 Welcome to lecture 21 on measure and integration. In the previous lecture, we had started looking at the properties of sequences of integrable functions and we started proving an important theorem called Lebesgue's dominated convergence theorem. Let us continue looking at that and after that we will start looking at the special case of integration on the real line and that will give us the notion of Lebesgue integral. Let us recall what we had started proving namely dominated convergence theorem, which says that if f n is a sequence of measurable functions such that there exists a function g belonging to l 1, so that all the f n's are dominated by this integrable function g almost everywhere x for all n, then and if f n converges to f, then the function the limit function is integrable and integral of f is nothing but the limit of integrals of f n's. So, the theorem basically says that if f n is a sequence of measurable functions all of them dominated by a single integrable function g, then all the f n's become integrable of course and if f n's converge to f then f is integrable and integral of f is nothing but the limit of integrals of f n's. We had proved this theorem in this particular case when instead of this almost everywhere that mod f n's f n's are dominated by g everywhere and f n's converge to f everywhere. So, to extend this case to almost everywhere we have to do only a minor modification. Let us define the set n to be the set of all x belonging to x where mod f n x is not dominated by g, so g of x or union the set of all those points x belonging to x such that f n x does not converge to the function f of x. So, on n complement we have f n x is less than g x and f n x converges to f of x. So, for every x belonging to n complement and mu of the set n is equal to 0 because we are saying that this mod f n x is less than g x almost everywhere and f n x converges to f of x almost everywhere. So, the set where it does not hold there is a set n and that set as n has got measure 0. Now, let us consider the sequence indicator function of n complement times f n. So, this is the sequence of functions which are dominated by. So, this is the sequence of functions for n bigger than or equal to 1 and they satisfy the property. So, namely this indicator function of n complement f n mod of that is less than g for all x everywhere and the functions converge to the indicator function of n complement f. So, by the our earlier case what we get is the following namely that indicator function of n complement times f is l 1 is integrable and integral of f n limit n going to infinity indicator function of n complement times f n the integral of that converges to the integral of indicator function of n complement times f. So, that is by the earlier case when everything is true for all points. So, that means, so this is same, but note that mu of n is equal to 0. So, that implies that the integral of f over n d mu is equal to 0 and we already know that on n complement f is integrable. So, that together with this fact implies integral of mod integral of mod f d mu is finite implying that f belongs to l 1 and this equation which said that integral of f n over n complement converges to integral of f over n complement and mu of n being 0 together gives us the condition that integral of f n d mu converges to integral f d mu because integral f n d mu is same as integral of f n over n plus integral over n complement and integral over n complement converges to integral over n complement of f and on n both are 0. So, this gives as the required result. So, that is how from almost everywhere conditions are deduced from the fact that something holds everywhere. So, this dominated convergence theorem holds for whenever the sequence f n is dominated by g and f n converges to f almost everywhere then f limit function is integrable and integral f converges to integral of f n. So, as I said this is one of the important theorems which helps us to interchange the limit and the integral sign. Let us look at some more minor modifications of this theorem. One thing more we can even deduce that integral of mod f n minus f d mu also converges to 0. So, to deduce that part we just have to observe that mod f n minus f is less than or equal to twice of g and mod f n minus f goes to 0. So, again an application of dominated convergence theorem dominated convergence theorem which we proved just now implies that integral of mod f n minus f d mu goes to 0. So, that is another modification, another consequence of the dominated convergence theorem. Let us prove what I call as the series version of this theorem namely that if f n is a sequence of functions which are integrable and integrals of f n summation 1 to infinity. So, some of all the integrals of f n mod f n are finite. Then the conclusion is that the series f n x converges almost everywhere and if we denote the limit the sum being f of x then that function is integrable and integral f is equal to summation of integral f n. So, essentially this theorem says that if the summations of mod f n are finite then this the series f n x is this almost everywhere is convergent almost everywhere and integral of f is equal to. So, integral of f is equal to integral of summation of integral f n. So, that is again interchange of limit essentially. So, let us see how from dominated convergence theorem we can get this. So, let us define, see to show that this series is convergent almost everywhere we likely show that so we show is absolutely convergent. So, for that let us define say g n of x to be equal to summation mod f k of x k going from 1 to n, the partial sums of the absolute values 1 to n. So, let us observe that this sequence g n. So, note g n is a sequence of non-negative measurable functions and g n are increasing to some function. So, that is they are going to increase to k equal to 1 to infinity mod of f k of x and let us call that as g x. They are increasing to the function g of x. So, that implies by monotone convergence theorem we have integral of g n d mu must converge to integral of g d mu. But integral of g n d mu, so that is the same as saying integral g d mu is equal to limit n going to infinity of integral g n d mu. But what is integral of g n? g n is sum of absolute values of f k 1 to n. So, by linearity property this is nothing but limit of n going to infinity of summation 1 to n k equal to 1 to n of integral mod f k d mu. And this limit is nothing but 1 to infinity and that is given to be finite. So, let us write that, so this is equal to summation k equal to 1 to infinity of integral mod f k d mu, which is given to be finite. So, hence what we get is thus g is integrable, g belongs to g is a integrable function. So, saying that g is integrable implies recall that if a function is integrable and g is a non-negative function, so g of x is finite almost everywhere. That is non-negative function, which is an integrable function must be finite almost everywhere. So, we get that g is finite almost everywhere and what is the function g? g is nothing but the limit of the absolute values of f k x. So, that means proves that the series, so hence sigma k equal to 1 to infinity mod f k of x is finite almost everywhere x. And once a series is absolutely convergent, it is also convergent, so that implies that sigma k equal to 1 to infinity f k x is finite for almost everywhere x. So, let us denote this limit by f of x. So, this is f of x and so as observed earlier, so note. So, f of x, we can write also f of x as the limit n going to infinity of summation k equal to 1 to n f k x. And if these functions are called as something say f n, sorry phi n, then note that mod phi n, if we call this as phi n, then what is mod phi n? Mod phi n is absolute value of 1 to n f k x, absolute value of that and that is less than or equal to summation 1 to n mod of f k 1 to n and that is nothing but our g n, which is less than or equal to g. So, this partial sums, which we have called as phi n are all dominated by g and phi n is converged to f, so by dominated convergence theorem, so what we have got is, so we have got, so all the phi n's are less than or equal to g and phi n's converge to f almost everywhere, so that implies by dominated convergence theorem that integral of phi n's d mu must converge to integral of f d mu, but this is nothing but this phi n, this is what we called as phi n that is summation 1 to n, so this is nothing but summation of 1 to n of integral k equal to 1 to n of integral f k d mu must converge to integral f d mu and that is same as saying that integral f d mu is equal to summation k equal to 1 to n of integral f k d mu. So, that proves the theorem namely, if f k is a sequence of functions which are integrable and the sum of the integrals is finite, then the series f n x n 1 to infinity itself is convergent almost everywhere and the limit function is integrable and integral of the limit function is equal to summation of integrals of f n's. So, this will refer to as the series version of dominated convergence theorem. There is another interpretation of the dominated convergence theorem, when the underlying measure space is a finite measure space, then one has that if x s mu is a finite measure space and f n is a sequence of measurable functions, so that all of them are dominated by a single constant m almost everywhere and f n x converges to f of x, then integral f n's converge to integral f. So, this is a particular case of dominated convergence theorem, when the underlying measure space is a finite measure space and the only thing to observe here is that because, let us see how does this follow from our is a dominated convergence theorem. So, we are given that mod f n x is less than or equal to m for almost everywhere x. Now, look at the constant function m, so look at the function g of x, which is equal to m for every x belonging to x. So, the constant function is measurable, so note that g is a non-negative measurable function, because it is constant function, note that so its integral g d mu is equal to the constant function m d mu and that is equal to m times the measure of the whole space x, which is finite. So, what we are saying is on finite measure space is a constant function is always integrable, so implies g is l 1, so f n x bounded by m, so that constant function and that is a integrable function. Once we have that and f n x converges to f of x almost everywhere, so now by dominated convergence theorem is applicable and that implies integral f d mu is equal to integral f n d mu limit n going to infinity. So, the main thing is on finite measure space is a constant function becomes integrable because of this reason. So, this is what is called bounded convergence theorem and it is quite useful when underlying measure space is a finite measure space. So, let us look at what we have proved till now, we have looked at the space of integrable functions and proved linearity property and an important theorem called dominated convergence theorem. So, if you recall for non-negative measureable functions we had two theorems one was monotone convergence theorem namely that was a theorem when f n is a sequence of non-negative measureable functions increasing to a function f then integral of f is equal to limit of integral. So, that means interchange of limit and integration is possible by monotone convergence theorem whenever the sequence f n is monotonically increasing and a sequence of non-negative measureable functions. The second theorem which involved sequences of measureable functions was again for non-negative measureable functions and that was called Fatou's lemma. So, there we do not emphasize we do not require that the sequence f n be non-negative and measurable. We only want the sequence f n to be a sequence of measureable non-negative measureable functions they need not be increasing. So, for the such a sequence we had that integral of the limit inferior of the sequence f n is less than or equal to limit inferior of the integrals of f n. So, that was Fatou's lemma and now we have the third theorem dominated convergence theorem which again helps you to interchange the notion of the notion of integral and the limiting operation under the condition that all the f n's are dominated by a single integrable function. So, these are the three important theorems which help us to interchange limit and the integral signs. Let us at this stage emphasize one more point about this technique of integration and so basically for integral we started with simple functions and then we go to non-negative functions and then we defined it for integrable functions. So, this process of step by step defining the integral can be is useful in proving many results and I call it as the simple function technique. So, this is a technique which is used very often to prove some results about integrable functions and non-negative measurable functions. So, what is the technique? Let me outline that and then I will give a illustration of this. Suppose you want to show that a certain property say let us all that property a star holds for all integrable functions. So, to prove that the property holds for all integrable function the technique is as follows basically that show that this property star holds for all non-negative simple measurable functions. So, if you want to show a property holds for all integrable functions first show that it holds for the class of non-negative simple measurable functions. And next show that star holds for non-negative measurable integrable functions by using the fact that non-negative measurable functions are limits of increasing limit of simple measurable functions. So, and use there one uses normally the monotone convergence theorem. So, using monotone convergence theorem one extends the property star from simple measurable functions to non-negative measurable functions or non-negative integrable functions. And then keeping in mind that for a function f it can be split into positive part and negative part. So, f can be written as f plus minus f minus and if a property holds for non-negative functions so about integrable. So, f for f plus that will hold for f minus that will hold and then conclude from there that it holds for f also. So, this is what I call as the simple function technique to prove results about integrable functions. To give an illustration of this let us look at the following result. Let us take a measurable space x s mu which is sigma finite measure space and let us look at a function f which is integrable on this measure space and is non-negative. So, we have got a sigma finite measure space and f is a non-negative integrable function on this measure space. Let us define nu of E for every set in the sigma algebra. Let us define nu of E to be integral of f d mu over the set E integral of f over the set E is denoted by nu of E for every set E in the sigma algebra S. Then we have already shown that this nu the set function nu is in fact a finite measure on S. So, this we have already proved but what we want to prove now is that further if g is any integrable function on the measure space x as nu this nu is the nu measure. So, if g is integrable on x with respect to nu then the product function f into g is integrable with respect to mu and this relation holds integral f d mu. So, integral of f with respect to nu is equal to integral of f into g with respect to mu. So, what we have done is by using a fixing a function f which is non-negative we have defined a nu measure on the measurable space by nu of E to be integral of E f d mu and we are saying if we want to integrate a function with respect to a function g with respect to this nu measure then it is same as integrating the function the product function f g with respect to the old measure mu. So, let us see how the simple function technique is used to prove this result. So, let us start. So, we want to show that for every g belonging to l 1 of x s nu integral of g d nu can be represented as integral f g d mu. So, where recall we defined nu of E to be equal to integral of f d mu over E. So, this is the property we want to prove. So, this is the property star we want to prove for every function g. So, as we said let us first check this property. So, step one let us say g is a l 1 function let us say g is a function which is indicator function of E. Let us take g as the indicator function of E, E belonging to s. In that case the integral g d mu the left hand side is nothing but nu of E because g is the indicator function. So, this integral of nu of E which by definition is equal to integral chi E of f d mu and so chi E is g. So, this is equal to integral g f d mu. So, what does it says? It says that the required property star holds when g is the indicator function. Now, let us take a non-negative simple function. So, that is step two let us take g is sigma A i chi E of E i i equal to some 1 to n where E i belong to s and our claim is that this property holds for this g also. So, we are saying that next step is to verify that the required property. So, integral of g d nu by definition is equal to integral of sigma A i indicator function of E i d nu and what is that? By generative property of the integral it is sigma i equal to 1 to n of A i that is the scalar times nu of E i because integral of the indicator function is the measure. So, that is equal to sigma i equal to 1 to n A i and nu of A i by definition is integral chi E i of f d mu that is the definition of nu of E i is this. So, which I can write it again as sigma i equal to 1 to n of A i we can take it out. So, let us so this is and again by the linearity property. So, that is integral A i chi E i times f d mu, but this is nothing but my function g. So, this is integral of g f d mu. So, what we are saying that if g is summation A i chi E i then using linearity property this is same as integral goes in. So, that is A i integral of the indicator function of E i that is nu of E i and nu of E i by definition is integral over E i of f d mu and again using linearity property of the integral I can shift it outside. So, it is integral of summation A i chi E i times f d mu which is g. So, it says so the required property holds. So, star holds for non-negative simple functions g. So, that is what I said the simple function technique and now let us try to prove it that this property also holds when g is a non-negative measurable function. So, let us look at now g. So, let g on x to be measurable. Then we know by the property of measurable functions implies there exists a sequence S n of non-negative there is sequence S n of non-negative simple measurable functions such that S n increases to the function g. So, then by Lebesgue, by Morton Convergence theorem integral of g d nu with respect to nu must be equal to limit n going to infinity integral S n d nu. But for non-negative simple functions just now we proved this the star holds that means this can be written as limit. So, by step 2 I can write this as integral of S n times f d mu. So, with the integration of a non-negative simple measurable function with respect to nu can be converted into the non-negative simple measurable function multiplied by f d mu. And now at this stage we observe that if S n is increasing to g then S n times f will be increasing to g times f and all are non-negative simple measurable all are non-negative measurable functions. So, once again Morton Convergence theorem is applicable and this limit is nothing but integral of g f d mu that is so once again we have used. So, this step this step was by our step 2 that the property holds for non-negative simple measurable functions integral with respect to nu is integral with respect to mu of the product function. And now once again we are applying Morton Convergence theorem. So, first integral of g is equal to limit of integral S n d nu by Morton Convergence theorem. And now by our earlier step this is equal to integral of S n f d mu again by Morton Convergence theorem it goes back. So, that means so this implies that star holds for non-negative measurable functions. And now let us come to the last part namely so let us so final step 3 is let g belong to L 1 X S nu g be a integrable function. Now then what is g equal to g plus minus g minus where g plus is a non-negative measurable function g minus is a non-negative measurable function and by step 2. So, by step 2 we know that integral g plus d mu is equal to integral of g plus sorry d nu. So, let me write it again integral g plus d nu is equal to integral g plus of f d mu and integral of g minus d nu is equal to integral g minus f d mu. So, that is by step 2 and now because g is L 1 so that implies g is equal to g plus g minus. So, g plus is in L 1 of nu and g minus also belongs to L 1 of nu a function g is integrable if and only if it is positive part and negative parts are integrable. So, that means these quantities they are all finite these are all finite quantities. So, that means what and f is non-negative so that implies integral of g f d mu is equal to integral of g plus f d mu minus integral g minus f d mu by definition of the positive part and the negative part of the function g f f is non-negative. So, the positive part of the function g f is same as g plus times f and the negative part is nothing but g minus f and all both of these are finite quantities. So, that means implies that g f is L 1 and by these two this is same as integral of g d nu. So, for a so this the step 3 for a g which is integrable we have deduced that this property is true. So, this is step 3 so this is what I call the simple function technique. So, let me go back and show you once again what we have done. So, we wanted to show that so this is property star we wanted to show for every function g which is L 1. So, first step is so this is my step 1 that look it at the functions g which are indicator functions. So, I want to verify this for the indicator function g to be the indicator function. So, when g is the indicator function so this left hand side is integral of over e of the constant function 1. So, this is equal to integral d nu is nu of e which by definition is integral f over e so which I can write as integral f e so that is true. So, step 1 is to verify the required thing holds for characteristic function and step 2 by using the property that the integral is linear we show it is true for every nonnegative simple function. So, take g a nonnegative simple measurable function and apply so g is equal to integral of nonnegative A i indicator function A i and interchange and show it required property holds. So, the step 2 was that the required property holds for nonnegative simple for nonnegative simple measurable functions and then using an application of monotone convergence theorem. So, that is step 3 that if g is a nonnegative measurable function then we know it is a limit of nonnegative simple measurable functions increasing limit. So, an application of monotone convergence theorem together with the earlier step gives us that integral of g d nu is equal to integral of g f d nu. So, that is the next step to show that it holds for nonnegative measurable functions and once that is done the final step that it holds for all integrable functions is via splitting the function g into the positive part minus the negative part and g integrable means both are integrable and for each one of them the required claim star holds. So, by putting them together we get that the required claim holds property star holds for all functions g so which are l 1. So, this is what I normally call as the simple function technique. So, while proving results about integrable functions one uses quite often the simple function technique and while proving some properties about subsets of sets recall we had the sigma algebra monotone class theorem technique. So, for proving properties about sets one uses monotone class sigma algebra monotone class technique and for proving the results about integrals one normally uses what is called the simple function technique. So, with this we have defined and proved general properties about integral of functions on sigma finite measure spaces. Now, we will try we will specialize this property this construction when x is real line. So, we want to specialize this thing for the real line. So, let us see what we get. So, while you will be looking at the special case when x is real line the sigma algebra is l that of Lebesgue measurable sets and the measure mu will be the lambda the Lebesgue measure. So, we will be working with the measure space x s mu which is same as real line Lebesgue measurable sets and Lebesgue measure. So, the space of all integrable functions on this measure space R, L and lambda is called the space of all Lebesgue integrable functions and is also denoted by L 1 of R or L 1 of lambda. So, this is the space of all Lebesgue integrable functions. So, we want to study this space of Lebesgue integrable functions in some more detail. Let us first agree to call integral f d lambda to be the Lebesgue integral of the function f. So, whenever f is integrable or non-negative integral f d lambda will be called the Lebesgue integral of f. Sometimes we have to look at functions which are defined on subsets of E. So, for any subset E which is Lebesgue measurable L 1 of E will denote the measure space of all integrable functions on the measure space E. So, the underlying set is E, L intersection E is the collection of all Lebesgue measurable sets inside E and lambda is the Lebesgue measure restricted to subsets of L intersection, the sigma algebra L intersection E. Of particular interest for the time being is going to be the set when E is a close bounded interval A B. So, we will start looking at the space L 1 of A B that is the space of all Lebesgue integrable functions defined on the interval close bounded interval A B. We also have the space R A B namely the space of all Riemann integrable functions on A B. So, we want to compare these two spaces on one hand we have got the space of Lebesgue integrable functions on A B, on the other hand we have got the space of Riemann integrable functions on A B and we want to see the relation or establish a relationship between the two. That was one of the starting points for our discussion of the subject namely the space of Riemann integrable functions had some difficulties, some problems, some drawbacks and for which we want you to extend the notion to a larger class and this is the larger class L 1 of A B. So, what we are going to show is R A B the space of all Riemann integrable functions is a subset of L 1 of A B and the notion of Riemann integral is same as the notion of Lebesgue integrable for Riemann integrable functions. So, that is called the relation between the Riemann integral and the Lebesgue integral. So, to be more specific we want to prove the following theorem namely if f is defined on a close bounded interval A B is Riemann integrable function then f is also Lebesgue integrable and the Lebesgue integral is same as the Riemann integral of the function f. So, this is what we wanted to want to prove. So, let us start looking at how do we prove this. So, the proof of the theorem so we are given that the function f belongs to R A B it is a Riemann integrable function. That means so let us recall how is the Riemann integral of a function defined it is defined via limits of upper sums and lower sums of partitions. So, implies there exists a sequence P n of refinement partitions with norm of P n going to 0 as n goes to infinity partitions n going to infinity with the upper sums of P n with respect to f limit of that is same as integral the Riemann integral of f is same as the limit of the lower sums L P n of f. So, that is the meaning of saying that a function f is Riemann integrable. So, we can find Riemann integrable implies there exists a sequence of partitions P n which are refinement partitions refinement means P n plus 1 is obtained from P n by adding one more point and norm of this partitions the maximum length of the sub intervals goes to 0 and integrable means that the upper sums and the lower sums both converge to the same value and that is the Riemann integral of the function f. So, this is the property of saying that f is Riemann integrable. Now, from here let us look at what is U P n f upper sum. So, let us write down the partition P n as something. So, let us say P n looks like A. So, interval is A to B. So, A the point x naught less than x 1 less than x n which is equal to B. So, let us say that is the partition P n. So, in the picture it will look like here is A here is B. So, this is x 0, this is x n and here is x 1, x 2 and so on. So, to construct the upper sums what one does? To construct the upper sums one looks at the maximum value of the function in this interval and the minimum values in this interval. So, let us write m k to be the maximum value of the function in the interval x, say x i minus k. So, let us say x k minus 1 to x k. I am just trying to make the intervals disjoint maximum in this interval. Of maximum in this interval, maximum of f of x, maximum in of f of x. Similarly, m k let us write it is the minimum in the interval x k minus 1 to x k of f of x. Only at the end points you have to make it close, but that is not going to matter much. So, that then we define what is U P n f that is essentially looks like summation of the maximum value into the indicator function of that sub interval and the lower sum with respect to P n f looks like summation small m k the minimum value of the function in that sub interval x k minus 1 and x k. So, let us do one thing. This is upper sums and these are not. So, let us let me so I am sorry this is not the upper sum. Let us call this when in the interval x k minus 1 to x k the value is capital m k. Let us call that as the function phi k and when you are taking the minimum value in that interval and summing up let us call that as psi k. So, these are functions because they are linear combinations of indicator functions and the upper sums and lower sums are nothing but the upper sum P n f is nothing but the Riemann integral a to b of phi k x dx and the lower sum P n f is equal to the integral of Riemann integral of this function psi k of x dx that these functions phi k and psi k which are linear combinations of indicator functions are in fact, non-negative measurable functions on the measure space AB the interval AB. So, note so that is the observation that we should note and then let us note down. So, note phi k and psi k are measurable functions are measurable non-negative sorry are simple measurable functions say that phi k is less than or equal to at every point x is less than or equal to f of x is less than or equal to psi k of x and as far as the integral is concerned the integral a to b of f x dx is between the upper sum and the lower sum. So, that is the phi k was maximum sorry this should be bigger than or equal to like this because phi k is taken as the supremum. So, this is so this is upper sum P k of f and that is bigger than or equal to the upper sum sorry bigger than or equal to the lower sum with respect to P k of f and in the limit both of them are converging. So, here is the second observation is that the upper sum with respect to the partition of f is same as so what was it? So, that was equal to sigma m k into the length of the interval x k minus x k minus 1. So, that is the upper sum that is also the Riemann integral. In fact, this is also equal to so length. So, this is the length of so you can write this as the length so m k times the length of x k minus 1 and x k which is same as the Lebesgue integral of the function phi k d lambda. So, this is this is the important observation that we should keep in mind that the building blocks for Riemann integral which are these step functions are also Lebesgue integrable and the Riemann integral of the step functions phi k and psi k are same as the Lebesgue integrals of phi k and psi k. So, similarly the lower sum p k f is equal to integral of psi k d lambda and now essentially the idea is to put them together. So, because phi k and psi k they are between these two. So, let us look at integral of so look at the sequence so consider the sequence psi k minus phi k minus psi k. Recall phi k is bigger than f x is less than psi k. So, phi k minus psi k is non-negative for every k and saying that the upper sums and lower sums converge to the same value is saying that the integral of phi k minus psi k phi k minus psi k d lambda that goes to 0. So, that goes to 0. So, because the phi k d lambda is the upper sum this is the lower sum and that goes to 0. So, that implies that limit so that means this implies that the limiting function f is trapped in between so that means limit phi k x is equal to limit psi k x almost everywhere. Why is that? So, that we can reduce from the fact that applying Fatou's lemma. So, to reduce this look at the limit inferior of phi k minus psi k integral d lambda will less than or equal to limit inferior of integral phi k minus psi k and that is 0. So, this is 0. So, that means and this is and so this says that integral of a non-negative function is 0. So, the function must be 0 almost everywhere and that is same as saying this must be 0 almost everywhere and f is trapped in between. So, that implies that limit phi k x limit of phi k x is equal to f of x is equal to limit psi k x for almost everywhere x. So, that proves is equal to so we are falling short of time. So, that means that the function f is measurable. So, we will continue the proof of this tomorrow in the next lecture. So, our aim is to prove that the space of Riemann integrable functions is inside the space of Lebesgue integrable functions and the Riemann integral is same as the Lebesgue integral. So, we will continue the proof in the next lecture. Thank you.