 Welcome to lecture 23 on measure and integration. In the previous lecture, we had started looking at the space of Lebesgue integrable functions on the interval a, b. We had defined the notion of L 1 metric on it and we had proved that under the L 1 metric, L 1 a, b is a complete metric space. We will continue the study of this space L 1 a, b a bit more and today we will show that the space of continuous functions on the interval a, b is dense inside the space of integrable functions under the L 1 metric. Let us just recall the proof of the fact that the space L 1 a, b under the L 1 metric is complete. So, we defined the notion of L 1 metric as follows. For the functions f and g in L 1 of a, b, we defined the distance between f and g to be the L 1 norm of f minus g. So, this one indicates what is called the L 1 norm which we had defined last time. So, this is precisely equal to the L 1 norm or the distance between f and g is equal to integral of mod of f minus g d lambda over the interval a, b. We showed that if you identify functions almost everywhere, that means if you do not distinguish between functions f and g which are equal almost everywhere L 1 of a, b, then this becomes a metric and the space L 1 a, b is a complete metric space under this metric. I just want to go through the main steps of the proof once again to emphasize something important as follows. So, let us go through the steps again to show that this L 1 of a, b is complete. What we have to show is given a Cauchy sequence f n in L 1 of a, b. We have to show that there exists a function f in L 1 of a, b such that f n converges to f in the L 1 norm. So, to do that we said it is enough to show that to show that the Cauchy sequence f n converges in L 1 metric. It is enough to show that there is a subsequence of f n which is convergent in L 1. So, this is a general fact about metric space is namely in any metric space given a Cauchy sequence converges if and only if there is a subsequence of it which is convergent. So, this is the fact about metric space is we are going to use here. So, to prove that L 1 of a, b is complete given the Cauchy sequence f n we will try to construct a subsequence of f n which is convergent in L 1 norm. So, as a first step using the Cauchyness property of f n we construct a subsequence f n k of f n such that the L 1 norm of f n minus f n j is less than 1 over 2 to the power j for n bigger than or equal to n j. This was done basically the Cauchyness says that the distance between f n and f m goes to 0 as n and m go to infinity. So, after some stage the difference between f n and f m can be made as small as you want. So, by using induction we construct this subsequence say that f n minus f n j L 1 norm is less than 1 over 2 to the power j. So, what we wanted to note down that in this step one we have not used anywhere the fact that we the functions are defined over the interval a, b or real line we are just use a general fact about Cauchy sequences. In step two we said that look at the Cauchy sequence f n k that we have just constructed this has the property that the L 1 norm summation of the L 1 norms of f n 1 that is the first term plus the consecutive difference is the norm of f n j plus 1 minus f n j is a convergent series. This follows from step one because in the step one the difference between f n and f n j. So, f n n plus 1 j minus f n j is less than 1 over 2 to the power j. So, that clearly says that this sum of the norms will be less than 1 over 2 summation 1 over 2 to the power j which is finite. So, again this follows from step one and we are not using anywhere the fact that our sequence is our underlying space is the real line or the interval. Now, in step three we want you to conclude that the function f n 1 x plus summation of f n j plus 1 x minus f n j x exists almost everywhere and if you recall the proof of this was from the fact using the series form of the Lebesgue's dominated convergence theorem namely whenever you have given a series of L 1 functions and if the L 1 norms are finite then the functions series itself is convergent almost everywhere. So, again here we do not use the fact that we are over the real line. So, this step three also is valid and hence as a consequence of that theorem of Lebesgue's dominated convergence theorem in the series form we get that f is L 1 and the integral of f is equal to integral of sum of the corresponding integrals and as a consequence of this it follows that f n j converges to f in L 1. So, what we are saying is in all these steps we have not used anywhere the fact that we are working over the real number system. So, this proof carries over to any measure space complete measure space x as mu and that means we can replace the real line by any set x and the sigma algebra Lebesgue measurable sets by a sigma algebra of subsets of x and a measure mu such that x s mu is a complete measure space and we can define the space of mu integrable functions we can define L 1 of x the space of integrable functions and the notion of the L 1 norm makes sense for any function f on the measures on the space x if it is mu integrable we can define the L 1 norm of this. So, what we are saying is that the L 1 norm makes sense on make sense for any L 1 metric makes sense on any on the space of Lebesgue on the space of integrable functions on any measure space x as mu which is complete and as we have seen just now in the proof of the theorem we do not use anywhere the fact that we are over the real line we use general statements about metric spaces or we use the series form of the Lebesgue dominated converse theorem. So, as a result I am saying that the same proof which we have worked out that saying L 1 of a b is complete works very well for the space L 1 of x s mu where x s mu is any measure space. So, that gives us the Rie's Fisher theorem for a complete measure space x s mu saying that the space of integrable functions on a complete measure space under the L 1 metric is always complete and as a. So, that is one observation and now let us go over to the fact we want you to prove that L 1 a b which is complete is in fact the completion of the space R a b of Riemann integrable functions on a b. So, for to do that what we have already observed that R a b is a subset of L 1 of a b we had proved the theorem that any function which is Riemann integrable is also Lebesgue integrable and the Riemann integral is same as the Lebesgue integral. So, R a b is a subset of L 1 of a b L 1 of a b is complete to show that this is the completion of our a b we want to show that R a b is a dense subset of L 1 of a b in the L 1 metric. So, the denseness of R a b is to be proved in L 1 of a b in fact we will prove something much stronger remember that every continuous function on the interval a b is also Riemann integrable. So, the space C a b of continuous functions on the interval a b is a subset of the space of Riemann integrable functions and we will show that C a b itself is dense in L 1 of a b. That means for any function f in L 1 of a b and any number epsilon bigger than 0 we want to show that there exists a function g belonging to C a b a continuous function such that the norm of f minus g is less than epsilon. So, that will prove that C a b is complete in L 1 of a b. So, we will do it in steps step 1 is that given that function f in L 1 of a b which we want to approximate by a continuous function it is enough to prove the theorem for functions in L 1 a b such that f is bigger than or f is a non-negative. And that is because if f belongs to L 1 of a b then we know that f can be written as f plus the positive part minus the negative part of the function. And f belongs to L 1 of a b if and only if both f plus and f minus belong to L 1 of a b. So, in case so if non-negative functions in L 1 of a b can be approximated. So, if there is a function g 1 belonging to C a b and a function g 2 belonging to C a b continuous functions such that the norm of f plus minus the continuous function g 1 L 1 norm is less than epsilon and norm of f minus g 1 is also less than epsilon. Then this will imply that the norm of f minus g 1 minus g 2 L 1 which will be equal to norm of f plus minus f minus minus g 1 g 2 and that will be less than or equal to norm of f plus minus g 1 plus norm of using the triangle inequality property of the norm. So, f minus minus g 2 and that will be less than epsilon plus epsilon. So, we will have a so if you call this function as g. So, what we are saying is that if non-negative functions in L 1 can be approximated by continuous functions then any function f in L 1 can be approximated because f can be split as a difference of two non-negative functions in L 1. So, this is step one namely that it is enough to prove the theorem for functions which are integrable and which are non-negative. So, this is the first observation that showing that C a b is dense in L 1 of a b we can assume that the function f in L 1 a b is a non-negative function. So, there is a first simplification or the first step. The second step says that for a non-negative function f in L 1 a b. So, observation is that for a non-negative integrable function there exist a non-negative simple measurable functions in L 1 a b such that the norm of f minus s is less than epsilon. So, what we are saying is that if f is a non-negative integrable function then it can be approximated by a non-negative simple measurable function which is integrable integrable. So, let us prove this step how does we do that. So, we are given that f is non-negative and f belongs to L 1 of a b. Now, because f is non-negative and it is integrable, so f is non-negative measurable. So, f bigger than or equal to 0, f measurable implies there exists a sequence s n of non-negative simple measurable functions such that s n increases to f, but then s n is less than or equal to f and all are non-negative. So, that implies that s n also belongs to L 1 of a b. So, because s n is dominated by f they are non-negative functions, so that implies as we have seen earlier that s n also will belong to L 1 of a b and also because s n is increasing to f. So, integral of f d mu can be written as limit n going to infinity integral of s n d mu. That is by the definition of the integral for a non-negative measurable function the integral is the limit of the approximating sequence of non-negative simple measurable functions. But note that each s n integral of s n is less than or equal to integral of f. So, we can write that actually as absolute value of f n f minus s n d mu that will be equal to integral of f d mu minus integral of s n d mu because f minus s n is non-negative, so its absolute value is same as f minus s n. So, that integral is equal to this and that goes to 0. So, that means we have got a sequence of simple measurable functions non-negative simple measurable functions which are in L 1 and the, so this is the L 1 norm. So, integral of mod s integral of mod f minus s n goes to 0 that means the norm of f minus s n L 1 norm goes to 0. So, once it happens that means for any epsilon we can choose a n naught such that f minus s n naught will be less than epsilon. So, implies for every epsilon bigger than 0 there is a n naught such that norm of f minus s n naught will be less than epsilon. So, that proves the second step that close to a integrable function f which is non-negative there is a non-negative simple measurable function close to it close in the sense of L 1 norm. So, as step 3 so that means what that means in order to approximate f by a continuous function we can approximate it is enough to approximate non-negative simple measurable functions in L 1 by a continuous function because f can be approximated by a simple non-negative simple measurable function in L 1 and if non-negative simple measurable function can be approximated then will be through. So, what we have shown till now is that the non-negative simple measurable functions in L 1 are dense in L 1 that itself is of interest a result of interest is independent result of interest. That means for a integrable functions the non-negative simple functions are dense close to it and using positive negative part this will give you that in the space of L 1 of a b if you look at the simple integrable functions they are dense in L 1 norm. So, for us our theorem so it is enough to prove that for a simple non-negative simple integrable function L 1 of a b there exists a function g continuous function close to it. So, that is what we have to prove to prove the our theorem. So, let us we have got a simple S which is simple integrable non-negative function. That means what? That means this S will look like sigma a i indicator function of sets a i i equal to some 1 to n because it is simple non-negative. So, a i's will be bigger than or equal to 0 and this sets a i's are subsets of a b. Of course, they are Lebesgue measurable they are disjoint. So, a i intersection a j is empty and the union of a i's is equal to a b i equal to 1 to n. So, saying that S is a non-negative function which is non-negative simple measurable function which is in L 1. So, non-negative simple means it is of this type and obviously, this becomes integrable. So, this must be of this form. So, what we want to show is that to show there exists a continuous function g on a b such that norm of S minus g is less than epsilon. So, this is what we have to show. Now, S is a linear combination of indicator functions of a i. So, our claim is that enough to show enough to show. So, to prove that this claim it is enough to show that for every a inside a b of course, a Lebesgue measurable a Lebesgue measurable there exists a function g which is a C a b such that the L 1 norm of the indicator function of a minus g is less than epsilon saying this is enough because if this is true for every i. So, the reason is because if true then what we will do for every i will approximate the indicator function of a i. So, for every i find g i belonging to C a b such that the L 1 norm of the indicator function of a i minus g i is less than epsilon. So, we will do that for every i equal to 1 to up to n then this will imply that if I define g to be equal to sigma a i of g i a i times the function g i then this function is a continuous function because it is a finite linear combination of continuous function. So, this is a continuous function and the L 1 norm of sigma a i 1 to n of indicator function of a i minus this g and this is s. So, the norm of s minus g L 1 norm will be less than or equal to sigma mod a i norm of L 1 norm of indicator function of a i minus g i. So, that is by triangle inequality where g is a sum of sigma a i g i. So, I can say it is less than or equal to absolute value 1 to n of mod a i times this and each one of them. So, this is less than sigma i equal to 1 to n mod a i and this each one of them is less than epsilon. So, it is less than this which is a small quantity. So, we can modify this epsilon suitably. So, we will have that if so what we are saying is to prove that for every non-negative simple measurable function s there is a continuous function g close to it in the L 1 metric. It is enough to show that for every indicator function of a set A in L 1 in A B there is a function continuous function close to it. So, that is what we have to prove. So, let us do that. So, let A B a subset of A B and A Lebesgue measurable to show there is a continuous function close to it. So, for that to prove that let us make an observation here to show that there exists a continuous function g close to it. So, that means C of A B such that norm of indicator function of A minus g is less than epsilon. So, this is what we have to show. So, for that let us observe one thing. So, note which you can call it as a lemma which we have already proved while dealing with the Lebesgue measure, but let us recall the proof of this once again. Then look at the Lebesgue measure of the set A. A is a subset of A B. So, it is a finite quantity and the Lebesgue measure what is it equal to? The Lebesgue measure of the set A because A is Lebesgue measurable it is same as the Lebesgue outer measure of A by definition and that is equal to the infimum of given the set A cover it by disjoint intervals i j some finite number of them i j as are pair wise disjoint and look at the sums of these intervals lengths of these intervals i j 1 to infinity and take the infimum of all such things. So, lambda star of A which is same as the Lebesgue measure of A is nothing but the infimum of the sums of the lengths of those intervals i j's which cover A and we can assume that they are pair wise disjoint. So, that is what is being that is the definition and now this being finite because it is inside A B. So, implies for every epsilon bigger than 0 I can find intervals i n's there is this intervals i n belonging i n's intervals n equal to 1 2 and so on such that the set A is contained in union of i n's n equal to 1 to infinity and i n's are pair wise disjoint. So, i n intersection i m is equal to empty and it is a infimum. So, I can make lambda of lambda star of A which is same as lambda of A plus epsilon should be bigger than summation of lambda of A plus epsilon cannot be the infimum. So, that must be bigger than summation of lambda of i n's. So, that is why the definition that lambda star of A or lambda of A is the infimum of certain things. So, lambda of A plus epsilon cannot be the infimum. So, it must be bigger than some terms over which you are taking the infimum. So, this is true. So, that means and this is finite. So, that implies. So, here is the observation that this sigma lambda of i n's n equal to 1 to infinity is finite because lambda star of A which is lambda of A which is finite is inside A B. So, this is finite. So, we can choose. So, this is a series which is convergent. We can choose a stage say n naught such that the sum from n naught plus 1 to infinity of lambda i n is small. So, let us say it is less than say again epsilon. Now, define. So, then let us look at the set A minus union i n n equal to 1 to n naught. So, this is a subset of A is contained in. So, recall A is contained in union of i n's. So, I can say this is subset of union of n equal to 1 to infinity of i n's because A is contained in this minus union n equal to 1 to n naught of i n's. But that means this is equal to union of i n's n equal to n naught plus 1 to infinity. So, this is set is same as this. So, that implies by the sub additive property that lambda of A minus the union of the intervals i n from 1 to n naught will be less than summation of lambda i n n equal to n naught plus 1 to infinity and that we know is less than. So, here is the property 1. So, by 1 this is less than epsilon. So, the difference of the measure, Lebesgue measure of the difference of A minus this finite union of disjoint intervals is less than epsilon. Also, look at the sets union of 1 to n naught the intervals i n minus A that is contained in union of n equal to 1 to infinity i n because instead of n naught we will take it to infinity minus A and everything is finite. So, this implies all I have got finite Lebesgue measure. So, this implies that the Lebesgue measure of union of i n's n equal to 1 to n naught minus A. So, Lebesgue measure of this set will be less than or equal to the Lebesgue measure of union of i n's n equal to 1 to infinity minus Lebesgue measure of A which is equal to summation less than or equal to summation of lambda of i n n equal to 1 to infinity minus lambda of A and that we know is less than epsilon because that is how we constructed the sequence i n i n's cover A and the difference between the summation of lambda i n's minus lambda of A is less than epsilon. So, we have got that the Lebesgue measure of Lebesgue measure of A minus the finite union of intervals 1 to n naught is less than epsilon and Lebesgue measure of the finite union minus A is less than epsilon. So, that together they imply that the Lebesgue measure of A symmetric difference between this set which is a finite union of intervals disjoint intervals 1 to n naught is less than 2 epsilon. So, to make things look very nice what we could have done is given an epsilon we could have selected here when we got this tail of the series we could have made it less than epsilon by 2 and in the beginning also when we got when we took the outer measure being finite and covered by. So, we could have made that inequality less than epsilon by 2 in the starting itself. So, saying lambda of A is finite. So, there is a sequence of intervals covering say that lambda star of A plus epsilon by 2 instead of epsilon. Similarly, in the second one we also could have made epsilon by 2 then we have gotten this as epsilon by 2 and we have gotten this by epsilon by 2. So, we have gotten it epsilon that is only a cosmetic change in our proof, but the basic fact is what we are saying is. So, what we have shown is the following that if so this is the important thing that we have shown. So, we have shown that if A is contained in A to B given epsilon bigger than 0 there exists disjoint interval I 1 some I n naught such that this Lebesgue measure of the set A symmetric difference between this union of the intervals I n's 1 to n naught is less than epsilon. So, this is this we had proved earlier also for general measures. So, I have repeated the proof for the Lebesgue measure because it is good to revise things anyway. So, using this, but now let us look at what does this statement last statement mean that means this thing is nothing but integral of the indicator function of A minus the indicator function of sets A I n's which are pair wise disjoint 1 to n naught absolute value of this. So, this quantity is precisely equal to this. So, what we are saying is the indicator function of A delta B is absolute value of indicator function of A minus indicator function of B. So, this is a general fact. So, I am using that here. So, Lebesgue measure of a set is the integral of the indicator function and the indicator function of the symmetric difference is the indicator function of the difference of absolute value of the difference. So, that is less than epsilon. So, as a consequence what we are saying is that given a set A inside the interval A B, we can find finite number of disjoint intervals I n say that this property is true. But now note that these are disjoint intervals. So, this is equal to the indicator function of A minus the summation indicator functions of I n's n equal to 1 naught and naught B lambda. So, that is less than epsilon. So, what we have proved that if you look at the indicator function of a set A inside the interval A B, then there are disjoint intervals I n such that the indicator function of A minus the sums of the indicator function of these intervals I n's will be less than epsilon. Now, what we want you to approximate? So, this is same as the L 1 norm. So, we have got the L 1 norm of the indicator function of A minus the function which is a sum of the indicator functions of I n's 1 to n naught that is less than epsilon L 1 norm. Now, what our aim was to approximate the indicator function of A by a continuous function and we are saying that the indicator function of A is close to sum of indicator functions of intervals. So, what does that mean? That means if I can approximate each one of these functions which are indicator functions of intervals inside the interval A B and a finite number of them by a continuous function then I am through. So, the next step is to show that. So, let us write it as further. We claim that if I is a interval, if say I is a interval inside the interval A B then there exists a continuous function G C A B such that with the property that norm of the indicator function of the interval I minus the continuous function is less than epsilon. So, once we are able to prove this fact for each interval I n will have a continuous function take the sums of those. So, they will approximate the sum of the indicator functions of the intervals and that in turn approximates the indicator function of A and our proof will be complete. So, what we want to show is that given a interval inside. So, given the interval I, so here is A and here is B and we are given a interval inside it. So, I just want to draw a picture and show what is the proof going to be. So, here is the interval I inside it. Let us say it is C to D. So, the indicator function of A indicator function. So, we have got the indicator function of C to D. So, let us look at. So, this is going to be the indicator function of. So, this height is 1. So, let us just draw this is 1. So, the indicator function of the interval C to D looks like it is 0 here in A to C and C to D it is going to be 1 and D to B it is going to be 0 again here. It does not matter what are the values at the point C and D. Now, we want to approximate this by a continuous function. So, it is obvious what we should be doing to make this function continuous and such that the area below the graph of this function does not exceed too much. So, let us take a point here which is C minus 1 by n for any n and let us take a point here which is D plus 1 by n. So, take this point and now what we do is we. So, take the function. So, I am going to define a function g n. What is the function g n? It is 0 in A to C minus 1 over n in this portion it is 0 and in the portion between C minus 1 over n to C it is going to be the line joining. So, that is the line segment. So, I am describing the graph of this. So, the line segment and then it is 1 in the interval C to D. So, it is 1 in the interval C to D and then again from D where the discontinuity is coming I join it by the point D plus 1 over n. So, it is again the line joining that between D plus 1 over D to D plus 1 over n and 0 remaining. So, what I am saying is if you are given the indicator function of a sub interval of A B then I can always make it continuous. I can approximate by continuous function. So, what will be the extra thing we will be adding? We will be adding the areas of these two rectangles. So, if I define this as a continuous function g n then what is the l 1 difference? So, the integral of mod of the indicator function of the interval I which is C to D minus this continuous function g n d lambda will be equal to the areas of these two rectangles. So, height is 1. So, it is 2 by n because each triangle because this length is 1 by n, this height is 1. So, half base into height so it is n by 2 actually. So, multiply by 1 by 2. So, that is 1 by n and that goes to 0 as n goes to infinity. So, that will prove the fact that close to the indicator function of… So, that will prove the fact that the close to indicator function, so close to the indicator function of an interval there is a continuous function and hence close to the indicator function of any set inside the interval A B there is a continuous function And that will end, finite linear combinations of the indicator functions are the simple functions. So, that will prove the fact that for a non-negative simple function, there is a continuous function close to it. So, step 4, which was that if, is a non-negative simple function of this form, then close to it. So, this is a linear combination. So, indicator function a i chi a i, I am just repeating the last step again. So, because each one of them can be approximated, if each one of them can be indicator function of a intervals can be approximated, then the simple function can be approximated. So, the problem came to approximating indicator functions of sets inside the interval a b and for that, we said we will use a lemma, which says that given epsilon bigger than 0, there exists a set f inside a b, which is a finite disjoint union of intervals, say that the Lebesgue measure of asymmetric difference f is less than epsilon. So, using that, we will get that indicator function of a minus indicator function of this set f, which is a finite disjoint union of intervals. And so, problem produces to approximating indicator functions of intervals inside the given interval a b and for that, we just extrapolate by piecewise linear functions and get the required thing. So, indicator function of a interval, there is a continuous function close to it. So, for that, we just now said look at the graph of, so here is a to b. So, look at some point close to it, c minus 1 over n or c minus 2 by n and take this linear, piecewise linear function, which is a continuous function and that will approximate in L 1 norm the given function, the indicator function of a interval. So, that will prove the fact namely that close to L 1 function, there is a continuous function. I just want to discuss the proof of this theorem. In effect, we have almost used all the results on measures and on integration that we have proved till now. So, and this is the technique, which we use to prove things about integrable functions. So, what we wanted to show was that given a function f, which is integrable on the interval a b, there is a continuous function on the interval a b such that the L 1 norm of the difference is small. That means, every integrable function on the interval a b can be approximated by a continuous function. So, what was the first step of our first step in our proof was that since every function f can be written as the positive part minus the negative part and f integrable implies if and only if the positive part and the negative part are integrable and we want to approximate f by a continuous function, which is the difference of two non-negative integrable functions f plus and f minus. If we can approximate each one of them separately, then we can combine them to approximate the function f. So, the first step was namely we can assume without any loss of generality that our integrable function is non-negative that is one. So, once so the next step is so if we take the function f to be non-negative and integrable, how is the integral of non-negative functions defined? They are defined by looking at limits of increasing sequences of simple measurable functions and taking their integrals. So, we go back to the definition of integral of a non-negative function. So, f non-negative integrable implies f is non-negative measurable. So, as a consequence or the definition there exists a sequence of non-negative simple measurable functions such that call that S n say that the sequence S n of simple non-negative simple measurable functions increases to the function f, but if S n is increasing to f that means S n is less than or equal to f and f is non-negative. So, that implies each S n is non-negative and integrable. So, S n is a sequence of simple non-negative integrable functions on the interval a b and the integral of S n increases to integral of f that means that is equivalent to saying that S n converges to f in L 1 norm. So, given f a non-negative integrable function we have a sequence of non-negative simple integrable functions in converging to it in the L 1 norm. So, this is a fact that simple functions in L 1 a b are dense, but we want to go a step further. So, look at a simple function now. So, to approximate f which is non-negative we have got a sequence of simple functions converging to it in L 1 that means close to f which is non-negative integrable there is a simple integrable function. So, if you can integrate if we can approximate simple non-negative simple integrable functions by a continuous function then we are through, but a non-negative simple function is a finite linear combination non-negative linear combination of indicator functions of sets. So, and if we can approximate each indicator function by a continuous function then the corresponding linear combination of those continuous functions will approximate the simple function. So, the next step is that to show that close simple integrable functions can be approximated by a continuous function it is enough to show that the indicator function of a set a inside a b is can be approximated by a continuous function. So, it now comes down to saying that look at a set a which is Lebesgue measurable inside the interval a b and we want to approximate the indicator function of this set by a continuous function. And here comes the property of the Lebesgue measure that the Lebesgue measure of a set a inside a b that means it is a finite set of finite Lebesgue measure. We can approximate this set by finite disjoint union of intervals and what does that approximation mean? It means that given a set a inside the interval a b which is Lebesgue measurable there exists a set which is a finite disjoint union of intervals such that the Lebesgue measure of the set a and the symmetric difference of this finite disjoint union is small is less than say epsilon. And but saying that the Lebesgue measure of the symmetric difference a with a finite disjoint union is small is same as saying that the L 1 norm of the indicator function minus the difference between the L 1 norm of the indicator function and the linear combination of the indicator functions of those disjoint intervals is small. And we want to approximate the indicator function by a continuous function. So, and close to it is a finite linear combination of indicator functions of intervals. So, the problem reduces to approximating the indicator function of an interval inside the given interval a b by a continuous function and that is achieved by just piece wise making the indicator function piece wise linear. So, this is the step in effect we have used almost all the theory in proving this theorem. So, this theorem proves that C a b is dense. So, that proves the fact that C a b is dense in L 1 of a b and C a b is a subset of R a b. So, that will prove that R a b is dense in L 1 of a b and hence as the result we get that L 1 of a b is the completion of the space of R a b of of course, of C a b. So, that proves the theorem. So, very important result that L 1 of any interval a a b is complete and as we observed L 1 of any subset also will be complete if you look at the proof and it is a completion of R a b. Let us look at some more properties of these functions. So, let us look at let us look at a function f in L 1 of R. So, f is a integrable function for real numbers h and k let us define f h f lower h of x namely equal to f of x plus h. So, the value of this new function f lower h at x is the value of f at the translated point x plus h. So, this we call as a translation of the function f and similarly let us define the function phi which is defined as phi of x to be f of k times x plus h for any x that means you multiply the number x by k and add h to it translate and then take the value of this. So, claim is both these functions f h and phi are again integrable and the integral of this function phi is absolute value of k times the integral of f and the integral of the translated function f h is same as the integral of the original function f. So, that means the space L 1 if you make a translation or a magnification then these are again leave the functions in the space L 1 and with these properties. So, this can be easily proved on the lines that we have proved just now. .. So, let us try to prove that integral of f h x d lambda x is equal to integral of f x d lambda x. So, we want to prove for every f in L 1. So, again we will be using that simple function technique. So, we want to prove for every f in L 1. So, note we can assume. So, step 1 show true for f non-negative. So, it is true for f non-negative because once it is true for f non-negative I can look at the positive part and the negative part. So, first show it is for and how do you show it is true for non-negative? So, step 2 show true for a simple function s belonging to L 1, s 1, s non-negative simple because every function can be approximated by non-negative simple functions and non-negative simple functions are indicator functions of sets. So, step 3 show that show for f equal to the indicator function show for indicator function and what is that? That means we want to show that lambda. So, indicator function of a x plus h d x d lambda x is equal to integral of indicator function of a d lambda. But that is same as saying showing that lambda of a plus h or a minus h does not matter is equal to lambda of a for every h and that is the property of the Lebesgue measure that it is translation invariant. So, what we are saying is this property is true for indicator functions. So, it will be true for non-negative simple functions. So, by taking limits it will be true for non-negative integrable functions and then by positive and negative part it will be true for all functions in L 1. And a similar result will work for second inequality, second identity namely if I multiply so f of k x let us just look at d lambda x is equal to mod k times mod f of mod k times f of x d lambda x. So, that again by the same technique let us look at what happens when it is a indicator function. So, lambda of multiplication k times a set E what is it equal to and we look at this by outer measures one can show that this is same as mod k times lambda of E. So, we show it for all sets E which are Lebesgue measurable then this is equal to the indicator function and then finally, linear combination of indicator functions and so on. So, as the reader to verify this by the simple function technique. So, these are properties of Lebesgue integrable functions. So, what we have done is that we specialize the space of integrable functions on the real line and reduce some nice properties and one of the properties was that the space of Riemann integrable functions is dense in the space of Lebesgue integrable functions. And so in one sense this is very nice and so this completes the process of extension of measures and defining integrals with respect to measures and their properties. In the next few lectures, we will start looking at how does one construct what are called product measure spaces and how does one integrate on product measure spaces. So, this is an important part of measure theory that means measure and integrations on product spaces. We will start looking at in the next lecture. Thank you.