 Welcome to lecture 15 on measure and integration. In the previous lecture, we had defined what is called a measurable function on a measurable space X s. Then, we had looked at some equivalent ways of looking at measurable functions. We looked at examples of what are called simple measurable functions. They are nothing but finite linear combinations of indicator functions of subsets of Zx. The simple measurable functions are sort of the core of the class of all measurable functions. We showed that some of simple measurable functions, product of measurable, simple measurable functions and maximum and minimum of simple measurable functions are simple measurable functions. And today we will start looking at some general properties of measurable functions f and then we will characterize measurable functions in terms of simple measurable functions. So, today's talk is going to be mainly on properties of measurable functions. . So, let us recall for simple function, we define what is called the positive part and the negative part of a function. So, this can be defined for any function. Let f be a function defined on x, taking extended real valued functions, extended real values r star. Then we define what is called the positive part of the function. So, that is denoted by f plus of x. This is again a function on the space x and it is defined as f plus of x is equal to f of x, if f of x is bigger than or equal to 0 and it is defined as 0, if f of x is less than 0. Essentially, what we are saying is, look at the graph of the function f of x, as long as the graph remains above the x axis, keep the function as it is. So, f of x is kept as it is, if f of x is bigger than or equal to 0 and as soon as the graph goes below the x axis, we define its value to be equal to 0. So, this is called the positive part of the function. So, note that for any function f, the positive part of the function is again a function on the space x, but it takes only non-negative values. And similarly, we can define the negative part of the function to be a function on x again such that it is denoted by f minus of x. So, f minus of x is equal to minus of f of x. Keep in mind, we are putting a negative sign here, if f of x is less than or equal to 0. So, that means as soon as the function goes, is on x axis or below the x axis, we reflect it against x axis and put its value as minus of f of x. So, f of x with a negative sign, so if f of x is negative, so this will always be a non-negative quantity. And it is function is defined as 0, if f of x is bigger than 0. Essentially, once again it is looking at the graph of the function as long as the graph of the function is above the x axis, we put it value equal to 0 and it is minus of f of x, if f of x is less than or equal to 0. So, these are called the positive part and the negative part of the function. And it is quite obvious from the definition that f can be written as f plus minus of f minus. So, and we want to prove that if f, extended real value is a measurable function, we want to show that in that case, the positive part and negative part are both measurable functions. And conversely, if positive part and negative part are measurable, then the function f is measurable. So, namely saying a function f on x taking extended real valued is measurable if and only if both f plus and f minus are measurable functions. And the proof is rather simple. So, let us assume first f is measurable, then for any point c in R, look at the inverse image of the closed interval c to plus infinity. So, f plus inverse image closed interval c to infinity, so that we know is because f of x, f plus of x is equal to f of x, if f of x is bigger than 0. So, if this value c is bigger than 0, then f plus of x will be equal to f of x. So, this inverse image of the closed interval c to infinity under f plus is nothing but the inverse image of f of the interval c to infinity, if c is bigger than or equal to 0, because in that case, f of x is always going to be positive. And it is equal to the inverse image of the interval 0 to infinity if c is negative, because then we do not want to look at the remaining part. So, in either case, because f is measurable, both these sets are in the sigma algebra. So, f plus inverse of c to infinity belongs to the sigma algebra, if f is measurable. So, that proves that f plus is measurable. A similar argument will prove that f minus is also measurable. So, essentially what we are saying is the f plus inverse, that is the inverse image of the interval c to infinity under f plus can be represented as inverse image of an interval of f under f of some interval. And both are f measurable implies that both whenever that f inverse of interval is a set in the sigma algebra. So, we have shown that if f is measurable, then f plus and f inverse are f minus are both measurable. Let us prove the converse part. Suppose both f plus and f minus are measurable and let us take a point c belonging to R, then we should show that the inverse image f inverse of c to infinity belongs to the sigma algebra s for every c in R. So, let us fix a c and look at this. And we have to interpret this in terms of inverse images of some intervals in terms of f plus and f minus. So, now let us observe that f, the inverse image of the interval c to infinity can be decomposed into two parts. Let me f inverse of c infinity intersection 0 to infinity. So look at the intersection of c to infinity with 0 to infinity and the intersection of this interval c to infinity minus infinity to 0. So, this interval c to infinity is decomposed into two parts. It is intersection with minus infinity to open interval 0 and it is intersection with close interval from 0 to infinity. So, f inverse of c infinity c to infinity is nothing but the inverse image of the interval c infinity intersection with 0 to the part of the interval which lies in the positive part and inverse image of the part of the interval which lies in the negative part. That means in the first part we are looking at whenever the function is in 0 to infinity that means function is non negative. So, the first inverse image is nothing but the inverse image of this same interval under f plus. And similarly, the second one is the inverse image of the interval c infinity intersection with minus infinity to 0 with respect to f inverse. So, the part of the function which lies in 0 to infinity is written as the inverse image under f plus of an interval and the other part is written as the inverse image under f minus. Since both f and f minus f plus and f minus are measurable functions. So, these two sets f plus inverse of that interval and f minus minus inverse image under that interval. These both are sets in the sigma algebra s. So, their union is also in the sigma algebra s. So, that means f inverse of c to infinity is in the sigma algebra for every c belonging to r and that proves that this is a f is a measurable function. So, we have shown that a function f is measurable if and only if its positive part and negative part both are measurable functions. .. So, next let us look at some more properties of measurable functions. We want to give a characterization of measurable functions in terms of simple functions. So, we start with a non-negative function. Let f be a non-negative function on defined on x. So, taking values in 0 to infinity, we want to show that this function f is measurable if and only if there exists a sequence as an of simple monotonically increasing functions. Again, I infect non-negative. We can also say they are non-negative, simple, measurable, non-negative sequence of functions which are monotonically increasing to the function f. That means if a function f is non-negative and measurable, this can happen if and only if s can be written as a limit of simple functions which are non-negative and the sequence s n is increasing. .. So, let us prove this fact. So, let us start with, so let f is from x to r and we are given a sequence such that there exists a sequence s n of simple non-negative functions s n increasing to f. That means f of x is limit of n going to infinity of s n of x for every x belonging to x and these s n's are monotonically increasing. So, this is written as this. So, to show non-negative simple functions and each simple of course, measurable. So, to show that f is measurable. Now, so let us let c belong to r and let us look at the inverse image of the interval c to plus infinity. So, what is that? That is all x belonging to x say that f of x is bigger than c. So, or it will be easier if we look at the other set. So, let me instead of this, let me look at the set which is f inverse of minus infinity to c. You will soon see why I am taking this instead of the earlier set why I am taking this because this proof becomes slightly simpler. So, what is this? This is all x belonging to x such that f of x is less than or equal to c and what is f of x? Recall, so just now we said f of x is limit n going to infinity of s n of x and s n x is increasing. So, if the limit of s n x which is f of x is less than or equal to c that means each s n has to be less than or equal to c because even if 1 goes above c then the limit has to be bigger than c. So, this condition implies that s n x is less than or equal to c for every n bigger than or equal to 1. So, implies that the set f inverse minus infinity to c. If x is such that f of x is less than or equal to c that implies s n of x is less than or equal to c. So, that means this is contained in this is for every n. So, this is contained in intersection n equal to 1 to infinity of all x such that s n of x is less than or equal to c and if conversely if so note if s n of x is less than or equal to c for every n then this automatically implies that f of x because f of x is a limit that also is less than or equal to c. So, hence what we are saying is this is actually an equality. So, hence f inverse of minus infinity to c can be written as intersection n equal to 1 to infinity of x such that s n of x is less than or equal to c and that is same as n equal to 1 to infinity. So, this is s n inverse of the interval minus infinity to c and s n is being simple measurable functions. Each one of them is an element each one of these sets is an element in the sigma algebra s. So, implies that f inverse of minus infinity to c belongs to the sigma algebra s. So, that means hence f is measurable. So, what we have shown is if there exists a so what we have shown is if f can be written as a limit of an increasing sequence of simple measurable non-negative simple measurable functions then f is measurable. Let us look at the converse part of it which is going to be slightly not so obvious. So, conversely let f from x to is a non-negative function. So, 0 to plus infinity be measurable. So, we want to show to construct a sequence s n of functions say that s n each s n is simple non-negative s n is increasing to f. So, this is all we want to do. So, this is this construction is intuitively very obvious, but needs to be explained. So, let us look at in a picture. So, here let us draw a picture of the function. So, let us draw. So, this is x axis and this is the values the real number they are all values are non-negative. So, the graph is going to be above the x axis. So, this is going to be the graph of the function f of x and the range of the function is a subset of 0 to infinity. So, what we are going to do is we are going to partition the range first into smaller intervals. So, to do that let me draw a picture on a slightly bigger piece so that we are able to look at. So, this is the graph of the function f of x and this is x. So, let us mark off, let us put a point n here. So, this is the point 0. So, from 0 to n and then from n to onwards. So, we are dividing the range of the function. The range is a subset of 0 to infinity. So, we have divided the range. So, I am writing 0 to plus infinity as equal to 0 to n, open and union to the close n to n to plus infinity. So, I have divided the range into two parts and now what we do is the portion 0 to n. So, this portion from 0 to n I am going to divide for every n into smaller pieces of length 2 to the power n. So, cut it into pieces so that the length of each piece is nothing but 1 over 2 to the power n. So, this is the length of each piece. So, let us call the interval. So, this is my general interval. So, the upper point here we will denote it by say k minus 1 by 2 to the power n and the lower part as k by 2 to the power n. So, I am going to write this is equal to and we are going to look at open at the bottom, close at the bottom, open at the top. So, I am going to write as a union of intervals of the type k minus 1 by 2 to the power n comma k by 2 to the power n union that other part we will leave it as it is n to plus infinity. This union how many such small pieces will be there total length from 0 to n each has got sub interval has got length 2 to the power n. So, this starts with k equal to 1 and goes up to n times 2 to the power n. So, this is what we do. Now, for every n I want a function s n of x. So, for any x look at the value of f of x. So, let us look at those values. So, for every for x belonging to x, f of x belongs to 0 to infinity. So, that means and so this union is a disjoint union. So, let us write f of x belongs to that partition that we have designed. So, k equal to 1 to n times 2 to the power n of the interval k minus 1 by 2 to the power n to k by 2 to the power n open and this union n to plus infinity. Now, if f of x belongs to this and this is a disjoint union, f of x will belong to only one of them. So, implies either f of x belongs to n to plus infinity or f of x will belong to one of the sub intervals. Let us call it as k minus 1 to the power n over k by 2 to the power n. So, these are the two possibilities and now we want to define a function s n. So, define s n of x we want to define. So, what should be the value of it? See, if the value is bigger than n, if f of x is bigger than n, then let us keep the value to be equal to n if this happens, if f of x belongs to n to plus infinity. Let us define it. So, if f of x is inside this k minus 1 by 2 to the power n to k by 2 to the power n, let us take the lower value. So, let us define s n x to be equal to 2 to the power n if f of x belongs to k minus 1 by 2 to the power n to k by 2 to the power n. So, this is how we are going to define the function s n. So, let us look at from the picture point of view. So, what I am going to do is, because this is the range of the function, so f of x is going to be somewhere. So, if f of x is above n, above this line, above this line n, say for example, that is happening here, this is the function, then for all these points, my s n is going to be, s n is going to be, so it is going to be, s n is going to be this constant. As soon as the function crosses n, the value of s n is going to be that part. And what is the other possibility? That f of x lies in one of these intervals. So, let us say f of x is here. So, this is my f of x. Then what do I want? My value of s n should be such that the difference between s n and f should be small and this is the smallness we have created. So, I will define my s n to be the lower value. So, this is going to be my s n. So, as soon as the function is inside this strip, wherever the function is inside the strip, the value is this lower value. If it crosses n, then that is the constant value n. So, the function s n x is defined to be equal to this. So, this is s n of x is n. If f of x is bigger than or equal to n and if it is strictly less than n, then it will belong to one of those small pieces and that is defined as the lower value of that interval as the value of the function s n of x. .. So, our claim is that this is the required sequence. So, claim this is the required sequence. Let us first observe what is s n? s n is defined as this. So, I can write s n of x to be equal to n times on this interval. So, it is n times the indicator function of the interval n 2 plus infinity of x. If the point belongs here, this number will be 1. So, the value will be n plus. If it lies in that interval k minus 1 by 2 to the power n to k by 2 to the power n, the value is this. So, it is this value times the indicator function of that set. So, it is summation k equal to 1 to 2 to the power n times 2 to the power n of the indicator function of. So, what is that set? That is nothing but the f inverse image of k minus 1 by 2 to the power n and k by 2 to the power n open of x. So, from the picture from the earlier formula, we get that my function is this function. So, it is clear from this. It is so implies s n is non-negative simple measurable. Why it is non-negative? Because n is the value taken or either n or the sorry in multiplied this by k minus 1 by 2 to the power n. That value we forgot to multiply. So, either it takes the value k minus 1 by 2 to the power n or n all are non-negative numbers. So, this is a non-negative function and it is a linear combination, finite linear combination of characteristic functions of sorry this is not n to infinity. So, this is f inverse. So, let me just be careful. So, it is n if f of x belongs to this. That means x belongs to f inverse of n to plus infinity and f of x belongs to this interval means x belongs to f inverse of this interval. So, that proves that s n is a. Now, why are these sets measurable? It is a linear combinations of indicator functions of sets and this set is measurable because f is measurable. This set is measurable because f is measurable. So, f is a measurability of f implies that inverse images of intervals are elements in the sigma algebra. So, these are elements of the sigma algebra and hence s n of x is a linear combination of indicator functions of sets in the sigma algebra s. So, it is a simple measurable function. Now, let us prove that this is s n is. So, claim we want to show. So, s n is increasing. So, let us fix x belonging to x to show s n plus 1 of x is bigger than or equal to s n of x for every n. So, let us look at that. So, why is that true? Now, look what is the value of. So, either f of x is bigger than or equal to n plus 1 is let us look at slightly differently. So, I want to prove that s n is increasing. So, to prove the increasing part, let us fix n. So, we want to look at what is s n plus 1 of x. So, that is going to be dependent upon whether. So, either it is n plus 1 or it is going to be some k minus 1 over 2 to the power n plus 1 for some k between 1 and n plus 1 times 1 over 2 to the power n plus 1. So, what we are saying is s n plus 1 x either it will be bigger it will be equal to n plus 1 that will be the case if f of x is bigger than or equal to n plus 1 or it will be equal to 1 of lower values of 1 of the sub intervals at the n plus 1th stage. In the n plus 1th stage, we will be dividing the interval into 2 n plus 1 parts. Let me draw this picture slightly here to understand what is happening. So, here is let us say j by j minus 1 by 2 to the power n and that is j to the power 2 to the power n. At the nth stage, f of x is somewhere in between. Now, at the next stage what we are doing we are going to divide this into 2 equal intervals. So, that this part is this part is something divided by 2 to the power n plus 1 and this part is something divided by 2 to the power. So, what will be this? So, this will be 2 times j minus 1 divided by 2 to the power n plus 1 and this part would be 2 j divided by 2 to the power n plus 1 and that is the middle line in between. Now, my s n plus 1 x depending on f x, if f of x is here, if this is f of x then s n plus 1 is this is the value of s n plus 1. So, this is the value of s n plus 1 and if f is here then this is the value of s n plus 1. So, either f of x will be here or it will be here. So, if f of x is here it lies in that interval of length through to the power n plus 1 the value is a lower end point. So, value of s n is here, but in that case what is the value of this is the value of s n plus 1. So, this is the value of s n plus 1 x and what is the value of s n? That is always going to be equal to this value and if f of x is this then this is the value of s n plus 1 x either will be here or it will be here and s n plus 1 of x will always be here. So, this value is less than or equal to this value. So, that analysis let us just write. So, it is equal to this. So, in either case so s n plus 1 of x either it will be n plus 1. So, either it will be n plus 1 or it will be which is bigger than n which is equal to s n of x. If not if it is below then the value is if it is in one of those intervals then the value is going to be some k minus 1 over 2 to the power n plus 1 which is always going to be equal to 2 to the power n the lower value here. So, let us it is difficult to write those symbols. So, that lower value this k minus 1. So, this is k minus 1. So, that is less than or equal to j minus for some j and that will be equal to j minus 1 over 2 to the power n which will be equal to s n of x. Geometrically it is quite clear what is happening. So, either if f of x is here in between here then the value of s n plus 1 is this value and if f of x is here, s n plus 1 is this value which is the value of s n also. So, in either case so this implies hence s n x is increasing and let us prove that s n x converges to f of x that the limit is equal to f of x. So, if x is fixed then either f of x is equal to plus infinity that is one possibility, but then if this is plus infinity what is s n of x that is always f of x is always bigger than n for any n. So, s n x is going to be equal to n which goes to plus infinity as n goes to infinity. So, if f of x is plus infinity then s n x is equal to n for every n and hence it goes to plus infinity or what is the second possibility f of x is not infinity that means it is a real number. So, it will lie between some n it will be less than or equal to less than or equal to some n. So, it will be some 0 less than or equal to n. So, then in that case so there exists a n such that this is happening. So, that also will be less than n plus 1 and so on. So, and what is s n of x in that case? In that case s n of x is going to be some k minus 1 over 2 to the power n. If this is less than this then f of x will belong to one of the intervals k minus 1 over 2 to the power n and k by 2 to the power n. So, implying that is the same. So, what is the difference between s n and this? s n is the lower value f is something in between. So, that implies that the absolute value of f of x or actually f of x is bigger. So, minus s n of x will be less than k minus 1 by 2 to the power n for some n. And that means if this is happening for some n. So, let us say for some n naught then for every n bigger than or equal to n naught f of x minus s n of x is less than, it will be less than those k minus 1 by 2 to the power n for some k and some n. So, it will be less than, sorry not k minus 1 over 2 to the power. The difference will be at the most, both lie in the interval of length 2 to the power n naught. So, it will be less than 1 over 2 to the power n 2 to the power n for every n bigger than n naught. And that implies that s n x converges to f of x. So, let me just go over to the construction once again to understand, because this is a important construction. So, it says that I want to, given a function f which is non-negative measurable, we want to construct a sequence of simple functions which are non-negative, which are increasing and they converge to f of x. So, what we do? We divide the range. So, this is the range of the function is a subset of it. So, we divide it into a partition the range. So, partition into 0 to n. So, this is 0 and this is n union n to infinity upwards. So, this is and the portion 0 to n is divided into sub intervals each of length 1 over 2 to the power n. So, this will look like k minus 1 by 2 to the power n, k by 2 to the power n, k equal to 1 from 1 to how many such intervals will be there? Each of length 2 to the power n, total length is n. So, n times 2 to the power n. So, this is we have partitioned range. Now, given a point x f of x, either it will be beyond n or given an n, either it will be beyond n or it will be between 0 to n. If it is beyond, then we define s n of x to be equal to n. So, if the value of f of x is bigger than n, then we define it to be equal to n. And if it is not, then it will be between the intervals 0 to n. So, it will fall into one of the sub intervals says somewhere here, in some k minus 1 by 2 to the power n to k by 2 to the power n. So, we define that lower value of that interval that is k minus. So, this is k and this is k minus 1. So, the lower value to be equal to the value of the function s and x. So, this sequence is increasing. See for any point x, f of x and s n will be at the most difference of 1 over 2 to the power n for n large enough or if not, then s n will go to infinity. So, that is the idea that it converges and increasing once again comes from the fact that we are taking the lower value at every stage. So, at any stage, either s n plus 1 is bigger than n plus 1. In that case, it will be bigger than n also. So, s n plus 1 will be bigger than s n. If not, it will be in one of those sub intervals or length 2 to the power n plus 1. But how did you get those? So, this total length is 1 over 2 to the power n. So, when we want to divide the next stage from s n to s n plus 1, we divide it into two equal parts. So, if f of x is here, then s n plus 1 is the lower value here or if s n plus 1 here, it is a lower value. So, in other case, s n is always going to be the lower value. So, that says it is increasing and converging. So, that proves the theorem that what we have shown is the following. Given a function f x to 0 to plus infinity measurable, there exists a sequence s n or non-negative functions which are simple and measurable increasing to f. So, that is what we have proved. .. So, let us come back to the theorem which said that this is one of the key theorems in the notion of for the concept of measurable functions. That every non-negative measurable function can be approximated, can be obtained as a limit of non-negative simple measurable functions and these non-negative simple functions can be selected to be an increasing sequence. So, you can approximate a non-negative function as a limit of increasing sequence of non-negative simple measurable functions. So, this immediately gives us a corollary for functions which are not non-negative. But, let us before that, let me just observe that this in the proof, if the function f is bounded, then the sequence can be chosen to be s n to be uniformly increasing to f. Not only it converges point wise to f, you can actually claim that it converges to f uniformly. So, to prove that it converges to f uniformly, we just observe because the function is bounded. So, let us just observe that if the function is bounded, that means what? That means f is a bounded function. So, it is graph. So, there is going to be n, so that the graph of the function always stays below this. So, once n is fixed, that means f of x is always going to belong to 0 to n for some n and for that f of x, so let us say n naught. So, n naught is the bound for the function. So, then f n is minus s n of x is going to be less than 1 over 2 to the power n naught for every n bigger than n naught. So, this works for all. So, given epsilon bigger than 0, I can select n naught, so that this is true. That means the same epsilon works for every x. That means the sequence s n converges uniformly to f of x. So, this is an observation, which we do not may not be needing it, but it is good to observe that if f is a bounded measurable function. So, this theorem says that if f is a bounded measurable function which is non-negative, then there is an increasing sequence of non-negative simple functions uniformly converging uniformly increasing to f. So, this is the case for when the function is non-negative. In the general case, for a general measurable function, we can look at the positive part and the negative part of the function, approximate the positive part by a sequence, approximate the negative part by a sequence of simple functions and then look at the difference of the two and that will give us a sequence of simple measurable functions converging to f. They will not any longer be monotone. So, as a consequence, we have that if f is not necessarily a non-negative function, if f is a measurable function, then there exists a sequence of simple functions converging to it. So, what we are saying is if f is measurable, then I can write f is equal to f plus minus f minus f minus and we just now observed f measurable implies both of them are measurable and for this there is a sequence s n which is increasing to of simple measurable functions non-negative increasing to this. There is a sequence another sequence call it as say s n dash which is again non-negative simple measurable functions increasing to f dash. So, if I look at this plus this, then that sorry this minus this minus then this will converge to. So, call this as your new sequence. So, this is a call it as phi n. So, phi n is a sequence of because difference of simple measurable functions is measurable. So, this is a s n is a simple measurable function, s n dash is a simple measurable function. So, phi n is a simple measurable function s n converges to f plus s n dash converges to f minus. So, the difference will converge to the difference which is f only thing is. So, this f i n converge to f, but we cannot say phi n are increasing any more this each one of them is increasing, but the difference may not be increasing. So, that proves that for a general measurable function is measurable if and only if there is a sequence of simple measurable functions converging to f. .. So, now let us look at some more general properties of measurable functions. Let us take, so we are going to look at the various properties given two functions f and g which are measurable given a scalar whether sum of the measurable functions is measurable or not whether the product of measurable functions is measurable or not whether scalar multiple of a measurable function is measurable or not. So, let us list all the properties which are true. So, first says if f is measurable and alpha is a scalar and alpha times f is also a measurable function. So, for that this alpha could actually be any extended real number also depending upon because we are taking only. So, keep in mind I am taking only real valued functions for the time being f and g are both real valued functions which are measurable and alpha is a real number. So, the claim is alpha f which is again a real valued function is measurable. Now, for this we can apply our sequential criteria because f is measurable. So, there is a sequence of simple measurable functions converging to it and so look at the sequence. So, let us look at the proof of this. So, that says f measurable implies there exists a sequence S n of simple measurable functions converging to f, but that implies by the properties of sequences alpha S n converges to f because S n is simple measurable a constant times a simple measurable function is again a measurable. So, this is a sequence of simple measurable functions converging to f. So, implies by the previous theorem f is measurable and the same proof works for some of two functions. Let us say f and g are two measurable functions we want to prove that f plus g is measurable. So, f measurable implies there is a sequence S n of simple measurable functions converging to it g is measurable. So, there is a sequence S n dash of simple measurable functions converging to it. So, that implies that S n plus S n dash converges to f plus g and this is once again this is a sum of simple for every n this is a sum of simple measurable functions. So, this is again a simple measurable function. So, we got a sequence of simple measurable functions which converges to f plus g. So, implies that f plus g is measurable. So, that implies f plus. So, we have proved the next step namely if f and g are measurable then f plus g is also measurable. Let us look at the next property if f is measurable then mod f is also measurable. Why is mod f measurable? You can look at two different ways now we have got enough techniques to conclude this. See either we can write mod f is equal to f plus plus f minus. This is a observation which will play a role later on also. This is a positive part of the function this is a negative part of the function f measurable implies both f plus f minus measurable implies f plus plus f minus f minus measurable and this is precisely my f. So, that is one way of looking at it or you can also look at from sequence point of view f measurable implies there is a sequence S n of simple measurable functions converging to f. But then a simple argument which works for sequences which what I have already seen that mod of S n converges to mod f and observation if S n is simple then mod f also is mod S n is also simple for every n. So, the sequence of simple measurable functions converging to f mod f that means mod f is measurable. So, either you can look at sequences or you can look at the positive part and negative part either one will help you to conclude that if f is measurable then mod f is also measurable. Let me look at another property of measurable functions namely that if we are seeing this property for simple functions that if E is a set in the sigma algebra S and f is measurable then product of f times indicator function of E is also a measurable function. So, once again we can take the help of the criteria just now proved f measurable implies that there exists a sequence of simple measurable functions converging to f at every point. But once that is true if S n converges to f then that implies look at chi E times S n that will converge to chi E times f because this remains multiplying by a function. So, this converges to simple properties of sequences and now observe that this is if S n is simple measurable function then the indicator function of E times S n is also a simple measurable function. So, that converges to indicator function of E times. So, that implies that indicator function of E times f is measurable. In fact, we can go a step further and prove that you can multiply by the same argument supposing f and G are measurable. For f we have got a sequence S n of simple measurable functions converging to f we have got a sequence S n of simple measurable functions converging to S n dash converging to G. So, that implies if I multiply S n dash that converges to f times G and product of simple measurable functions we have already seen is again a simple measurable function. So, a sequence of simple measurable function converging to f of G that means implies that f G is measurable. So, product of measurable functions is also measurable. I think we will close here today and look at the sequences of measurable functions next time. So, today what we have proved we have looked at the important criteria characterization of measurable functions. A function f defined on a set x taking extended real valued functions is measurable if and only if keep in mind it is a characterization. So, f measurable f a function defined on x is measurable if and only if we can find a sequence of simple functions converging to it. And if f is non-negative we can find this sequence of simple functions S n which is increasing and converging to f. If in addition we know that f is a bounded measurable function then you can have the sequence of simple functions S n which converges uniformly to f. So, that is the important criteria we have seen some applications today we will see more applications later on also. And then we looked at the algebra of measurable functions we proved that if f is measurable then scalar times f is also a measurable function. If f and G are measurable then f plus G is also measurable f into G is measurable mod f is also measurable. This is for the real valued functions in case the functions are extended real valued while defining f plus G and f into G you have to be slightly careful because f may take the value plus infinity at a point and G may take the value minus infinity then how will you define f plus G. So, for such kind of problematic sets we can separate them out. So, separate out a set A on which f of x is plus infinity or G of x is equal to minus infinity or f of x is minus infinity and G of x is the plus infinity. So, on this set A we may not be able to define what is f plus G, but outside that we can define f plus G and this set where f is plus infinity and G is equal to minus infinity or other way round is a measurable set is in the sigma algebra. So, we can change the values we can define f plus G to be equal to anything we like and still that f plus G will be a measurable function. So, modifications of the algebra of measurable functions the properties still remain true when the functions are extended real valued. We will continue the study of sequences of measurable functions next lecture. Thank you.