 Welcome to lecture number 16 on measure and integration. If you recall in the previous lecture, we had started looking at the notion of measurable functions on measure spaces. We will continue that study of measurable functions and their properties and then we will specialize measurable functions on measure spaces. And then we will look at the space of measurable functions on when x is real line and the sigma algebras are the Borel sigma algebra or Lebesgue sigma algebra. And if there is a time, we will start looking at the integration of non-negative simple measurable functions. So, let us recall what we had been doing. We had been looking at properties of measurable functions. So, that is what we will continue doing. Then we will look at measurable functions on measure spaces, look at Borel and Lebesgue measurable functions and then look at integral of non-negative simple measurable functions. So, let us just recall what we had done regarding measurable functions. We had said that if x s is a measurable space and f is a function from x to r star, then saying that f is measurable, f measurable, it is same as saying the inverse image of every interval i belongs to s for every interval i contained in s. And there are equivalent ways of defining measurable functions in terms of special intervals like this is same as if and only if f inverse of the intervals of the type c to infinity belong to s for every c belonging to r and so on. So, and then we looked at what is called the algebra of measurable functions. We proved that if f 1 and f 2 are measurable, then f 1 f 2 measurable implies f 1 plus f 2 is measurable implies f 1 into f f 2 is measurable and so on. So, today we will look at the properties of sequences of measurable functions. So, we want to prove the following namely look at a sequence f n of measurable functions, then look at the function what is called the maximum of f n's. So, this is a function denoted by v n equal to 1 to infinity f n of x is defined as the maximum of f n of x n bigger than or equal to 1. So, this is called the maximum of the sequence f n and similarly we have the notion of a minimum of f n's which is noted by veg 1 to infinity f n x equal to minimum. So, this is extra, so that is the definition. So, claim is that if f n is a sequence of measurable functions, then the maximum and the minimum are also measurable functions. So, let us prove this. So, f n is defined on x to r star and f n is measurable for every function bigger than or equal to 1 and we define maximum n equal to 1 to infinity of f n of x to be equal to maximum of f n x n bigger than or equal to 1. So, this is the definition of this maximum and we want to prove to show that this function v n equal to 1 to infinity f n, this is measurable. So, to prove that, we can use any one of those conditions which we have defined earlier for measurable. So, let us look at maximum n equal to 1 to infinity f n of inverse of the interval, let us look at say c to infinity. So, that means what? That means this is all x belonging to x such that n equal to 1 to infinity f n of x is bigger than or equal to c. So, to prove this, now we have to convert somehow this relation into individual f n's because each individual f n is measurable. So, that seems, this is saying the maximum is bigger than or equal to c. That means at least one of them crosses over c. So, that is one way of doing it, but let us look at the equivalent criteria. Namely, let us look at the sets f n inverse, the maximum is less than or equal to c. So, that is equal to minus infinity to c. Let us look at that. This is same as x belonging to x such that the maximum value f n x is less than c. Now, if maximum of something is less than c, then each one of them has to be less than c. So, that is the reason instead of using this kind of intervals, it is more convenient for the maximum to use this kind of intervals because then this set can be written as intersection n equal to 1 to infinity of x belonging to x such that f n x is less than c. So, this may not be exactly true. So, let us take it this close because maximum less than or equal to c, then every one of them will be less than or equal to c. So, we look at the intervals of the type minus infinity c close and that is intersection of the sets and each one of the sets belong to the sigma algebra s because each f n is given to be measurable. So, it is the intersection of elements in the sigma algebra. So, this set also belongs to the sigma algebra s. So, that means this proves the fact that the maximum of f n is a measurable function. A similar proof will work for the minimum. So, let us look at the wedge n equal to 1 to infinity f n that is defined as the minimum of f n x for n bigger than or equal to 1. So, claim is that this is a measurable function. Once again for any c belonging to R, let us look at the minimum of f n n equal to 1 to infinity inverse of some type of intervals we want to show it belongs to s. So, let us try looking at minimum of this is bigger than c. So, let us try this. So, this is all x belonging to x says that the minimum of f n x n bigger than or equal to 1 is bigger than or equal to c. So, if the minimum of certain numbers is bigger than or equal to c then each one of them has to be bigger than or equal to c, because even if one is smaller then the minimum will become smaller. So, this is equal to intersection of n equal to 1 to infinity of all x belonging to x is that f n x is bigger than or equal to c. So, that implies and this is nothing but intersection n equal to 1 to infinity of f n inverse of c to plus infinity and each f n being measurable this is the measurable set. So, implies that this is a set in s. So, we have proved that if f n is a sequence of measurable functions then we look at the maximum or the minimum of the sequence of this measurable functions then that both of them are again measurable functions. So, as a consequence of this we prove that the limit of measurable functions is also a measurable function. So, that is our next aim to prove that the limit of measurable functions is also a measurable function. So, for that let us understand what is the limit of a function. So, let us look at a sequence f n the sequence f n of functions each f n is defined from x to star. So, to define the notion of limit of f n at a point x what we do is for every x belonging to x, let us look at the maximum or the supremum of f n x for n greater than or equal to some stage m. So, this number the supremum that depends on m and then we take the infimum over all m's. So, this is supremum first take the supremum and then take the infimum. This gives you a function this is called limit superior of f n at the point x. So, this is called the limit superior of the function sequence of functions f n x at the point x. Similarly limit inferior of f n x is limit inferior is defined as you first take the infimums of f n x for n greater than or equal to some stage m and then look at the supremum for all m bigger than or equal to 1. This is called the limit superior and you must have seen in your elementary analysis classes that limit superior of f n x is always bigger than or equal to limit inferior of f n x. So, this inequality always holds and the sequence f n f n x converges to f x if and only if limit superior f n x is equal to f of x. So, these are elementary effects from basic analysis about when is the sequence of real numbers convergent. So, it says that a sequence for any sequence of real numbers or extended real numbers you can define the concept of limit superior and also you can define the concept of limit inferior. Limit superior is defined by looking at the supremums of the sequence a n from some stage m onwards and then this supremum depends on m. So, look at the infimum of all these supremums. So, that is called the limit superior and similarly limit inferior is defined as first taking the infimums of the sequence f n x from some stage m onwards and then looking at the supremums of these numbers which depend on m. And one proves that the limit superior of a sequence is always bigger than or equal to limit inferior and the sequence is convergent if and only if the limit superior is equal to limit inferior. So, in case you have not come across these concepts strongly says that you pick up a book on elementary analysis and revise the concepts of limit superior and limit inferior. So, we are going to use that fact now here to prove that f is measurable. So, what is f? f of x is nothing but limit superior and limit inferior. So, only thing to show is that the limit superior and limit inferior are both measurable functions, but limit superior of f n is nothing but first taking the supremums of f n x from n bigger than or equal to m and then taking the infimums m bigger than or equal to 1. And just now we have shown that if f n is a sequence of functions then the supremums the maximums are also measurable functions. So, this is a measurable function, infimums of measurable functions that is a measurable function. So, this implies that limit superior f n is measurable and similarly limit inferior f n is measurable and saying that f n converges to f is same as saying this f is equal to limit superior f n or also equal to limit inferior of f n. So, that proves the fact that f n converges to f for every point x implies and f n measurable implies f is a measurable function. So, limits of measurable functions are also measurable functions. So, this proves the theorem that the class of all measurable functions is nice, it is closed under taking point wise limits. Now, let us observe that most of these properties hold for extended real valued functions also when properly defined. So, because only thing to observe is the following that if f and g are extended real valued functions and you want to define f plus g, care has to be taken because f of you will like to define it as f of x plus g of x. But the problem comes the problem comes if f of x is equal to plus infinity and g of x is equal to minus infinity then what will be this number that is not defined or f of x is equal to minus infinity and g of x is equal to plus infinity even then the problem comes this number is not defined. So, what one does is to meaning suitably defined means look at all the points call this set as a where all x belonging to x where either of these things happen where f x equal to plus infinity g x is equal to minus infinity or f x equal to minus infinity and g x equal to plus infinity. Now, one observes that this set a is in the sigma algebra because f x is equal to plus infinity belong to sigma algebra intersection with that belongs to sigma algebra. So, all these sets all this set a is in the sigma algebra. So, a is the set on which the problem can come. So, what one does is one defines f plus g of x to be equal to f x plus g x if x does not belong to a and if x belongs to a you can define it any number alpha if x belongs to a. So, with this definition it is easy to observe. So, let me leave it as exercise for you to show that if I define it this way with alpha any value on the set a then f plus g is s measurable. So, that is what I mean by saying that the above results most of these properties hold for extended real valued functions also when those functions are appropriately defined. So, will not go much into detail of this one can easily verify these things. So, let us look at f and g to be another property of measurable functions is the following. Let us look at two functions f and g which are measurable. Then the following holds the look at the sets x belonging to x say that where f x is bigger than g x or the set x belonging to x where f x is strictly less than g x or x belonging to x where f x is equal to g x and similarly where f x is bigger than or equal to or f x is less than or equal to. So, all these type of sets claim is are in the sigma algebra s. So, let us look at to prove of one of them and others will follow similarly. So, f and g are functions x to r star measurable. Let us look at the set x belonging to x such that f of x is strictly less than g of x. So, our aim is to show that this belongs to the sigma algebra s and since we are given f and g are both measurable, we are given the property that f x less than or equal to some real number is belongs to the sigma algebra and similarly g x less than or equal to a real number belong to the sigma algebra. So, objective is try to interpret this set in terms of union, intersections of something of sets of the type where f x is less than something and g x is less than something. So, for that we observe that for any x if f of x is less than g x then there must be a rational number in between them. So, for every x belonging to x there exist a rational r such that f x is less than r is less than g x. So, here we are using the effect that rational are dense on the real line. So, with this property we can write x belonging to x say that f x is less than g x. So, this implies that x belonging to x say that for some rational f x is less than r is less than g x for some and conversely if for some r this is true then obviously f x is equal to g x. So, claim is this is equal to union over r belonging to rational numbers. So, this is the only crucial point in this that the set f x less than g x can be written as a union over all rationals such that f x strictly less than r strictly less than g x. Now observe this set is an intersection. So, I can write it as r belonging to q this set is where f x is less than r and g x is bigger than r. So, it is x belonging to x say that f x less than r intersection with the set x belonging to x say that g x is bigger than r. So, what we have done? We have interpreted the set x belonging to x say that f x less than g x as let us just look at the set again. So, this is union over r that is equal to union over r belonging to q. What is this set? This is f inverse of f x less than r that is less than r means it is minus infinity to r. The second set is the in nothing but g inverse of g x bigger than r. So, it is r plus infinity. So, this is so the set f x less than g x is written as union over rational intersections of these two sets. Now f and g being measurable this set belongs to the sigma algebra g being measurable this set belongs to the sigma algebra it is intersection. So, the whole set belongs to the sigma algebra intersection belongs to the sigma algebra and rationals are countable. So, this is a countable union of elements in the sigma algebra. So, this belongs to the sigma algebra. So, what we have shown is the set x belonging to x there f x strictly less than g x belongs to the sigma algebra s. So, that proves the first property of the theorem. Now, if we take just the complement of this set. So, also implies that x belonging to x such that f x less than g x the complement of this set what will be that. So, that is all x belonging to x such that f x is bigger than or equal to g x. So, that set also belongs to the sigma algebra and similarly the argument which will similarly we said that f x less than g x belongs to it. So, let us write x belonging to x say that f x is strictly bigger than g x is also in the sigma algebra because by similar arguments I can write this as the union over all rationals of f inverse of. So, f x bigger than so that will be r plus infinity in intersection with g inverse of minus infinity to r. By similar argument where we had f x less than again we can enter f x is bigger than g x. So, there must be a rational in between. So, that must be true and that will imply that this belongs to the sigma algebra. So, saying that f x bigger than g x belongs to the sigma algebra is and if you take the complement of this that is nothing but x belonging to x such that f x less than or equal to g x. So, that also belongs to the sigma algebra because this is the complement of the set in the sigma algebra. So, measurable sets have nice properties namely if f and g are measurable then operations involving measurable sets measurable functions give you again sets in the sigma algebra. So, these are nice properties and we will see use of these properties soon. So, with this we complete the study of measurable functions on measurable functions on measurable spaces. Next we are going to look at measurable functions which are defined on measure spaces. So, they play also a play role later on. So, we want to define we want to look at functions f which are defined on a set x taking extended real values and on x there is a sigma algebra. So, this is a sigma algebra S and a measure mu given on S. So, let us first define what is the meaning of the notion of almost everywhere. So, we say that a property p about the points of x is set to hold almost everywhere with respect to the measure mu. If you look at the set of points x for which the property p does not hold at x. So, look at all those points of x say that the property p does not hold at the point x. So, this is a subset and we want this subset belongs to the sigma algebra and mu of e is equal to 0. So, what we are saying is except for a set of measure 0 the property holds. So, that is why we give it a name that the property p holds almost everywhere with respect to the measure mu. Let me illustrate this with some examples. Let us take a function f and we look at the statement that f is 0 almost everywhere. So, f is a function which is extended function defined on the set x and we want to say that this function is 0 almost everywhere. So, look at the set of points where f is not 0. So, what will the statement mean? That set where f is non-zero should be element in the sigma algebra and its measure should be 0. So, x belonging to x say that f of x is not 0 that should be element in the sigma algebra and its measure of can be defined only when the set is in the sigma algebra. So, mu of that set is equal to 0. So, the statement that f is equal to 0 almost everywhere will mean mu of measure of the set of points where f x is not 0 is 0. Let us look at another illustration of this. The statement that f is finite almost everywhere what will that mean? So, that will mean look at the set of points where f is not finite that means what f is extended real value to function. So, it can take the value plus infinity or minus infinity. So, the set of points x belonging to x say that mod f x is equal to plus infinity. So, that is same as where either f of x is plus infinity or f of x is equal to minus infinity that set is in the sigma algebra and mu of that set is equal to 0. So, saying a function f is finite almost everywhere means the set of points where it can take the values plus infinity or minus infinity is a set of measure 0. Similarly, let us look at two functions f and g and let us look at the statement that f is strictly bigger than g almost everywhere. So, f is strictly bigger than g almost everywhere what will that statement mean? That means the set of points where f x is not strictly bigger than and that is the same as the set of points where f x is less than or equal to g x. So, the compliment of that statement is f x less than or equal to g x these set of points have got measure 0. So, saying that f x is strictly bigger than g almost everywhere with respect to mu means mu of the set where this statement is not true and that is f x less than or equal to g x is 0. This concept of almost everywhere is quite useful when looking at measurable functions. So, let us prove the property that if f and g are two extended evaluate functions say that f x is equal to g x almost everywhere mu then, measurability of one of the functions f implies the measurability of the other function g. So, f is measurable, s measurable if and only if g is s measurable. So, let us prove this property that if two functions are equal almost everywhere then the measurability is not changed one in measurability of one. So, let us look at f is from x to r star g is also from x to r star and we know that the set of points x belonging to x says that f x not equal to g x this set has mu measure equal to 0. So, let us suppose f is s measurable to show g is s measurable. So, to show that g is s measurable let us look at g inverse of any interval i. So, g inverse for every interval i i a interval in r star then g inverse of i we want to show that. So, claim is that this belongs to s for every interval i and now we have to transform this property this set into something regarding g. So, let us regarding f. So, let us look at g inverse of i is same as all x belonging to x such that g x belongs to i. So, this is a subset of the set x. So, what I can do? I can write this set as so g inverse of i I can write as intersection of so x belonging to x say that g x belongs to i and intersected with the set a also union. So, intersect with a complement. So, g inverse of i I have intersected with a and a complement. So, it is a union of these two sets x belonging to x say that g of x belongs to i intersection a complement. Now, let us look at the first set. So, this is g x belonging to y intersection a and what was the set a? What is the set a? Where a is the set x belonging to x where f x is not equal to g x f x. So, let us observe we are given that mu of a equal to 0. So, that automatically implies that a belongs to a belongs to the sigma algebra s and that automatically implies that the set x belonging to x say that g x belongs to i intersection a also belongs to the sigma algebra. So, this set also belongs to the sigma algebra. Why? Because this is a subset of a and a is a set of measure 0. So, this is a set of measure 0 and we have already assumed our measure spaces are complete. So, because of so the implies this is so because the measure space x s mu is complete. So, we have made the assumption that we are working on complete measure spaces. So, that gives that shows the importance of complete measure spaces. So, this set belongs to a and the other part a complement on a complement f is equal to g. So, I can replace it by a complement. So, the set x belonging to x say that g x belongs to i intersection a complement is same as the set x belonging to x where f x belongs to i intersection a complement because on a complement f is equal to g. So, that means what? So, g inverse of i is written as g inverse of i intersection a. So, let us just rewrite this statement again. So, what we are saying is g inverse of i can be written as g inverse of i intersection a union g inverse of i intersection a complement and that is same as g inverse of i intersection a union f inverse of i intersection a complement because on a complement f is same as g and this is a set of measure 0 mu of this set is equal to 0. So, implies that this set g inverse of i intersection a belongs to the sigma algebra and this set f is measurable. So, implies this set is in the sigma algebra a is in the sigma algebra. So, a complement in the sigma algebra intersection in the sigma algebra. So, this element belongs to the sigma algebra. So, and this is a union of two elements in the sigma algebra. So, this implies that g inverse of i also belongs to the sigma algebra s. So, we have shown f measurable f equal to g almost everywhere mu implies g measurable. So, that is the importance of measurable functions equal almost everywhere, but keep in mind we have used the fact that underlying measure space is a complete measure space. So, measurable functions that means this says that if f is measurable you can change its values on a set of mu measure 0 and still the function will remain measurable. So, another interpretation of this result is if f is measurable and you change its values on a set of measure 0 and call that function as g. So, that is measurable. So, that is a quite an important fact. Another application of this concept of almost everywhere is the following. Look at the sequence f n, a sequence of measurable functions converging to a function f almost everywhere. That is the set of points where f x is not equal to the limit has got this set, this set has got measure 0. Then the claim is then the f is also measurable. So, just now we proved that if a sequence f n of measurable functions converges to f then f is measurable and now we are saying that if f n are defined on a complete measure space and f n converges to f almost everywhere even then this property remains true. So, basically the idea is same as before. So, let us just look at how does one write a proof of this statement. So, we have got a complete measure space x s mu measure space. We have got a sequence f n of functions f n is measurable f n is converged to f almost everywhere mu. So, that means look at the set A x belonging to x such that f n of x does not converge to f of x. Then what is given to us is that this set A belongs to the sigma algebra and mu of A is equal to 0. So, now let us look at, so we want to show that f is measurable to show f is measurable. So, once again look at for any interval i look at f inverse of i. So, I can write as f inverse of i intersection A union f inverse of i intersection A complement. Now on A, so we are given A is a set of measure 0. So, this is a subset of A. So, that has set of measure 0. So, this implies that f inverse of i intersection A belongs to the sigma algebra S because this is a set of measure 0 and our underlying measure space is complete. On this portion A complement f n is converging to f. So, this f I can write as limit n going to infinity of f n inverse of i intersection A complement. So, this set is same as this. Now, we know that f n on A, f n converge to f on A complement. So, that is a measurable set. So, this is a element in the sigma algebra because on A complement f n is converging. So, if we restrict ourselves to A complement then that must be element in the sigma algebra. So, both belong to sigma algebra. So, this belongs to the sigma algebra S. So, the concept of almost everywhere when dealing with the complete measure space is we can exploit that property. So, this implies that if f n is a sequence of measurable functions converging to a function f almost everywhere then the limit also is a measurable function. So, this emphasizes the property of something holding almost everywhere. Now, let us specialize the case when our underlying measure space is set is the real line. Then we have got two sigma algebras when x is equal to real line. Then we have got two sigma algebras one is the Borel sigma algebra and other is the sigma algebra of Lebesgue measurable sets. We have shown that the sigma algebra of Borel subsets is a subclass of Lebesgue measurable sets. So, when we are looking at functions defined on real line taking values as extended real numbers there are two possibilities to analyze whether the function is measurable with respect to the Borel sigma algebra or measurable with respect to the Lebesgue sigma algebra. So, there are two notions of measurable as far as real line is concerned and so we will separate them out. So, we will say a function is Lebesgue measurable if the inverse image of every interval in r star is a Lebesgue measurable set. So, if the inverse image of every interval in r star is an element in is a Lebesgue measurable set then we will say that function is Lebesgue measurable and we will say a function is Borel measurable if for every interval in r star we will pull back its image its pre image in r is a Borel set in r. So, here is a difference Lebesgue measurable requires that the inverse image is in the Lebesgue sigma algebra sigma algebra of Lebesgue measurable sets and f inverse of i in B r says it is the inverse image is everywhere is always a Borel set in r. So, obvious because Borel subsets form a subset of r. So, it obviously clear that every Borel measurable function is also a Lebesgue measurable function because inverse image of every interval if it is in B r and B r is a subset of l r. So, every Borel measurable function is also a Lebesgue measurable function. For example, let us look at a function which is continuous. So, if f r to r is continuous function then it is going to be a Borel function. So, let us prove that every continuous function is a Borel measurable function and hence also Lebesgue measurable. So, f is a function which is defined from x sorry x is real line. So, f is a function defined from r to r and f is continuous. Claim that f is Borel measurable that is f inverse of any set E belongs to B r for every set say E belonging to B r and continuity of a function can be expressed in terms of open sets. So, let us look at the class A of all subsets E belonging to B r such that this property is true f inverse of E belongs to B r. So, what you have to show? To show saying that f inverse of E belongs to B r for every E in B r it is an equivalent to saying to show that this A is equal to B r and that is where we are going to use our sigma algebra technique. So, to show that A is equal to B r note that A is already a subclass of B r because of we are picking up sets in B r. So, to show that A is equal to B r we have to show that B r is inside A. So, for that we will show two steps one open sets are contained in A and second will show that A is a sigma algebra because once A is a sigma algebra and includes open sets it must include the smallest sigma algebra generated by open sets that is B r. So, B r will be inside A and will be through. So, to prove these two facts are quite obvious because of the given condition so fun open sets belong to A. So, let U contained in R be open f continuous implies that f inverse of U is open and hence this means f inverse of U belongs to B r. So, what we have shown is if U is open then f inverse of U is in B r. So, that proves so implies that the open sets are inside A and A is a sigma algebra that is more set forward. So, let us observe empty set is equal to f inverse of empty set and so and r is equal to f inverse of r. So, both belong to A because empty set and the whole space they are equal to this. So, this is obvious empty set and the whole space belong. Second property if E belongs to A that implies f inverse of E belongs to B r and that implies f inverse of E complement belongs to B r and that implies because this set is same as f inverse of E complement that belongs to B r. So, what we have shown is if E belongs to A then f inverse of E complement belongs to B r and that implies that E complement belongs to A. So, that means E complement. So, the class A includes empty set includes the whole space it is closed under complements. So, finally, let us show that it is also closed under countable unions. So, that means let us take sets E n belong to A n bigger than or equal to 1. Then look at that means what we are given that f inverse of E n belongs to the sigma algebra B r because the property E n belongs to A means the inverse image is in B r. So, what that implies B r is a sigma algebra that implies union 1 to infinity f inverse of E n belongs to B r and now a simple observation that this set is same as f inverse of union E n n equal to 1 to infinity that belongs to B r. So, if E n belongs to A then f inverse of the union belongs to B r. So, that means union of n equal to 1 to infinity E n belongs to B r belongs to A. If E n belongs to A then f inverse of the union belongs to B r that means the union belongs to A. Hence, we have shown that A is a sigma algebra of subsets of A and it includes open sets. So, it must include B r and hence this is equal. So, that proves that every continuous function from r to r is Borel measurable and hence it is also Lebesgue measurable. So, all topological nice functions, continuous functions become Lebesgue measurable on the real line. Let us look at some more properties. So, we showed that every Borel function is Lebesgue measurable. So, there exist functions first of all r to r star which are not Lebesgue measurable. So, to prove that we can only simply observe that there are sets. Let us go back and recall that for a function f from r to say r star, let us look at f equal to indicator function of a set A where A is a subset of x. So, recall chi of A is Lebesgue measurable if and only if A belongs to L of r. So, if you can produce a set which is not Lebesgue measurable, then the indicator function will not be Lebesgue measurable. So, the answer to this question does that this non-Lebesgue measurable functions depends upon whether there is non-Lebesgue measurable sets. If you recall, we had proved the fact that the non-Lebesgue measurable sets exist. That question is related to basic set theory. So, if we assume exume of choice, then we constructed non-Lebesgue measurable sets. So, assuming exume of choice, one can claim that there exist functions which are not Lebesgue measurable. By the same reasoning, one can ask the question do there exist functions which are Borel measurable, but not which are Lebesgue measurable, but not Borel measurable because every Borel measurable is Lebesgue measurable. So, that means to for that by the same logic again, if we pick up a set A which is Lebesgue measurable, but not a Borel set, then the indicator function of that set is going to be a function which is going to be Lebesgue measurable, but not Borel measurable. So, these two questions that whether there exist non-Lebesgue measurable, non-Lebesgue measurable functions and whether there exist functions which are Lebesgue measurable, but not Borel get tied up with the fact that the Lebesgue measurable subsets is a proper subset of power set of R and B R is a proper subset of the Lebesgue measurable sets. So, with that we conclude the study of properties of measurable functions. So, basically let me just recall the measurable functions or functions defined on the underlying set X with properties that the inverse image of every set E in the Borel sigma algebra of extended real numbers. The inverse image is again in the sigma algebra on the domain space that is S. So, this is a property about the inverse images of sets being in the sigma algebra S and we will see that how this property plays a role in our further study of study of integration. So, we will do in the next lecture we will start the notion of integration for measurable functions. Thank you.