 Welcome to lecture number 22 on measure and integration. If you recall, in the previous lecture, we had started looking at the properties of Lebesgue measure and of Lebesgue integrable functions. We started looking at analyzing, when does a function which is Lebesgue integrable on the interval a, b and its relation with Riemann integral of the functions on the interval a, b. Let us, we had started looking at the proof of the theorem namely that, if f is a function defined on an interval a, b to R which is Riemann integrable, then we wanted to show that it is also Lebesgue integrable and the Riemann integral is same as the Lebesgue integral. We will continue the proof of that theorem and then go on to analyze some more properties of the space of Lebesgue integrable functions on the interval a, b. So, today's lecture is going to be mainly concerned with Lebesgue integral and its properties. So, the theorem we wanted to prove was that, if f is a function defined on an interval a, b to R and it is Riemann integrable, then it is also Lebesgue integrable and the Riemann integral of the function is same as its Lebesgue integral. So, to prove this theorem, we started with this idea that, since f is Riemann integrable, there exist sequences psi n and phi n of step functions on the interval a, b such that this sequence psi n is monotonically increasing and the sequence phi n is monotonically decreasing and the function f is between these two sequences phi n and psi n for all points x belonging to a, b and the Riemann integral of psi n converges to the same value as the Riemann integral of f and it is the same as the limit of the Riemann integrals of the step functions phi n. So, let us recall these steps which we had proved last time. So, what we had given is f is a function defined on an interval a, b to R and f is Riemann integrable. So, what does the Riemann integrability imply? The Riemann integrability of the function implies the following namely, there exists a sequence P n of refinement partitions of the interval a, b such that the norm of these partitions goes to 0 and the upper sums of f with respect to these partitions that decreases and the upper sums and the lower sums of f with respect to P n increases and the common value is the integral. So, limit n going to infinity of upper sums is same as Riemann integral of f and that is same as the limit of the lower sums. So, this is because f is Riemann integrable. Now, let us see what are the upper sums and what are the lower sums. We need to analyze them slightly more carefully to look at. So, this is, let us draw a picture of the function say the function looks like this. So, this is a and this is b and we get the partition. With respect to a partition, let us say this is the general interval say x i minus 1 and x i. So, in this interval look at what is the smallest value of the function. So, what is the smallest value of the function? That is this. So, look at this height and look at the largest value of the function in that interval. So, largest value of the function is somewhere here. So, look at that height. So, lower sums consist of the areas of these rectangles with height as the blue line and the upper sums consist of the sums of all the areas which are the green lines. So, mathematically what this means is the following. So, let us write mathematically what it means. So, mathematically these things mean the following namely. So, look at consider the function. So, let us write. So, consider let us define define. So, let us say p p is the partition say p n is the partition which looks like a equal to x 0 less than x i minus 1 less than x i x n equal to b. So, let us say that is the partition. So, define let us say m 1 to be the supremum of the function f x x belonging to a that is less than or equal to x 1 x 1. And let us write m k to be the supremum of the function in the general interval. So, f x for x between x i minus 1 and x i. So, keep in mind here I am taking left open and right close here at the end point is both sides close. So, these intervals are disjoint intervals. So, what I am doing is in the first part I am looking at this, then I am looking at left open, right close, left open, right close, left open, right close. So, I am partitioning the interval a b according to the partition points p n and then looking at the supremums in the respective intervals. Similarly, let us write m k to be equal to infimum of f x in x x i minus 1 less than or equal to x i and m 1 to be the infimum in the first interval. So, that is f x x in x 0 x 1. So, what is this value? This m 1 and m k's are corresponding to the height which is the green line. So, that is the maximum value of the function in the interval x i minus 1 to x i and the blue ones correspond to small m k's. So, once we have done this mathematically, let us define the required functions. So, let us define phi n is the function which is summation m i indicator function of x i minus 1 to x i i equal to 2 to n and in the first one, let us put that value as m 1, the indicator function of A x 1. Similarly, let us write psi n to be sigma i equal to 2 to n capital m i the maximum value in the interval x i minus 1 to x i and capital m 1 in the first interval. So, that is indicator function of A to x 1. So, these are the functions we defined. So, they are corresponding to the function which is phi n, small phi n will look like the minimum values like this and it will look like this. And the capital and psi n's, they will look like maximum values. So, they will look like this and look like this and look like this. So, quite clearly, these functions phi n and psi n are step functions and phi n of x is less than or equal to f of x and less than or equal to psi n of x. So, as a first step, the Riemann integrability of the function f over the interval a, b gives us a sequence of functions phi n and psi n, where each phi n is a step function, each psi n is a step function and phi n is less than or equal to f of x is less than or equal to psi n of x. And since our requirement was that Riemann integrability implies that the sequence p n of partitions is a sequence of refinement partitions. So, that implies that the sequence psi n will be a decreasing sequence and phi n will be an increasing sequence. So, let us write that observation namely that, so we have phi n's. So, each phi n psi n is a step function, phi n's are increasing and psi n's are decreasing. Moreover, further, let us look at the Riemann integral of the function phi n x dx a to b. So, because phi n is constant on each sub interval of the partition, this is nothing but equal to m 1 times the length of the first interval. So, that is x 1 minus a plus the sums of those rectangles, so that is small m k times x k minus x k minus 1 k equal to 1 to n. And similarly, the Riemann integral of the function psi n x dx is equal to capital M 1 in the first interval times the width of the interval that is x 1 minus a plus k equal to 1 to n the areas of the other rectangles, so that is capital M k times x k minus x k minus 1. So, those are the Riemann integrals. And by the definition of the upper and the lower sums, this is precisely the lower sum of f with respect to p n and this is precisely the upper sum of the function f with respect to the partition p n. And saying that the function is Riemann integrable implies that the upper sums and the lower sums converge to the same value. So, this and the Riemann integrability implies that the Riemann integral a to b of phi n x dx limit n going to infinity is same as the Riemann integral a to b f x dx and that is same as the Riemann limit of the upper sums, so that is limit n going to infinity the Riemann integral a to b of psi n x dx. So, that proves our first step. So, as a consequence of the Riemann integrability of the function f, we have constructed two sequences of step functions psi n and phi n, where psi n is monotonically increasing and phi n is monotonically decreasing and limit of both the integrals of both of them converge to the Riemann integral of f. Now, let us observe at this stage. So, in some sense the step functions and the Riemann integrals are the building blocks for the Riemann integral. Now let us observe that the lower sum of the function f with respect to the partition which was the Riemann sum which was the Riemann integral of phi n is also nothing but the Lebesgue integral of the function phi n with respect to the Lebesgue integral, because this is this function phi n is a simple function, is a simple measurable function and its integral is nothing but this integral. So, the observation is that for every n integral a to b of phi n x dx, the Riemann integral is same as the Lebesgue integral of the function phi n with respect to the Lebesgue integral over the interval a b and this observation is simply by the fact that phi n is a step function. So, hence it is a simple measurable function that is integral is nothing but the value times the Lebesgue measure of the portion on which that value is taken and that being here sub intervals. So, it is same as the Riemann integral and similarly the Riemann integral a to b of psi n x dx is equal to integral over a b of psi n d lambda. So, this is the observation which is going to play an important role for us. So, now let us define, let us consider the, because we have got the observation that the function f n, the function phi n is always dominated by f x is less than or equal to psi n. So, let us look at the function psi n minus phi n. So, look at the function. So, consider the sequence psi n minus phi n. So, look at the sequence of these step functions. These are step functions as well as they are simple measurable functions and they are non-negative. So, they are non-negative simple measurable functions and by Fertu's lemma. So, we are going to apply Fertu's lemma to this. Consider this sequence and apply Fertu's lemma. So, what will that give me? So, that will give me that limit inferior of psi n minus phi n d lambda limit n going to infinity integral over a b. So, the Lebesgue integral of the limit inferior is less than or equal to limit inferior of the integral psi n minus phi n d lambda. So, that is by Fertu's lemma. But, let us observe here that integral of psi n over a b minus integral of phi n over a b, the limit inferior of that is equal to 0, because the Lebesgue integrals of psi n are same as the Riemann integrals of psi n and Lebesgue integrals of phi n are same as the Riemann integrals of phi n and those Riemann integrals converge to the Riemann integral of f. So, this right hand side is equal to 0. So, that means that the limit inferior of psi n minus phi n, that function is a non-negative function and its integral, Lebesgue integral is equal to 0. So, that we know implies that the function must be 0 almost everywhere. So, this implies that limit inferior of the sequence, which is psi n minus phi n x must be equal to 0, almost everywhere x. But, we know that psi n's are increasing, psi n's are decreasing and phi n's are increasing. So, this limit inferior exists. So, that implies that limit n going to infinity of psi n's is equal to limit n going to infinity of phi n x almost everywhere x. But, we know since f of x is between phi n and psi n of x, this implies along with the earlier fact, this implies that limit n going to infinity of psi n x must be actually equal to f of x, actually equal to limit n going to infinity of psi n of phi n of x for almost all x. So, this must happen. But, that implies because each psi n is a measurable function, each phi n is a measurable function, this implies f is measurable. But, recall f was remain integrable function. So, obviously it is a bounded function. So, implies that f is Lebesgue integrable because f bounded. So, by boundedness of f, every bounded measurable function and this is defined on AB, which is a finite measure space. So, that must be integrable. That we had observed earlier. So, f is Riemann integrable. So, only we have to prove now. So, claim that integral f d lambda over AB is equal to integral a to b of f x d x. So, this is the only thing left to be shown. So, what we have shown till now is f is Lebesgue integrable. So, this integral exists. Now, let us observe one thing. Not only f is Lebesgue integrable, it is a limit of a sequence of functions phi n's. Phi n's converge to f of x. So, the sequence phi n is converging to f of x and psi n's are decreasing to f of x. So, whichever fact we require, you can use. So, let us use the fact that psi n's are limit of psi n's converge to f of x. They are decreasing and each psi n is a integrable function. So, we can apply Lebesgue's dominated convergence theorem to conclude. So, by dominated convergence theorem, since psi n x converges to f of x, almost everywhere x and psi n's are decreasing and integrable psi n's decreasing psi n's integrable. So, implies by Lebesgue's dominated convergence theorem that integral of f d lambda over A B must be equal to limit n going to infinity of the Lebesgue integral of psi n d lambda. So, that is the observation we have already made that the Lebesgue integral of the step function psi n is same as the Riemann integral. So, n going to infinity. So, that is A to B of psi n x d x. And this Riemann integral we have observed is the upper sum and which limit is equal to… So, this is the limit of the upper sums of f with respect to p n and that converge to integral A to B f x d x. So, that will prove that integral, Lebesgue integral of f is same as the Riemann integral of f over d lambda. So, that proves the theorem. So, let us go back and recall the proof. What are the first step? As a first step in the proof, using the Riemann integrability of the function, we construct two sequences of step functions, psi n and psi n, so that f is trapped in between them. And the upper sums are nothing but the Riemann integrals of psi n and the lower sums are nothing but the phi n. So, that means the Riemann integral of psi n converges to the Riemann integral of f and which is also equal to… So, here is the equality sign equal to the Riemann integral of phi n. So, that is this construction is purely from the fact that the function f is Riemann integrable. And the second observation is that each phi n and psi n being a step function is also measurable and Lebesgue integrable. And the Lebesgue integral of phi n is same as the Riemann integral of phi n and the Lebesgue integral of psi n is same as the Riemann integral of psi n. So, that is the second observation we will make. So, these two Riemann integrals of the step functions are same as the Lebesgue integrals. And now because of the earlier consequence that the Riemann integrals of phi n and psi n can work through the Riemann integral of f, look at the difference phi n minus psi n. So, look at that sequence of measurable functions phi n minus psi n that is a measurable non-negative measurable function. And an application of Fatou's lemma will give us that this psi n and phi n both must converge to the same value and that is f of x almost everywhere. So, that will imply that f is measurable and being a bounded measurable function on a finite measure space it becomes integrable. And now the psi n's are decreasing to f of x. So, an application of Lebesgue's dominated convergence theorem gives that the Riemann integral of f is same as the limits of the Riemann integrals of psi n's which are equal to the Lebesgue integrals of psi n and that dominated convergence theorem gives you that is the Lebesgue integral of f over a b. So, that proves the theorem completely namely. So, this is the step we wanted to this was the beginning of our lectures namely we wanted to say that to remove the defects of to remove the defects of Riemann integral namely that the fundamental theorem of calculus may not hold and that the space of Riemann integrable functions may not be complete under what is called the L 1 metric we wanted to extend the notion of integral from Riemann integrable functions to a bigger class. So, we have constructed here a class of functions on a b which are called the Lebesgue integrable functions on a b and we have shown that if the class of Riemann integrable functions is a subset of the class of Lebesgue integrable functions and the notion of Lebesgue integral extends the notion of Riemann integral beyond the class of Riemann integrable functions. So, that was that is the first step in our extension theory. So, Lebesgue integral is an extension of the Riemann integral. So, as a next step we want to look at that. So, the space of Riemann integrable functions are inside the class of Lebesgue integrable functions and there is here is a observation which one can observe from the proof of this theorem that if a function is Riemann integrable then it must be continuous almost everywhere. So, to conclude this observation from the proof itself basically what we have to look at is that the function f is the limit of those step functions. There is a limit of the step functions. So, if we leave aside those partition points. So, remember we concluded that f is limit of those step functions phi n's and psi n's. So, with that we concluded namely that here. So, we concluded that this is the fact that we proved in our theorem that the limit of the step functions phi n and psi n is f of x. So, that almost everywhere f is limit of psi n's and psi n's are piecewise continuous functions. So, the points where psi n's may not be continuous are possibly the points of the partition points of p n's. So, if we pull together all the points partition points they will be at the most countably many and if we remove them along with this almost everywhere set. So, outside a set of measures 0 this function f will become continuous. So, I am just indicating the possibility of a proof of that. So, those interested probably should look into the text book and in fact the converse of this theorem is also true namely that if f is a bounded continuous function which is bounded function which is continuous almost everywhere then f is Riemann integrable. So, that means there is a characterization of Riemann integrable functions in terms of continuity namely a function f is Riemann integrable if and only if it is continuous almost everywhere. So, we are not given the proof of this. So, those interested probably should look into the text book as mentioned above the in an introduction to measure and integration in which the complete proof is given. But what we have wanted to indicate that the proof of the theorem proof of one part of the theorem namely for a Riemann integrable function it should be continuous almost everywhere is already included in the proof of the fact this now we proved that Riemann integrable implies it is Lebesgue integrable. So, that is one part of the properties of the space l 1 a b that R a b the space of Riemann integrable functions is in l a b. Now, let us look at a metric on l 1 of a b. So, recall we have already shown that l 1 of a b is a vector space we showed that if f and g are integrable functions then f plus g is integrable f into g is integrable alpha times f is integrable. So, it is a linear space. So, it is a vector space over the field of real numbers. So, on this we are going to define a notion of a magnitude. So, for a function f in l 1 we define its called the l 1 norm of f to be the integral of the absolute value of f of x d lambda x. So, this is called the l 1 norm of the function f which is in l 1. So, clearly it is a number finite number because f is integrable. So, this right hand side exists and is finite and you are integrating a non-negative function. So, the first property namely the norm l 1 norm of a function is bigger than or equal to 0 for all functions f in l 1 of a b. So, that is obvious. The second property namely supposing this function f is 0 almost everywhere then clearly integral of the function will be equal to 0 that we have already observed. So, norm will be equal to 0 and conversely if the l 1 norm is equal to 0 then the function absolute value of f of x being a non-negative function its Leverk integral 0 that implies that the function must be 0 almost everywhere. So, the second property namely the norm is equal to 0 if and only if the function is 0 almost everywhere and the third property which is that if you multiply the function f by alpha the norm of alpha f is equal to the absolute value of alpha times norm of f and that is obvious because in the definition if you replace f by alpha f then this being a constant the property of the integral. So, the integral of alpha f is equal to mod alpha times integral mod f. So, that gives you the property that the l 1 norm of alpha f is equal to the absolute value of the constant with which you are multiplying alpha mod alpha times norm of f 1. And finally, the triangle inequality property namely if f and g are integrable functions then we have already shown that f into g is also a integrable function and integral of the absolute value of f plus g will be less than or equal to integral of absolute value of f plus integral of absolute value of g by the triangle inequality of the numbers. So, that will give us that the l 1 norm of f plus g is less than or equal to l 1 norm of f plus l 1 norm of g. So, these are the properties of this magnitude or this norm of f 1. At this stage I just wanted to point out that this is very much similar to the Euclidean metric or Euclidean norm on R n. See on R n for a vector x with components x 1, x n we normally define the norm to be equal to we can take the norm to be equal to sigma mod of x i i equal to 1 to n. So, this is what is called the l 1 norm on R n. And now, so what we are saying is that if x is a vector with n components then this must be the norm l 1 norm. Now, treat a function f defined on an interval a b to R. So, treat this treat as a vector with components with as many components as points in a b. So, I want to treat this f as a vector with components as many components as points in a b. So, for a point x in a b what is the x th component of f is nothing but f of x. So, you can treat f of x, x belonging to a b as a vector. So, if you treat that way then what is the l 1 norm of f you would like to define. So, keeping in mind, so take the absolute values of the components. So, take the absolute value of the components. So, this is the component take its absolute value and you want to sum it and the summation is nothing but integral d lambda. So, this is in that sense this is the perfect generalization of the ordinary l 1 norm on R n. So, the only problem in this is that we do not have the property that if l 1 norm is equal to 0 if and only if f is equal to 0. So, only we have got that the l 1 norm is equal to 0 if and only if f of x is equal to 0 almost everywhere. But, let us keep in mind that the l 1 norm does not change if we change the function on a null set on a set of measure Lebesgue measure 0. So, with that observation in mind from now onwards what we do is in l 1 of a b we identify functions which are equal almost everywhere. So, if two functions are equal almost everywhere if f and g in l 1 are same are equal almost everywhere lambda then we treat these functions to be as same. So, with that understanding this becomes a metric d f g which is equal to norm of f minus g l 1 norm of f minus g becomes a metric on l 1 of a b and is called the l 1 metric. And the fact we wanted to the observation I want to point out is R a b is a subset of l 1 of a b and so R a b as a subspace with l 1 metric is not complete in the l 1 metric. So, that is an observation which you should read from that textbook already mentioned. So, this was one of the defects of Riemann integral that motivated the development of Lebesgue integral namely the space R a b under the l 1 metric is not a complete metric. So, this is not a complete metric under the l 1 metric. So, but there is a theorem namely every metric space can be completed. So, there is an abstract theorem in metric space is that if every metric space can be completed for example, the look at the set of rational numbers that is not complete under the usual distance of absolute value as a distance and its completion is the real numbers and that is the important property of real numbers that it is complete under that metric. So, similarly R a b under the l 1 metric is not complete and there is a abstract theorem that R a b can be completed it can be put inside a complete metric space. What we are going to show is that that completion is nothing but l 1 of a b. So, we are going to prove that l 1 of a b the space of all Lebesgue integrable functions on a b can be realized as the completion of the space R a b under the l 1 metric. And this we will be doing in two steps one we will show that in the l 1 metric the space l 1 a b is complete that is 1 and it to be a completion of R a b we will show that R a b sits inside l 1 of a b as a dense subset. So, R a b sits inside l 1 of a b as a dense subset in the l 1 metric and l 1 is complete. So, that will show that l 1 of a b is a realization of the completion of the space R a b like rationales are dense in the real line reals are dense real the field of real numbers is complete that is why we say that real numbers is the completion of the field of rational numbers. So, in the similar way we want to prove that l 1 of a b is the completion of the space of R a b. So, that means we want to prove that l 1 a b in the l 1 metric is complete and R a b is dense in it. So, the completion part is called the Ries-Fischer theorem. So, the Ries-Fischer theorem states that l 1 of a b is a complete metric space in the l 1 metric. So, let us look at a proof of this. So, the proof that to prove that it is a complete metric space what we have to show? We have to show that every Cauchy sequence f n in l 1 of a b converges to a value in l 1 of a b. Every Cauchy sequence in l 1 of a b is convergent and convergent to a point in l 1 of a b. So, that is what we have to show. So, for that let us show that. So, this is what we have to show that there exists some function f in l 1 of a b such that f n minus f l 1 norm converges to 0 as n goes to infinity. So, to prove this fact here is an observation what we will show is it is enough to show that there exists a subsequence of f n which is convergent in l 1 norm. So, what we will show is to show that f n is convergent in l 1 metric it is enough f n is a Cauchy sequence to show that a Cauchy sequence is convergent it is enough to show that there exists a subsequence of the Cauchy sequence which is convergent that will prove that the sequence is convergent. So, to do that so we have to produce a subsequence of f n which is convergent to a function in l 1. So, let us look at the first step of our construction to construct that subsequence we are going to use the Cauchyness of the sequence f n. So, f n is Cauchy. So, the first step is that saying that the sequence f n is Cauchy implies that I can pick up a subsequence f n k of f n such that the norm of f n minus f n k is less than 1 over 2 to the power k for all n bigger than or equal to n k. So, to do that we start with so what is Cauchyness means? So, saying that the sequence f n is Cauchy saying that the sequence f n is Cauchy that means it is same as for every epsilon bigger than 0 there exists some stage n naught such that norm of f n minus f m is less than epsilon for every n and m greater than or equal to n naught. So, that is Cauchyness. So, to start our construction, take epsilon equal to 1 and find n 1 such that norm of f n minus f m is less than epsilon equal to 1 for every n and m bigger than n 1. So, in particular when m is equal to n 1 that will give me so that is f n minus f n 1 will be less than 1 for every n bigger than or equal to n 1. So, that is the first stage. Now, suppose n 1 less than n 2 less than n k have been constructed have been found. So, now use Cauchyness. So, find n k such that so how do I find n k? So, I will find n k such that norm of f n minus f m will be less than 1 over 2 to the power k for every n and m less than n and m bigger than or equal to find n k plus 1. So, the next one we have to construct n k plus 1. So, I will find n k plus 1 such that n k plus 1 is greater than n k and this property holds. So, then for m equal to n k plus 1 I will have f n minus f m k plus 1 will be less than 1 over 2 to the power. So, I can put it k plus 1 also that does not matter k plus 1 for every n bigger than or equal to n k plus 1. So, by induction there exists n 1 less than n 2 less than n k less than so on such that norm of f n minus f n k is less than 1 over 2 to the power k for every n bigger than or equal to n k. So, that is step one we do that. So, once that is done we go to step two and the step two is showing that this subsequence that we have constructed has the property that the sum of the l 1 norms of f n 1 plus the l 1 norm of 1 to infinity of this is finite. So, that the sums of the l 1 norms of f n 1 plus the l 1 norms of f n j plus 1 minus f n j j 1 to infinity that is all finite and that is that follows from the effect that just now we looked at this. So, that means norm of f n k plus 1 minus f n k is less than 1 over 2 to the power k. So, if I specialize n to be equal to n k plus 1 then I have this property for every k. So, that implies that if I look at the summation, so norm of f n 1 plus sigma norm of f n k plus 1 minus f n k k equal to 1 to infinity will be less than or equal to norm of f n 1 plus sigma 1 to infinity 1 over 2 to the power j and that is finite. So, that will prove that the required property step two namely sums of norms of these quantities f n 1 and this is finite. Once that we have that now I can apply my series form of the Lebesgue's dominated convergence theorem. So, as a step three as a next step I want to conclude that if I look at the corresponding functions f n 1 x plus summation j 1 to infinity f n j plus 1 x minus f n j x then this series is convergent almost everywhere. So, that is precisely follows from the series form of the Lebesgue's dominated convergence theorem. So, if I look at this series then I know that L 1 norms of this series is finite. So, that is precisely the perfect situation where the series form of the Lebesgue's dominated convergence theorem applies and says that these functions they must the series must converge almost everywhere. So, this sum must exist almost everywhere and if you denote the sum by f of x then f of x equal to this exists almost everywhere and it also says that that function is actually a integrable function and integral of f is equal to integral of f n 1 plus integral of the sums of the corresponding series. So, this integral of f is equal to integral of f n 1 plus this series. So, we have located a function f and now was only to claim that the difference f minus f n j goes to 0 that the f is the limit of that subsequence. So, let us look at. So, note that if we look at f minus f n j that is precisely equal to summation k equal to let me n j. So, that means k equal to n j plus 1 of f n k minus f n k minus 1. So, the difference between f and f n j is nothing but the tail of the series with the terms f n k minus f n k minus 1. So, norm of f minus f n j l 1 norm is going to be less than or equal to summation norm of f n k minus f n k minus 1 from k from the stage n j plus 1 onwards and that is the tail of the convergent series geometric series 1 over 2 to the power j. So, that goes to 0 as j goes to infinity. So, that will complete the proof that hence the f minus f n j goes to 0 as j goes to infinity. So, that will prove that the subsequence f n j is convergent. So, let us just go back to the proof once again. Basically, the proof requires one observation namely to show that a sequence a Cauchy sequence is convergent. It is enough to produce a subsequence which is convergent and that subsequence is produced in such a way that by using the Cauchyness we produce that subsequence with the property that the norms of the consecutive terms of that subsequence are less than 1 over 2 to the power j. Once that is done, we apply the series form of the lab-examinated convergence to conclude that the sum of f n 1 x plus summation f n j plus 1 x minus f n j that exists almost everywhere that is a integrable function and the integrals converge. Once we have that, it is obvious that norm of f minus f n j goes to 0 because once we subtract that the remaining thing is a tail of the series 1 over 2 to the power j and that must go to 0. So, with that we prove the Ries-Wischer theorem. So, today we analyzed two important things namely one that the space of Riemann integrable functions is inside the space of Lebesgue integrable functions, Lebesgue integrable on the interval a, b and the notion of Lebesgue integral extends the notion of Riemann integrable. So, that was one property and the second property we observed that the space of Lebesgue integrable functions on the interval a, b under the L 1 metric is a complete metric space. So, and we will continue this analysis tomorrow and we will show that the space of Riemann integrable functions which is inside the space of Lebesgue integrable functions sits inside as a dense subset and that will prove that L 1 f a, b is the completion of the space of Riemann integrable functions. Thank you.