 Welcome to lecture number 14 on measure and integration. Today, we will start looking at functions on measurable spaces. These are called measurable functions. To start with, we will assume that we have a measurable space x, s. So, x is a set, s is a sigma algebra of subsets of the set x, and we have a function f defined on x, taking extended real values. So, the r star denotes the set extended real line. There is a set of all real numbers together with plus infinity and minus infinity, and with the possible operations that we had defined earlier. So, we will be looking at functions, which are extended real value defined on the set x. To start with, we want to prove the following. Namely, for this function f, the following statements are equivalent. Inverse image of the interval, open interval c closed at infinity. If you take the inverse image of any such interval, then that belongs to the sigma algebra s. We will show that this is equivalent to saying that the inverse image of the closed interval c to infinity belongs to s for every c, the real number. Also, this is equivalent to saying that the inverse image of the interval minus infinity to c, minus infinity closed, c open also belongs to the sigma algebra s. Then we will show that this is also equivalent to saying that the inverse images of all the intervals of the type minus infinity to c, c closed belongs to s for every c belonging to r. We will show that these four are equivalent to each other and also these are all equivalent to the following. Namely, the points f inverse of plus infinity and the set f inverse of minus infinity along with f inverse of every set e, e, a Borel set in r, they belong to s. We will show that for a function f defined on x, taking extended real values, extended real numbers as the values, these five conditions are equivalent. So, methodology is going to be, we will prove one is equivalent to two, two is equivalent to three, three is equivalent to four and any one of them is equivalent to five. Let us start proving these properties. The first property is, so we are given that f inverse of c to plus infinity belongs to s for every c belonging to r. We want to prove the same property for f inverse of, keep in mind what is f inverse? This is the f inverse of c plus infinity is the set of all points x belonging to x, so that f x belongs to c to plus infinity. So, this is a set of all points x in the domain which are mapped into the interval c to infinity. F inverse does not mean that the function is invertible or anything. This is a symbol used, it is a pullback of the points which go into this c to plus infinity. So, we want to look at f inverse of closed interval c to plus infinity and we want to show that this belongs to s. To show that, let us observe the simple set theoretic equality namely the closed interval c to plus infinity can be written as intersection of, look at the open interval c to c minus 1 by n to plus infinity and look at the intersection of all these intervals. So, keep in mind here is c and here is c minus 1 over n. So, if we take this open interval, so this open interval c minus 1 by n to plus infinity includes this closed interval c to infinity for every n. So, intersection also will include it and actually this is equal because given any point which is slightly bigger than c can be excluded by taking n sufficiently large. So, c minus 1 over n converges to c, that is the basic idea. So, this is a simple identity about intervals which should be easy to prove. Then this implies that the f inverse of c to plus infinity is equal to f inverse of intersection n equal to 1 to infinity of c minus 1 over n to plus infinity. Here is another simple observation that the inverse images of intersections are same as intersection of the inverse images. So, this is equal to f inverse of c minus 1 over n to plus infinity closed. We are given that whenever the interval is of the type c to plus infinity c open inverse image is in S. So, each one of these sets belongs to S. S is a sigma algebra, so intersection belongs to S, so this belongs to S. So, basically what we have done is the closed interval c to plus infinity is written as the intersection of open intervals c minus 1 over n to plus infinity and observing that the inverse images of intersections are intersections of inverse images, we get that f inverse of the closed interval c to plus infinity belongs to S. So, this implies, so we have proved 1 implies 2. Let us show that 2 also implies 1. So, what is the statement 2? Statement 2 says f inverse of the closed interval c to plus infinity belongs to S for every c belonging to R. So, that is the statement 2. So, we want to now prove the same thing for open intervals. So, the idea is the open interval c to plus infinity can be expressed as union of the closed intervals c plus 1 by n to plus infinity n equal to 1 to infinity and that is quite easy to verify. The interval c plus 1 by n is to infinity is inside the interval c to plus infinity. So, this union is inside it and conversely it is easy to check that because c plus 1 over n goes to c, this is actually equal to c to infinity. So, now once again observing that the inverse image of c to plus infinity is equal to f inverse of the union n equal to 1 to infinity c plus 1 over n to plus infinity and once again a simple observation that the inverse image of union is union of the inverse images. So, that gives us that this is f inverse of c plus 1 over n to plus infinity and we are given that each one of them belongs to S and this is a union of sets in S. S is a sigma algebra. So, implies that this set also f inverse of c to plus infinity also belongs to S. Hence, we have shown that 2 implies 1. So, 1 implies 2 and 2 implies 1. Thus, we have shown that the statement 1 implies statement 2 and the statement 2 implies 1 and if you see the proofs carefully in both of them, we have just tried to represent a closed interval as a intersection of open intervals and also on the 2 implies 1, we have tried to use the fact that you can represent an open interval as a union of closed intervals. Similar facts are used in proving the remaining statements. So, let us just prove the remaining statements namely 2 implies 3. .. So, the statement 2 is regarding closed intervals. So, we are given that f inverse of this belongs to S for every c belonging to R. So, that is the statement 2, which is given and we want to show that f inverse of minus infinity to c open belongs to S for every c belonging to R. And if you look carefully, this interval and this interval are related with each other, they are namely complements of each other. So, the given statement implies that this because this belongs to S. So, it is complement, so x minus f inverse of c 2 plus infinity also belongs to S. And this set, the complement of this is nothing but, so here is a small observation that the complement of the inverse image is nothing but the inverse image of the complement. So, this set is equal to f inverse of R star minus c 2 plus infinity and that is equal to f inverse of minus infinity to c open, because c is closed. So, this also belongs to S because S is a sigma algebra. So, if a set belongs, its complement belongs. So, this belongs to S. And see, all these statements are reversible. If this belongs, then its complement belongs. So, these are all if and only if statements. So, 2 implies 3 is obvious by taking complements. Let us prove 3 implies 4. So, the statement 3 implies, so what is given to us is f inverse of minus infinity to c open, that belongs to S for every c belonging to R. And from here, we want to conclude this closed interval. So, note once again, note that the closed interval to c is equal to, so let us, we want to include the point c inside. So, it is nothing but, look at minus infinity to c plus 1 over n, the open interval. So, here is c and here is c plus 1 over n. So, this interval, the closed interval is already inside c plus 1 over n for every n. So, if I take the intersection of all this, so that will give us the closed interval minus infinity to c. So, similar to the earlier argument, so this implies that f inverse of minus infinity to c, which is equal to, so f inverse of the intersection, that is intersection of the inverse images, f inverse of minus infinity to c plus 1 over n, open and each one of them belongs to S, so this belongs to S. So, 3 implies 4. So, if f inverse of minus infinity to c open belongs, then f inverse of minus infinity to c closed also belongs. So, 3 implies 4 and let us prove the converse statement namely 4 implies 1. So, we are given f inverse of minus infinity to c closed belong to S for every c belonging to R and we want to look at f inverse of the open interval c. So, that, so once again it is similar situation that is minus infinity to c, it is here and we want to look at the open interval. So, let us look at the union of intervals minus infinity to c minus 1 over n, n equal to 1 to infinity. So, there all here is c minus 1 over n. So, these are all inside it closed interval. So, the unions will give us this open interval. So, once again taking the inverse images minus infinity to c is equal to f inverse of the union, n equal to 1 to infinity and f inverse of the union is union of the inverse images. So, that gives us minus infinity to c minus 1 over n and all of them each one of them is given to be inside S. So, that implies that this belongs to S. So, 4 implies 3 also true. So, what we have shown till now is that all the first four statements are equivalent to each other. So, first statement was about intervals of the type minus infinity to c to infinity open, then the next one was c closed and then next was minus infinity to c. So, all inverse images of all these types of intervals are inside S, all the statements are equivalent to each other. Now, let us prove that this implies that f inverse of plus infinity and f inverse of minus infinity and f inverse of every Borel set is inside S. So, let us assume any one of 1 to 4. So, assume any one of the statements 1 to 4 hence all because they are all equivalent. So, we know that f inverse of a interval belongs to S whenever for every interval i, which looks like either c to plus infinity or looks like closed c to plus infinity or it looks like minus infinity to c open or minus infinity to c closed. So, for all these four type of intervals, any one of the first four statements implies they belong to S. Now, look at any other interval, supposing i is a open interval a to b, open interval a to b, then we can write this open interval a to b as minus infinity to b, open interval intersection with minus infinity to b, intersection with the open interval a to plus infinity and we know that inverse image of this interval belongs to the sigma algebra, inverse image of this belongs to the sigma algebra. So, that will give us that the inverse image of a b is equal to f inverse of minus infinity to b intersection f inverse of a to plus infinity and both belongs to the sigma algebra. So, this will belong to the sigma algebra S. So, what I am trying to say is that any one of the statements 1 to 4 imply that inverse image of every open interval also belongs and similarly, we can take actually a closed interval also. For example, a closed interval a b can be written as minus infinity to b, intersection a to plus infinity and similar argument will imply that in f inverse of this interval also belongs to S. So, if we assume any one of these statements 1 to 4 and that implies, imply that f inverse of every interval belongs to S for every interval. Now, recall that in the real line any open set, say open set in R is a countable union of open intervals. So, in R, say open set is u, then u can be written as union of i j is j equal to 1 to infinity i j is open. Actually, we can write it as this joint union of open intervals also. It is countable disjoint union of open intervals. So, f inverse of u will be equal to f inverse of union disjoint union of i j's and which is same as union of f inverse of i j's j equal to 1 to infinity. Actually, disjoint is not needed, but anyway this is okay. So, f inverse of every open interval belongs to S. So, this belongs to S. So, if we assume any one of the four conditions, then that implies that f inverse of every open set is in the sigma algebra. Now, this here is a sigma algebra technique. So, consider the class A of all sets belonging to S. All sets belonging, all subsets in B are such that f inverse of E belongs to S. Then just know what we showed, the open sets are inside A and it is easy to check that A is a sigma algebra and A is a sigma algebra. So, let us check that, that A is a sigma algebra. So, why A is a sigma algebra? So, clearly empty set and the whole space and the whole space R belong to A, because X and the empty set belong to S. Secondly, let us observe that if a set E belongs to A, that means that f inverse of E belongs to S and that implies that f inverse of E complement belongs to S, because S is a sigma algebra and that is same as f inverse of E complement belong to S. So, implies E complement belongs to A. So, A is closed under complements and finally, if E n's belong to A that implies and f inverse of E n belong to S. So, implies union of f inverse of E n's also belong to S, because S is a sigma algebra and hence that implies that union of inverse images is inverse image of the union. So, union E n also belong to S. So, implies union E n's belong to A. So, we have proved, verified that A is a sigma algebra. So, this is a sigma algebra including open sets. So, it must include the Borel sigma algebra inside, but it is already a subclass of Borel sets. So, A is equal to the class of Borel sets. That means, if we assume any one of those first four conditions in the statements that we just now stated, then that implies the statement that f inverse image of every Borel set is in the sigma algebra S. Let us verify that the inverse images of the points plus infinity and minus infinity are also. So, note that plus infinity can be written as intersection of n 2 plus infinity, n equal to 1 to infinity. So, f inverse of plus infinity is equal to intersection n equal to 1 to infinity f inverse of n 2 plus infinity. f inverse of this is equal to f inverse of the right hand side. Right hand side is an intersection. So, it is intersection of the inverse images and each one is an interval. So, that f inverse image, inverse image of each one of the intervals belong to S. So, intersection belongs to S. So, this belongs to S. And a similar argument for minus infinity will imply, because minus infinity can be written as intersection of n equal to n equal to minus 1 to infinity of minus infinity to minus n. So, inverse image of this will be intersection of inverse images and will imply that f inverse of minus infinity belongs to S. So, we have shown that if you assume any one of those four conditions stated above, then that implies that inverse image of the point plus infinity and inverse image of every Borel set belong to the sigma algebra S. The converse statement is obvious, because every interval is a Borel set. So, saying that statement 5 implies any one of the statements 4 above is obvious, because every interval is a Borel set. That is a special case. So, we have proved this theorem, namely for a function f defined on a set x, taking non-accented real valued functions, all these five conditions are equivalent to each other. And if you assume any one of them, then other varying will also hold. So, a function which satisfies any one of these conditions is called a measurable function. So, a measurable function on x, taking extended real valued is a function which satisfies any one of those five conditions as stated here. So, these are going to be important class of functions for us to deal with. Let us look at some examples. The first example is that of what is called the indicator function of a set. So, let us look at what is called the indicator function of a set x. So, let us take x, any set and a subset of x. So, we define a function called the chi of a. This is a Greek letter chi. So, lower suffix a, a function on x taking two values 0 or 1. So, this is called the characteristic function or the indicator function. So, this function takes the value at a point x. The value is 0, if x does not belong to a and at the point a, the value is 1, if x belongs to a. So, here is a set x, here is a set a. So, on a it gives the value 1 and on outside a it gives the value 0. So, it is a two valued function, it indicates. So, the points where it takes the value 1 is exactly the points in the set a. So, this is called the characteristic function. So, characteristic function or the indicator function of the set a. So, this is called the indicator function of the set a and the claim is, so we want to look at, so x is a set, s is a sigma algebra and we have got the indicator function a of the set a on x taking values. Of course, only two values, so we can consider it as a function in taking extended real values. So, we want to know, is it measurable? So, suppose the indicator function of a is measurable. So, that implies, if I look at chi a inverse of the singleton point 1, that belongs to s. But what is that value? So, what are the points where it takes the value 1? That is precisely a. So, that is the set a. So, a belongs to s. So, if the indicator function is measurable, then we get a belongs to s. Conversely, if a belongs to s, we claim that chi of a is measurable. So, for that look at chi a of chi a inverse of any interval i. So, what is that going to be? The inverse image of a interval is going to be equal to empty set, if 0 or 1 does not belong to the interval i, because then there is no point which goes to the interval and it is equal to, it is equal to a, if 0 does not belong to i and 1 belongs to i. And similarly, it is a compliment, if 0 belongs to i and if 0 belongs to i and 1 does not belong to i and is equal to x, if both 0 and 1 belong to i. So, in either case, either it is empty set or it is a set a or a compliment or x and all of these are elements of the sigma algebra s. So, inverse images of every interval is in s. So, hence the indicator function is a measurable function. So, what we have shown is that the indicator function is measurable. So, this indicator function which is defined as 1, if x belongs to a and 0, if x does not belong to a. So, the characteristic function is measurable if and only if the set a belongs to s. So, this is a simplest example of a measurable function. .. Consider a linear combination of the indicator functions. Suppose, s is a function defined on x, say that s of x is equal to a i times the indicator function of a set a i at evaluated at x i equal to 1 to n. So, look at sets a 1, a 2, a n, subsets of x. Look at their indicator functions and take a linear combination of them a i times the indicator function of a i. Such a function is called a simple function on x. Such a function is called a simple function on x. Our claim is that a simple function is measurable if and only if each one of the a i's belong to s. So, we want to prove a simple function s, which is sigma a i indicator function of a i i equal to 1. We want to show that this is measurable if and only if. Let us observe. So, note to check measurable we have to look at f inverse of an interval i. So, what is that going to be? That is the function s takes values a i's, small a i's on the set a i. So, this is the main thing to be observed that a finite linear combination of the indicator functions is a function, which takes only finite number of values namely a 1, a 2, a n and the value small a i is taken on the set capital A i. So, what will be s inverse of i? That will be union of those sets a i, union over i says that a i belongs to the interval i. Let us once again observe that s takes values a 1, a 2, a n. So, look at the inverse image of an interval i. So, look at those i's say that a i belongs to the interval i. So, the pullback of this will be the set a i. So, look at the unions of this a i. So, s inverse of i is union of a i's. So, if each a i belongs to s for every i, this will imply that s inverse of intervals belong to s. So, implies that s is measurable, because inverse image of every interval belongs to i. And this interval is in the extended real numbers. So, plus infinity and minus infinity are included in this. So, it is measurable. Conversely, if s is measurable, then look at s inverse of singleton a i. So, that will be equal to a i. Hence, measurability implies this belongs to s. And of course, here a slight care one has to take. We can assume that all the a i's are distinct, because if they are not distinct, we can put together those a i's into one box. So, that leads to our, so anyway, so that says that a simple function is measurable if and only if all the sets involved in the representation as a i times chi of a i are all measurable. So, as observed just now, they have given a simple function s. We can also write it in the form. We can represent as a i summation a i indicator function of a i, where all the a i's are distinct and this capital a i's are disjoint. Because if sets are not disjoint, we can put them together. If the same value is taken on two distinct sets, then we can put them together in one box and call that set as a new a i. So, this is sometimes called a standard representation of a simple function, where the a i's are all distinct and this capital a i's form a partition of the whole space x. So, a simple function is nothing but a finite linear combination of indicator functions. So, that is an example of measurable function. We will study some more properties of these simple functions or this class of simple measurable functions. So, let us start with s, s 1 and s 2 be simple measurable functions and alpha is a real number. So, first of all we want to observe that every constant function is a simple measurable function. What is a constant function? A constant function is nothing but a function which takes a single value everywhere on the set. So, we can think it as if the constant value is taken is c, then it is c times the indicator function of the whole space x. So, every constant function is simple measurable because it is a constant multiple of indicator function. So, alpha times a simple function is also a simple measurable function because of the fact that if a simple function s is equal to summation a i chi a i i equal to 1 to n, then alpha times s is equal to sigma alpha a i times chi of a i i equal to 1 to n. So, alpha s is again a simple function and only its values have changed, but the sets on which these values are taken remain the same. So, clearly it indicates that if s is measurable, then alpha s is also a measurable set. The next property we want to check is that if s 1 and s 2 are two simple measurable functions, then s 1 plus s 2 is also a simple measurable function. So, let us take a function s 1 which is sigma a i chi a i 1 to n. So, let us say s 1 has the representation sigma a i chi a i and s 2 has the representation j equal to 1 to m b j chi of b j. Now, whenever one is dealing with more than one simple function, the idea is try to bring the sets involved in the representation to be same. So, here we are assuming that the last one is a, let us say we have a standard representation, then union a i is equal to x and here union b j is is also equal to x. Then, we can write s 1 as each a i can be decomposed into union of the b j's. So, we can write i equal to 1 to n a i chi of a i intersection b j and union over j's. So, each a i can be intersected with union of b j's. Now, here is a simple fact, a simple observation that the indicator function of disjoint sets. So, here is the observation, if you have two sets a and b and they are disjoint, then a union b is equal to chi of a plus chi of b. So, this we will leave it for you to verify that the indicator function of the union of two sets is equal to sum of the indicator functions whenever the sets are disjoint. So, using that we can write s 1. So, this s 1 can be written as summation i equal to 1 to n a i summation over j equal to 1 to m chi of a i intersection b j. So, this is same as summation over i summation over j a i chi of a i intersection b j. Similarly, for the second simple function s 2, which had the representation b j chi of b j j equal to 1 to m, we can write this as summation over i summation over j b j chi of a i intersection b j. So, what we are saying is that whenever we are given two or finite number of simple functions, we can assume without loss of a generality that the indicator functions involved are of same sets. So, s 1 is equal to summation over i summation over j a i times indicator function of a i times b j indicator function of a i intersection b j. Similarly, s 2 can be written as summation over i summation over j b j indicator function of a i. So, then what is s 1 plus s 2? s 1 plus s 2 is nothing but summation over i summation over j of a i plus b j indicator function of a i intersection b j. That should be clear because if I take a point x, then if x belongs to a i intersection b j, then s 1 gives the value a i and s 2 gives the value b j. So, some will give the value a i plus b j. Outside the value is 0, so one does not have to bother. So, s 1 plus s 2 can be given the representation summation over i summation over j a i plus b j of this. And now, since a i belong to the sigma algebra, b j's belong to the sigma algebra. So, that implies a i intersection b j's also belong to the sigma algebra. So, s 1 plus s 2 is written as a linear combination of indicator function of sets, which are in the sigma algebra. So, that implies s 1 plus s 2 is measurable. So, this proves the property that the class of simple measurable functions is closed under addition. First property said it is closed under scalar multiplication. This says it is closed under addition. Next, let us take a fix a set any set E in the sigma algebra and multiply s with the set indicator function of E. Then claim is this is also a simple measurable function and that comes from a very simple observation. So, let us take a set E belonging to s and s is a simple function with the sigma a i indicator function of a i. Then, s times the indicator function of E is nothing but summation i equal to 1 to n a i indicator function of a i times indicator function of E. So, here is an observation that the product of indicator functions is nothing but the indicator function of the intersection. The product of indicator functions is equal to indicator function of the intersected set. So, if we use this, then the set, the function s times indicator function of E can be written as sigma a i indicator function of a i intersection E. So, it is again a linear combination of indicator functions of sets. a i is intersection E and since a i belong to the sigma algebra, E belong to the sigma algebra. So, this belongs to the sigma algebra. So, the function s multiplied by the indicator function is a linear combination of characteristic functions of sets which are in the sigma algebra. So, implies s indicator function of E is measurable. So, that proves our next property that and using this it is easy to check that the product of simple functions is also simple measurable functions is also a simple measurable function. So, for that let us take s 1 is sigma a i indicator function of a i and s 2 is sigma j equal to 1 to m b j chi of b j. Then, s 1 multiplied with s 2 is nothing but we can do distributive law. So, 1 to n a i of chi a i summation b j 1 to m chi of b j. So, we can write this as summation over i 1 to n a i summation over j 1 to m chi of a i chi of b j into that constant b j. So, let us write that b j to be here. So, anyway we need not have done that much because I have just said that is chi indicator function of a i times s 2 each one of them is simple. So, anyway this can be written as summation over i 1 to n summation over j equal to 1 to m a i b j chi of a i intersection b j. So, once again s 1 into s 2 is a linear combination of indicator function of sets where a i belong to s because s 1 is measurable b j's belong to the sigma algebra as s because s 2 is measurable. So, intersection is measurable. So, s 1 s 2 is measurable. So, the product of simple measurable functions is again measurable. Let us go a step further and let us define for given two simple functions s 1 and s 2. Let us define what is called s 1 maximum of these two function s 1 v s 2. So, what is s 1 v s 2 this is the function whose value at a point x is defined as the maximum of the numbers s 1 x and s 2 x. So, at every point x compare the values of s 1 and s 2 whichever is higher define the value to be that number. So, the claim is s 1 v s 2 is also a simple measurable function. Once again the technique is same as for the sum. So, let us write s 1 equal to sigma a i chi of a i i equal to 1 to n and let us assume s 1 is simple that means all the a i's are in the sigma algebra s. Similarly, s 2 is measurable. So, let us write s 2 as j 1 to m and b j chi of b j where b j's belong to s. Now, we bring them to common sets as before. So, let us write s 1 as sigma over i sigma over j a i chi of a i intersection b j and s 2 equal to sigma over i sigma over j b j chi of a i intersection b j. Then, s 1 maximum s 2 at any point x. So, we want to define what will be at any point x the value of this. So, look at a point x either it will be in one of the sets a i intersection b j. Then, s 1 will give the value a i s 2 will give the value b j and maximum of that has to be put. So, it is maximum of a i b j if x belongs to a i intersection b j. So, s 1 v s 2 is nothing but summation over i summation over j of this. This is once again a finite linear combination of characteristic function where the sets involved are in the sigma algebra. So, this will imply s 1 veg s 2 belong to s. A similar argument will imply that the corresponding minimum of the two simple functions measurable functions is also a measurable function. So, what is the minimum function? So, s 1 veg s 2 at a point x is defined as the minimum of s 1 x s 2 x that is called the minimum of the two functions and we want to show that also belongs to is a simple measurable function. Once again, if s 1 is defined as this and s 2 is defined as this then what is s 1 veg s 2. So, this can be written as simply sigma over i sigma over j minimum of a i b j indicator function of a i intersection b j. So, once we write it that way it becomes clear that the minimum also is a. So, implies that s 1 veg s 2 is a measurable function whenever s 1 and s 2 are measurable functions. So, not only the maximum the minimum also is a simple measurable function whenever s 1 and s 2 are measurable functions. .. Now, I will briefly prove that if s is simple measurable then mod s is also a simple measurable function. Well, that is there are many ways of looking at this. One can also look at if s is equal to sigma a i chi of a i 1 to n then what is mod s? Mod s is a function defined at x to be equal to it is mod of s of x. So, mod s is nothing but sigma mod of a i i equal to 1 to n indicator function of a i. So, this also is a measurable because it is if s is measurable each a i is a measurable set and mod s is a linear combination of indicator function of sets which are measurable. At this point it is worthwhile noting a few things about mod of a function. So, let us take any function f from x to r or r star. Let us look at function. Let us define f plus of x to be a function on x as follows. It is equal to f of x if f of x is greater than or equal to 0 and it is 0 if f of x is less than 0. So, what we are saying is look at the value of the function f of x. If it is bigger than or equal to 0 then you keep the value of the function as it is as soon as it goes below you cut it off by the value 0. So, if this is the function then f of x then what is f plus? So, f plus when it goes below you keep the value to be 0 because it is going below because it is up you keep it as it is now it is going below. So, keep the value to be 0 and now it is going up. So, this is the value of this is the function f plus. So, this is called the positive part of the function. So, this is called the positive part of the function. Similarly, we can define what is called the negative part of the function to be as follows. So, given a function f from x to r r star the negative part of the function f of x is defined as 0. If f of x is bigger than 0 as soon as it becomes positive we make it 0 and we make it equal to minus of f of x if f of x is less than or equal to 0. Keep in mind the negative. So, what it says look at if this is the graph of the function then what do we do? We look at the graph as soon as it is below we keep it as it is when it is going above when it is it is 0 if f is positive. So, on the positive part we keep it as when it is below we reflect it up. So, it is this this this and this and so on. So, note that so this is called the negative part part of f. So, let us observe the function f is written as the positive part minus the negative part. Every function can be represented as the positive part and the negative part and both these functions are both these functions are non-negative functions and mod of f can be written as f plus f minus. So, that is the mod f. You can also think of the positive part f plus can be also thought of as the maximum of f and function constant function 0 and f minus can be thought as maximum of minus f and 0. So, another ways of looking at it. So, if for a simple function saying that mod f is measurable can also be looked at because if s is measurable simple function is measurable the maximum of simple function and 0 is measurable the positive part is measurable the negative part is measurable and hence mod f will be also measurable. So, we will continue properties of measurable functions in our next lecture. In the next lecture we will prove an important theorem namely we will look at how sequences of measurable functions behave whether the limits of sequences of measurable functions are measurable or not. Thank you.