 Welcome to lecture number 19 on measure and integration. In the previous lecture, we had started looking at the properties of integral for non-negative measurable functions. We had looked at the linearity property of the integral for non-negative measurable functions and then we said we will start looking at the limiting properties of functions which are non-negative measurable and integrals of them. So, today we will prove some important theorems. We will start with proving what is called monotone convergence theorem and then we will prove for 2's lemma and then go over to define integral for general functions. So, let us look at what is called monotone convergence theorem. Monotone convergence theorem says that let fn be a sequence of functions in class L plus. That means fn is a sequence of non-negative measurable functions increasing to a function f of x at every point. That means f of x for every x in x is limit n going to infinity of fn of x. So, we are given a sequence fn of non-negative measurable functions which is increasing and the limit is f of x. Then the claim is the function f belongs to L plus. This we have already observed and the additional properties that the integral of the limit f d mu is same as limit of the integrals of f and d mu. That means whenever a sequence fn of non-negative measurable functions increases to f then integral of the limit is equal to limit of the integrals. So, this is the first important theorem about convergence of sequences of non-negative measurable functions and their integrals. Let us prove this property. So, we are given fn is a sequence each fn belongs to L plus is a non-negative measurable function for every n bigger than or equal to 1. That means that implies there exists a sequence will denote it by s and j of functions n bigger than or equal to 1 such that s and j are non-negative measurable simple functions for every n and for every j and s and j increases to f of f. Let us fix the notation which one we are going to vary. Let us say that the upper one will be fixed. This is going to fn as j goes to infinity. For every n fix s and j is a sequence of non-negative simple measurable functions increasing to fn's and fn's they increase to f. So, we want to show, we have already shown but we will show it again that this implies f belongs to L plus is a non-negative measurable function and integral f d mu is equal to limit n going to infinity integral fn d mu. So, to prove this, we are going to use this sequence s n and construct a new sequence of non-negative simple measurable functions out of it. So, what we will do is the following. Let us write that for n is equal to 1 that s11 s12 s1 j s1 say j this converges to f1. So, the upper index is going to give you. So, s21 s22 s2 j this increases to f2 and in general we will have sn1 sn2 snj will increase to fn and so on and this increases to f. So, let us observe that as we go from left to right this is increasing. So, everywhere left to right it is increasing and down to up that also is increasing. So, every sequence if you look at that, so this is an array of non-negative simple measurable functions, each row is increasing to the function on the right side and this is increasing upwards. So, let us out of this, I am going to define. Let us look at the function. Let me define from this a function. Let us define gn to be the function which is maximum of snj j between 1 and n. In a sense what I am doing is in this picture look at the one. So, let us say here is s1n and here is s2n and here is snn. So, I look at this column. So, s1 we are looking at the column sn1 sn. So, let us look at this column and call that maximum of this to be gn. So, what is gn? So, gn is the, so let me write again. So, gn is the maximum, so define gn equal to maximum of sjn j 1 to n. So, let us observe that each gn is a maximum of non-negative simple measurable functions. So, each gn is a non-negative simple measurable function for every n and gn is increasing because at the next stage n plus 1. So, all this, this is going to be bigger the next stage. If we look at gn plus 1, so that is going to be s1n plus 1, s2n plus 1 and so on, sn plus 1, n plus 1, n and sn plus 1, n. So, all this, this one is going to be bigger than everything on the left hand side and these are, we are looking at the maximum. So, in the, in the maximum of this is going to be bigger than or equal to maximum of this because at each the right hand side function is bigger than the left hand side function. So, this is going to give us that is gn is a increasing sequence of functions. So, let us write, let g be equal to limit n going to infinity of gn. So, g is, so all these gn's are increasing and they are going to increase to some function g. So, what we are going to show is g is equal to f. So, that is what we are going to check. So, let g be equal to, so then clearly by definition, g is a non-negative simple measurable function because it is a limit of an increasing sequence of non-negative simple measurable functions. So, g belongs to l plus. So, let us observe also each gn is less than or equal to fn. Each gn is less than or equal to fn for every n. So, that is because, see gn is the maximum of this. So, and the maximum of this, each one of them is less than f1, is less than f2, is less than fn. So, the maximum of these gn's is less going to be less than or equal to this fn for every n and fn is increasing to f. So, that will imply that, so gn is less than or equal to fn for every fn, fn is less than or equal to f. So, implies that gn is less than or equal to fn and is less than or equal to f for every n. So hence, and gn is increasing to g, so that implies g is less than or equal to f. So, that is one observation that the function g is less than or equal to f and we claim that the other way round is also true. So, claim is that f is also less than or equal to g. So, let us note that for every j between 1 and n, if I look at sjn, gn is the maximum of this. So, this is less than or equal to gn for every n. So, this is less than or equal to gn for every n and j between less than this. So, if we fix and gn is less than or equal to, so and this is less than or equal to g, so sjn is less than or equal to gn is less than or equal to g for every j between 1 and n and for every n. So, let us now fix j and let n go to infinity. So, as n goes to infinity, what happens? So, this converges to fn. So, note that as n goes to infinity, sjn goes to fj. So, from this and this, these two observations, sjn is less than or equal to g for every, so if we fix j and let n go to infinity, then n is going to cross over j and sjn as n goes to infinity converges to f of j. So, this implies f of j is less than or equal to g for every j. So, we get, so implies that fj is less than or equal to g for every j and fj's are increasing. So, this implies that f is also less than or equal to g. So, we have already shown g is less than or equal to f and now we are saying f is less than or equal to g. So, this implies that f is equal to g. So, hence, so one observation from here is, hence that g belongs to L plus, so f belongs to L plus. So, we have once again proved that if fn's are increasing to a function f and fn's are non-negative measurable, then f is also non-negative measurable. Now, note that integral of f d mu is same as integral of g d mu because f is equal to g and this is equal to limit n going to infinity of integral g n d mu because g n's are non-negative simple measurable increasing to g. So, by definition this is so, but each g n is less than or equal to f, if you recall. So, each g n is less than or equal to f. So, integral of g n will be less than or equal to integral of f. So, limit of integrals of fn's will be less than or equal to integral f. So, this is less than or equal to integral f d mu or we can even introduce in between. So, g n is less than or equal to fn, so it is less than or equal to limit n going to infinity integral fn d mu which is less than or equal to integral f d mu. So, what does this imply? Integral f d mu is less than or equal to limit fn integral of fn d mu and that is less than or equal to f d mu. That implies that integral of f d mu is equal to limit n going to infinity integral fn d mu. So, that proves the theorem completely namely integral of f d mu is equal to limit n going to infinity integral of f n d mu. This is a construction which is quite useful. This is the kind of analysis one has to carry out. Let us go through the proof again so that we understand what we are doing. Each f j or f n is a measurable function. I can look at a sequence s 1 1, s 1 2, s 1 j, s 1 n which is going to increase to f 1. Similarly, copper fix is fixed at 2, so s 2 1, s 2 2, s 2 j, s 2 n that increases to f 2 and so on. Each row is increasing to the function on the right side and the functions f 1, f 2, f n are increasing to the function f. What we do? We look at the maximum of this column. What is this column? This column is the maximum of the functions s 1 n, s 2 n and s n n. Call this as g n. This function is called g n. The observation is each g n is a maximum of non-negative simple measurable function, so it is non-negative simple measurable. Each g n is less than or equal to f n because we are going to be up to this corner only. Each g n is less than or equal to because s 1 n is less than f 1, s 2 n is less than f 2, f 1 is less than f 2 and so on. This says g n will be less than or equal to f n and each f n is less than or equal to f. Each g n is less than or equal to f n is less than f. If we write the limit of g, g n to be, so we write the limit of this to be equal to g, then g is less than or equal to f by this simple construction. Also, for any fixed j, let us look at s n j. Let us look at s n j, where j is fixed and n is going to vary. As n varies, what happens to these functions? For every fixed j, this sequence of functions is going to be s n j is less than or equal to g n and g n is less than or equal to f. We let g is less than or equal to f. s n j is less than or equal to g n for j between 1 and n, so that will give us that f is also less than or equal to g. That will prove the theorem that limit of increasing sequence of non-negative measurable functions, if f n is a sequence of non-negative measurable functions increasing to f, then integral of f d mu is equal to limit of n going to infinity integral f n d mu. This is called monotone convergence theorem. Monotone, because we are looking at monotonically increasing sequences f n and convergence, because we are looking at the convergence of the integrals of integral f n d mu. This proves monotone convergence theorem. Let us remark, we have proved the theorem monotone convergence theorem for when f n is an increasing sequence. Naturally, the question arises, will the similar result hold if I have a decreasing sequence f n of non-negative measurable functions? That result unfortunately is not true. So, here is an example which says that if f n is a sequence of functions which are non-negative measurable and they decrease to a function f, then integral of f need not be equal to integral of f n d mu. The example is on the Lebesgue measurable space. Look at x to be the real line, the sigma algebra to be the sigma algebra of Lebesgue measurable sets and mu to be the Lebesgue measure. Look at the function f n, which is the indicator function of the interval n to infinity. The claim is, this is actually a non-negative simple measurable function, each f n and f n is decreasing and decreasing to the identity function identically equal to 0. That is quite obvious to see. So, what is f n? So, here is n and we are looking at the interval n to infinity. So, we are looking at this interval and we are looking at the indicator function of n to infinity. So, the function is 0 and it is 1 here. So, the function is this height is 1. So, this is the function f n. It is 0 here up to here and then it starts and goes. So, that is the function f n. So, if we take n plus 1, so this is n plus 1. So, n plus 1 will be 0 here, but f n is equal to 1 here. So, clearly f n of x is bigger than or equal to f n plus 1 of x for every x. So, f n is a sequence in L plus and f n is decreasing and the claim is f n decreases to f of x, which is identically equal to 0 for every x and that because if I take any point x on the real line, then I can find some integer n, say n naught, which is on the right side of it. Then, so for every x belonging to real line fix, I can find a point n naught, a positive integer n naught. Of course, it will depend on x, such that n naught of x is bigger than x. So, that will imply that the indicator function of n naught to infinity or n naught to infinity, let us even n to infinity at x is going to be equal to 0 for every n bigger than or equal to n naught and that is my f n of x. So, f n of x is equal to 0 for every n bigger than n. So, that means f n of x converges to f of x, which is equal to 0. So, f n is a sequence of non-negative measurable functions, which is decreasing to f identically 0. But if we look at the integral of each f n, so what is the integral of each f n? So, integral of f n d lambda, so this is integral of the indicator function n to infinity d lambda. So, that is equal to lambda of n to plus infinity and that is equal to plus infinity for every n. So, integral of f n is equal to plus infinity for every n and integral of f d lambda is equal to f is 0, so it is 0. So, this implies that integral f n d lambda does not converge to integral f d lambda, whenever f n is a decreasing sequence of function non-negative simple non-negative even simple functions we have given example here. So, for decreasing sequences this result does not hold, so that gives importance to the monotone convergence theorem. That means, whenever a sequence f n of non-negative measurable functions is increasing, then integral f is equal to limit integral f n d mu for decreasing this need not hold. So, this is what we have shown just now by an example. So, however one can prove not an equality, but some kind of an inequality for a sequence of non-negative measurable functions. And that is also an important result, so let us prove the result which is called Fatou's lemma. It says let f n be a sequence of non-negative measurable functions, then the integral of limit inferior of f n d mu is less than or equal to limit inferior of the integrals f n d mu. So, this is only an inequality and it need not be an equality, so what we are saying is if f n is a sequence of non-negative measurable functions, then it is always true that the integral of the limit inferior of f n is less than or equal to limit inferior of the integrals. So, let us give a proof of this theorem. So, to prove this theorem, so let us just first recall what is, so f n is a sequence of non-negative measurable functions, so each f n is a non-negative measurable function. And we want to look at limit inferior of f n as n goes to infinity. This is a function, so let us observe how this function is defined. Limit inferior of f n at a point x is defined as you take the infimum from some stage onwards, so m bigger than or equal to n of f n of x. So, look at the numbers f n of x, f m of x for m bigger than or equal to n, so I am looking at the tail of the sequence f n of x from m onwards. So, this number infimum will depend on m, so let me take the supremum of this overall m. So, first take the infimum from some stage onwards and then take the supremum of these infimums. So, let us observe that this infimum, let us put a bracket here. So, observe, so let me call it as phi m to be the infimum from the stage n onwards. So, infimum of m bigger than or equal to n of f n of x, phi n of x to be defined as the infimum from the stage n onwards of f m of x, because it is the infimum of a sequence of functions which are non-negative measurable. So clearly, so note, so observation is that each phi n is also a non-negative measurable function, so it is a non-negative measurable function that is 1. And secondly, we are taking the infimum from some stage n onwards. So, if we increase, so the claim is this phi n is increasing, this is the increasing sequence, because phi n from the infimum from the stage n onwards is going to be less than or equal to the infimum from the stage n plus 1 onwards, because we will have more numbers for which you are taking infimum. So, infimum can infimum, when you take infimum over more numbers then infimum can decrease. So, infimum from the stage n onwards and the infimum from the stage n plus 1 onwards, so that says that the infimum from the stage n plus 1 onwards will be bigger than or equal to the infimum, so increasing that is phi n plus 1 is bigger than or equal to phi n of x for every n. So, it is a increasing sequence of non-negative measurable functions and its limit is nothing but the limit inferior. So, it is increasing and so it is increasing and limit n going to infinity of phi n is equal to limit inferior of f n, n going to infinity. So, the stage is set perfect for an application of monotone convergence theorem, phi n is a sequence of non-negative measurable functions phi n's are increasing. So, by monotone convergence theorem, so we can apply implies by monotone convergence theorem monotone convergence theorem by monotone convergence theorem that integral of limit n going to infinity of phi n d mu is equal to limit integral phi n d mu n going to infinity. So, this side is nothing left hand side is nothing but integral of limit inferior n going to infinity of f n d mu, so that is equal to limit of integral phi n's. Now, let us look at what is phi n? Phi n is the infimum from the stage n onwards, so each phi n is less than or equal to f n. So, that is observation from here by the definition of phi n, we have that each phi n is less than or equal to f n. So, integral of phi n will be less than or equal to integral of f n, so it will be less than or equal to limit inferior of n going to infinity. So, what we are observing here is because each phi n is less than or equal to f n, so this implies, so this is what we are using here that each phi n is less than or equal to f n, then the n limit n integrals of phi n's are increasing, so it limit exists. So, limit n going to infinity integral phi n d mu is less than or equal to, however integral of phi n f n's may not exist, so we can say that it will be less than or equal to limit inferior of integral f n's d mu. So, this is what is being used in this conclusion and that proves the theorem, what is called the Fatou's lemma. So, we have two important results for non-negative, so we have got two important results for sequences of non-negative measurable functions, one of them is called the monodon convergence theorem, which says if f n is a increasing sequence of functions, non-negative measurable functions increasing to a function f, then integrals of f n's will converge to integral of f. So, keep in mind monodon convergence theorem is for a non-negative sequence of non-negative measurable functions, which is increasing to f and in case the sequence of non-negative measurable functions is not increasing, then we have got Fatou's lemma, which says that for any sequence f n of non-negative measurable functions, the integral of the limit inferior of f n's is going to be less than or equal to is always less than or equal to limit inferior of the integrals. So, these are the two important theorems which help us to relate the limit of the integrals with the integral of the limits and we will see applications of this in rest of our course. So, with this we conclude the section on the definition of integral for non-negative simple, non-negative measurable functions. So, let us just recall what we have done. We started with defining the integral of non-negative simple measurable functions, the functions which look like linear combinations of indicator functions sigma a i indicator functions of a i. For them we define the integral to be nothing but a i time summation of a i times mu of a i. We showed it is independent of the representation and we proved various properties of the integral for non-negative simple measurable functions. And then we looked at the class of non-negative measurable functions, since every non-negative measurable function is a limit of some sequence of non-negative simple measurable functions increasing to that function f. So, we defined integral of the non-negative measurable function to be nothing but the limit of the integrals of that sequence of non-negative simple measurable functions increasing to it. And we showed this integral is independent of the limit of the sequence S n you select which increases to f. And then we proved various properties including monotone convergence theorem and Fatou's lemma. And now let us look at how can we define the integral for a function which is not necessarily non-negative. So, for that we will do the following. We will start with so let us look at a function f. So, keep in mind we have got a measure space x s mu which is complete f is a function which is defined on x taking extended real valued. And we want to define so to define integral f d mu, this is what we want to know what it should look like. And of course, we would like this integral to have nice properties. So, we would like to have it to be a integral to be a linear operation. Now recall, so let us recall the function f can be written as f plus minus f minus. We can split it into two parts the positive part and the negative part of the function where f plus and f minus both are non-negative functions. And if f is measurable of course, with respect to the sigma algebra s that is true if and only if both f plus f minus are measurable. So f can be written, so this is the clue how we should go about. So f can be written as a difference of two non-negative measurable functions if f is measurable. And integral of non-negative functions is defined, so integral of f plus is defined, integral of f minus is defined. And if our integration is going to be linear it is all but necessary that our integral. So we should define integral f d mu whatever be the way we define it, it should have the property this is f plus d mu minus integral of f minus d mu. This is what we would like to have. This is defined, this is defined, now the question is the difference defined. So the difference will be defined if both of these quantities are finite numbers. So the left hand side it is defined if f plus d mu is finite and integral f minus d mu is finite and integral f minus d mu is finite. So that means whenever f is a measurable function so that integral f plus d mu is finite, integral f minus d mu is finite we can define its integral to be equal to integral equal to integral of f plus d mu minus integral f minus d mu. So with that we let us define what is the integral, what is called a integrable function. So a measurable function f defined on x taking extended real values is said to be mu integrable of course mu is the measure underlying space which is fixed. So it is said to be mu integrable if both integral of f plus d mu f plus is a non-negative measurable function f minus is a non-negative measurable function. So by our earlier discussion both these numbers f plus d mu and f minus d mu are defined. So if they are both finite in that case we say that the function f is integrable and its integral is defined as integral f plus minus integral f minus. So integral of f is defined as integral of the positive part of the function minus the integral of the negative part of the function. So whenever a function f is defined on x we say f is integrable if both the positive part and the negative part have finite integrals and in that case we define the integral of f. So written as integral f d mu to be integral of f plus minus integral of f minus. So we will denote by the symbol capital L lower 1 x s mu to be the class of all mu integrable functions. So in case it is clear what is x and what is mu and what is s we can sometimes simply write it as L 1 of x or simply L 1 of mu. So if we understand what is the underlying measure space, so the space of integrable functions either it will be explicitly written as L 1 x s mu or sometimes simply as L 1 of x or L 1 of mu. So this is the class of all integrable functions that means all functions f say that integral f plus is finite, integral f minus is finite and in that case integral of f is defined as integral f plus minus integral of f minus. So for all integrable functions f belonging to L 1 of x we have integral f d mu. We will now study the properties of this integral. So the first property is let us fix functions f and g which are integrable and f and g to be which are f and g to be real number a and g to be real numbers. So if f and g are measurable functions at mod f of x is less than g of x for almost all x and g belongs to L 1 then the claim is f is in L 1. So this is a very simple property we want to check namely that if, so let us f and g are measurable functions on x and we are given that mod f of x is less than or equal to g of x almost everywhere mu. So from here first claim is that if g is L 1 of x then that implies that f is in L 1 of x. So to prove that let us observe, so note we are given f x is less than or equal to g x almost everywhere x. So let us define n to be the set of all points x belonging to x where mod f of x is not less than or equal to g of x where this property is not true. Then we know that n belongs to the sigma algebra and mu of n is equal to 0, n of n is equal to 0. Now note because mod of f is a non-negative, so mod f belongs to L plus, g is non-negative measurable, so g belongs to L plus and on is almost everywhere. So mod f of x is less than or equal to g of x on n complement, so that is what is given to us. Thus implies integral mod f x d mu x which we can write as integral over n mod f x d mu x plus integral over n complement mod f x d mu. Now let us observe that the set n has got measure 0, so this part is 0 and on n complement mod f is less than or equal to g. So this is equal to 0 plus integral over n complement of mod f x d mu x and on n complement f is less than or equal to g. So this is less than or equal to integral over n complement of g of x d mu x and that is less than or equal to integral over the whole space g of x d mu x which is finite. So what we have shown is that in case mod f x is less than or equal to g x then we have shown that the integral of mod f is finite. So this says, so hence integral mod f d mu is finite and now let us note that f plus is always less than or equal to mod f and f minus is also less than or equal to mod f. For any function the positive part is less than or equal to mod f, the negative part also is less than or equal to mod f. So that implies that integral f plus d mu and integral f minus d mu both of them are less than or equal to integral mod f and which is finite. So we have shown that the integral of f plus and integral of f minus both are finite whenever mod of f is less than or equal to, so what we have shown is whenever mod, so this property is true, this implies that integral of f plus and integral of f minus both are finite. So that implies, so implies f belongs to l 1 and further let us calculate what is integral of mod f d mu, mod f if you recall is nothing but f plus plus f minus f plus plus f minus, so that means this is equal to integral d mu plus integral of f minus d mu. So integral of mod f is nothing but integral of f plus plus integral of f minus d mu and both of them are finite, so that we have already observed. So I wanted to check that integral of mod f is less than or equal to integral of integral g d mu which we have already actually checked. So we have already checked that integral of f plus, so which is less than integral of mod f is less than, so mod f is less than integral g implies, so we do not have to do this, so is less than or equal to integral g d mu, so that follows directly from that mod f now we have shown is integrable is integral is finite, so this is less than or equal to integral of g. So this proves the first property namely if f and g are measurable functions and mod f x is less than or equal to g x for almost all x and if g is integrable then f is integrable. So what we are saying is if a function f of x which is measurable is dominated by a function g which is integrable then the function f also becomes integrable. Let us look at next the property that if f and g are equal almost everywhere and f is integrable then g is integrable and the integrable of the two are equal and that property is something similar to that we have just now shown a similar analysis will work. So let us look at we have got two functions f and g and f of x is equal to g of x almost everywhere mu. So let us write this at n x belonging to x where f x is not equal to g x then by the given condition mu of n not equal to that is equal to 0 and f x equal to g x for every x belonging to n complement. So now let us look at so we are given that the function f is f belonging to l 1. So implies we want to show that g belongs to l 1 and that is because if f x is equal to g x almost everywhere then that implies that mod f x is also less than or equal to mod g x almost everywhere. The sets where they are not equal so wherever they are equal so that is mod x is equal to g x. So because on then complement that will happen so this less than so that implies integral of f x is equal to g x. So we should say that f x is equal to g x almost everywhere so that implies f x is equal to g x mod f x is equal to mod x. So and just now we showed that whenever f and g are equal almost everywhere integral mod f is equal to integral mod g d mu. So either of them finite implies other is finite where given this is finite so implies mod g d mu is finite. So implies once again that g is l 1 and g is l 1 so integral of g d mu is equal to integral g plus d mu minus integral g minus d mu. But f is equal to g almost everywhere we ask the reader to verify this that means f of f plus must be equal to g plus and f minus must be equal to g minus almost everywhere. So once again this integral is equal to minus integral of f plus d mu minus integral of f minus d mu which is nothing but equal to integral of f d mu. So integral g is equal to integral f whenever f and g are equal to g almost everywhere. So these are simple properties of integrable functions that we have looked at. So if f is equal to g almost everywhere and one of them is integrable then the other is integrable and the two integrals are equal. Next let us check the property of linearity. So if f is l 1 then we want to check that alpha f is also in l 1 and alpha f of d mu is equal to alpha times integral of f d mu. So to check that property let us just observe one thing that saying that just now we looked at this kind of analysis namely if f belongs to l 1 of x it is same as if and only if mod f belongs to l 1 of x. So once again let us do this because this we are going to use it again and again. See saying that f belongs to l 1 this implies integral of f plus d mu is finite and integral of f minus d mu is finite. So that implies because what is mod f? mod f is equal to f plus plus f minus so that implies integral f plus d mu plus integral f minus d mu is finite and this is equal to integral of mod f d mu. So f belonging to l 1 implies integral of mod f is finite. Conversely, so let us look at the converse part that if mod f integral is finite so let us so conversely. So let us this is given to us that integral of mod f d mu is finite. So once again let us observe that f plus is less than or equal to mod f and f minus is less than or equal to mod f. So that implies all are non negative measurable functions so implies integral of f plus d mu is less than integral of mod f d mu which is finite and integral of f minus d mu is less than integral mod f d mu which is finite. So that implies that f belongs to l 1. So saying that a function is integrable is equal into saying that mod f which is a non negative measurable function has got finite integral. So this property will be used again and let us see how that property is used in our proposition. Now so a belongs to real line and f is l 1 so look at mod of a f. So mod of a f is less than or equal to mod a mod f all are non negative functions. So integral of mod a f is less than or equal to integral of this which is mod a times integral of mod f d mu which is finite. So that implies that a f is integrable function and so now we can write not only it is integrable of a f d mu you can write as integral of a f plus d mu minus integral of a f minus d mu. Now possibilities either a is equal to 0 in this case a f will be 0 and everything is 0 so no problem. If a is positive then this part is same as a times f plus d mu minus a times integral f minus d mu if a is positive and so this a comes out because of the property for non negative measurable integral of non negative measurable functions. So this will be finite in case a is less than 0 this becomes a of minus negative part again that thing is similarly for a minus so that proves the property that if f is integrable and a is a real number then a f is integrable and a comes out. So we will continue looking at the properties of integrable functions in the next lecture we will show that this integral is a linear operation on the space of integrable functions and various other properties of this space of integrable functions and integral on it. So we will continue this study in the next lecture. Thank you.