that integrals are linear. So, the difference subtract implies subtraction and similarly and also the corresponding identities for the other one interchanged thing. So, this is also equal to integral over y integral over x of F plus d nu d mu for non-negative that is true. So, we will subtract this from this. So, we will get integral of F d mu cross nu x cross y because integral of F plus minus integral of F minus is integral of F is equal to the iterated integral of F plus with respect to x y minus the iterated integral of F minus with respect to the same iterated integral. So, that will give you y integral of respect to x of F plus minus F minus. So, that is F x y d nu d mu. So, that will prove that for the integrable function the double integral is equal to the iterated integral one of them. The other proof is similar. So, basically this result that for integrable functions the corresponding interchange of integrals hold is basically coming from the previous result namely that the corresponding result holds for non-negative simple non-negative measurable functions. So, what we have proved is two Fumini's theorems Fumini's theorem one and Fumini's theorem two. Fumini's theorem one says that for non-negative measurable functions the double integral the integral over the product space can be computed by integrating one variable at a time and similarly this can also be done for functions which are integrable. So, we will continue this Fumini's theorem a bit more and then specialize it to for integrals for R 2, R 3 and so on. Thank you.