 Welcome to lecture 20 on measure and integration. In the previous lecture, we had started defining what is called the notion of a function to be an integrable function and then we started looking at some of the properties of integrable functions. Let us just recall what is a integrable function and then we will start looking at various properties of these integrable functions and we will prove one important theorem called dominated convergence theorem. So, if you recall, we said a measurable function f on x to an extend material valued measurable function f is said to be integrable with respect to mu and written as mu integrable if both the integral of the positive part of the function and the negative part of the function are finite. So, we say f is mu integrable if integral f plus d mu and integral f minus d mu both are finite numbers and in that case, we say the integral of f is equal to integral of f plus minus integral of f minus. So, let us just once again emphasize saying that a function is mu integrable is if and only if both f plus and f minus are having finite integrals and the integral of f is written as integral of f plus minus integral of f minus. The class of all integrable functions on the measure space X s mu is normally denoted by L1 capital L lower 1 X s mu or sometimes so we drop s and mu if they are clear from the context that what are the sigma algebras or what is the measure or sometimes we just emphasize mu because we know what is X and what is s. So, these are various notations used for denoting integrable functions L1 X s mu or L1 bracket X or L1 bracket mu. So, this is the space of all mu integrable functions. We will start in looking at the properties of these functions. The first important thing we observed was a function f is which is measurable is integrable if and only if mod f which is a non-negative function is integrable. That means to check whether a function a measurable function is integrable or not it is to enough to look at the integral of the function mod f and see whether that is finite or not. And in that case and this is always true and the for the integrable function integral of mod f integral mod of the integral of f d mu is less than or equal to integral of mod f d mu. So, this is a important criteria this is a equivalent definition equivalent way of defining integrability of a measurable function namely mod f is measurable. This is not equal so this is a this is wrong here it should be less than or equal to. So, integral of mod of integral f d mu is less than or equal to so this is a typing mistake here this should have been less than or equal to integral of mod f d mu. Let us recall some of the other properties that we had proved we said if f and g are measurable functions and mod f is less than or equal to g of x for almost all x with respect to mu and g is integrable then f is also integrable. That means if a function f of x is dominated by an integrable function then that measurable function automatically becomes integrable. We also proved the following property namely if two functions f and g are equal almost everywhere and one of them is integrable say f is integrable then the function g is also integrable and integral of f is equal to integral of g. So, that essentially says that the integral of the function does not change if the function is change if the values of the functions are change almost everywhere. So, f equal to g almost everywhere f and g measurable functions and one of them say f integrable implies g is integral integrable and the integral of the two are equal. We also proved the following property namely if f is a integrable function and alpha is a any real number then alpha time f is also integrable and the integral of alpha f is equal to alpha times the integral of f. So, we continue this study of properties of integrable functions and next we want to check the integrability property namely if f and g are integrable functions then we want to show that f plus g is also integrable and integral of f plus g is equal to integral f plus integral g. So, to prove this property let us look at what we are given. So, we are given that f and g are integrable functions that is integral mod f d mu is finite and integral of g d mu is also absolute value of g with respect to mu is also finite. So, to check whether the function f plus g is integrable or not we have to look at the integral of f plus g absolute value and show that integral of absolute value of f plus g is also finite, but that follows easily because absolute value of f plus g is always less than or equal to absolute value of f plus absolute value of g. So, and all are non-negative measurable functions. So, using the property of the integral for non-negative measurable functions this implies that integral of mod f plus g d mu is less than or equal to integral of mod f plus mod g d mu and that by linearity is same as integral mod f d mu plus integral mod g d mu and we are given that both of them are finite. So, this is finite. So, implies that f plus g is integrable. To compute the integral of f plus g we have to go back to the definition of the integral. So, f and g integrable so that implies integral of f plus d mu is finite, integral of f minus d mu is finite, integral of g plus the positive part of g is finite, integral of g minus d mu is finite and we have to show. So, to show integral f plus g plus d mu finite and integral f plus g minus d mu is finite. So, these two properties we have to show. To show this somehow we have to relate the positive part of f plus g with the positive part of f and positive part of g and similarly the negative part of f plus g with the negative part of f and negative part of g and that is done as follows. So, what we do? Look at f plus g by definition we can write it as f plus g positive part minus f plus g the negative part. So, that is by the definition of the positive part and the negative part of the function. Also f plus g we can also write it as decompose f into positive part and the negative part. So, that is f plus minus f minus and similarly we write g as g plus minus g minus. So, now from these two it follows that integral of f sorry not the integral from this it follows that f plus g positive part minus f plus g the negative part is equal to f plus minus f minus plus g plus minus g minus. So, from these two equations it follows this is so and now what we do is all the negative terms we shift on the other side of the equation. So, this implies that f plus g plus plus f minus plus g minus is equal to f plus plus g plus from here and this term on the other side will give me plus f plus g minus. So, we just rearrange the terms and now you observe that the left hand side is a non-negative function and the right hand side is a non-negative function. So, by the properties of integrals for non-negative functions this implies that integral of f plus g plus d mu plus integral of f minus d mu plus integral of g minus d mu. So, that is the integral of the left hand side is equal to integral of f plus d mu and plus integral g plus d mu plus integral f plus g minus d mu. So, from this equation by using the properties of integral for non-negative functions the linearity property integral of the left hand side is equal to integral of the right hand side and integral of the left hand side consists of integral of f plus g plus plus integral of f minus plus integral of g minus and that is equal to integral of f plus plus integral of g plus plus integral of f plus g minus. Now, we observe that in this equation all the terms are finite quantities or real numbers that is because f plus g we have already shown is integrable. So, this first integral of f plus g plus that is finite integral f minus is finite and similarly all the terms are non-negative real numbers. So, we can again manipulate them and treat shift terms on the left hand side and right hand side. So, what we will do is this term f plus g minus on the right hand side we will bring it on the left hand side and the terms f minus d mu integral integral g minus d mu we shift it on the right hand side. So, that gives us the property so shifting. So, implies that integral of f plus g plus d mu minus this term will give you integral f plus g minus d mu. So, this term we have shifted. So, and shift these two terms on the other side is equal to integral f plus d mu that is this term and this integral of f minus bringing on this side will give you integral f minus d mu plus integral of g plus d mu which is already there and integral of g minus from the left hand side will give you integral of g minus d mu. So, this rearrangement of the terms here once again give you that integral of f plus g plus d mu and the integral of the negative part of f plus g is equal to integral of f plus minus integral f minus plus integral g plus and now by the definition the left hand side is nothing but integral of f plus g d mu and the right hand side is integral f d mu plus integral g d mu. So, that proves the linearity property of the integral that if f and g are integrable functions not only f plus g is integrable integral of f plus g is equal to integral of f plus integral of g d mu. So, that is the linearity property of the integral. So, we have proved the basic properties of the integrals namely the integral of a function which is integrable of course, it is a finite quantity and it is linear namely if you take a function f multiplied by a scalar alpha then alpha times f is integrable and the integral of alpha f is equal to alpha times integral of f and similarly if f and g are integrable then f plus g is integrable and the integral of f plus g is equal to integral of f plus integral of g. Let us look at some more properties of this integral which are going to be useful later on. So, let us look at the next property. So, for a integrable function f so f in l 1 of mu let us look at we have already shown that if you mod f is a non-negative measurable function and if you multiply it by the indicator function of a set E then we already shown that this is again a non-negative measurable function and of course, this function is less than or equal to integral of mod f. So, mu of E is going to be always a finite quantity. So, the claim is mu is a measure in fact a finite measure and it has the property that mu of E equal to 0 implies mu of E equal to 0. So, whenever a set E has got mu measures 0 the measure of mu also is going to be equal to 0 and this basically follows from the properties of the integral for non-negative functions because if f is integrable then mod f is a non-negative measurable function and its integral is a finite quantity. So, mu of E is a finite measure for every E the function chi E times mod of f is less than or equal to mod f. So, this integral is going to be less than or equal to integral of mod f which is finite and obviously, for non-negative functions we have already proved this property that for if mu of E is 0 then the integral over E is equal to 0. So, this property follows from our earlier discussions. Let us look at the integral of for the integrable function not the integral of mod f but let us look at the integral of f times indicator function of E and if you recall we had already shown that if f is measurable and E is a set in the sigma algebra then chi E times f is a again a measurable function and just now we observed that this number is going to be a finite number because this is again a integrable function. Let us just observe this property once again that if f belongs to l 1 of mu and E is a set in the sigma algebra then this implies that chi E times f is a measurable function. So, that we have already seen because f is a measurable function indicator function of E is a measurable function. So, product of measurable function is measurable and we observed just now if you look at the absolute value of chi E of E times f that is same as indicator function of E because that is negative into absolute value of f. So, this implies that the integral of chi E times f absolute value d mu is less than or actually is equal to integral chi E of mod f d mu and which is less than or equal to integral mod f d mu which is finite. So, what does that imply? So, this implies that integral chi E d mu is a so this we are denoting it by nu tilde of E. So, this is a real number is a finite real number. So, that is the observation and we want to claim that mu of E is equal to 0 implies that nu tilde of E is also equal to 0. So, the claim is that this claim is that this property. So, let us prove this property. So, suppose mu of E is equal to 0 then what is nu tilde of E? Nu tilde of E by definition is integral chi E of f d mu which is same as which is same as the integral of chi E f plus d mu minus integral chi E f minus d mu and now we observe mu of E equal to 0 chi E of f plus is a non-negative function and for properties of non-negative functions imply if the set has got measure 0 then the integral of this is equal to 0. So, the first integral is equal to 0, second integral is equal to 0 by properties of integrals of non-negative measurable functions. So, implies so implies nu tilde of E equal to 0. So, what we are saying is mu of E equal to 0 implies nu tilde of E is also equal to 0. So, that is the property we are proving here, but keep in mind that nu tilde of E is defined as a real number for every E belonging to S, but it is not a non-negative number because f may not be a non-negative function. So, we cannot say nu tilde of E is a measure. We will look at this property a bit later, it may not be a measure, but it has some properties similar to a measure. Here is another important property. Let us look at again the same value nu tilde of E which is equal to integral of f over E d mu and suppose this is equal to 0 for every set E in the sigma algebra, then the claim is then this function f must be equal to 0 for almost all x belonging to mu. So, let us prove this property namely, so given nu tilde of E which is nothing but integral chi E times f d mu is equal to 0 for every E belonging to S. So, that is what is given to us and we want to show that if you take the set n which is x belonging to x such that f of x mod f of x bigger than 0, if we let this set n then note that this set n is a set in the sigma algebra and we want to show that nu of n. We want to show f is 0 almost everywhere and this is a set where f is not 0. So, we want to show this is equal to 0. So, this is the problem we want to show. Now let us look at consider the set say for example, let us look at let us write the set say a n to be the set where x belongs to x, say that f of x is bigger than 1 over n and similarly let us write b n to be the set of x belonging to x where f of x is less than minus 1 by n. So, now the claim is that the set n is nothing but union over a n, union over b n, n equal to 1 to infinity union of union n equal to 1 to infinity. That means all these sets a n's and b n's if you take the unions that is besides the set n where n is where n and what is the set n? n is a set where f of x is not equal to 0. So, if f of x is not equal to 0 then either f of x is positive or f of x is negative. So, if it is positive then it is going to be bigger than 1 over n for some n. So, if x is positive and bigger than 1 over n then it is going to belong to 1 over n or if f of x is not 0 and it is negative that means it is negative. So, it is going to be less than 1 over minus 1 over n for some n. So, it belong to b n. So, every set n every point x in n either belongs to a n or belongs to b n and obviously if x belongs to a n or b n then f of x is not equal to 0. So, it belongs to n. So, n is equal to this. So, n is written as a countable union of sets and all of these are sets in the sigma algebra S and we want to show this union has got measure 0. So, in case mu of n is not 0 that will mean for some n either a n has got positive measure or b n has got positive measure because otherwise mu of n will be less than or equal to sigma mu of a n plus sigma mu of b n all of them equal to 0. So, let us write what we are saying is the following to show that mu of n is equal to 0. So, suppose mu of n is bigger than 0 then that implies then this condition implies there exist some n naught such that either mu of a n naught is bigger than 0 or mu of b n naught is bigger than 0 because if not then mu of n will be equal to 0. So, let us look at these conditions. Suppose the first one holds if mu of a n naught is equal is bigger than 0 then look at the integral then integral of f then integral of f over the set a n naught let us look at. So, let us look at integral of f over the set a n naught. So, that is equal to integral. So, which is same as integral chi a n naught times f d mu. Now on the set f now a n naught f is bigger than 1 over n. So, this is bigger than obviously integral 1 over n times n naught times mu of a n naught. So, let us observe that on the set a n naught outside a n naught this function is equal to 0 indicator function of a n naught times f that is 0 and on a n naught f is bigger than 1 over n naught. So, this function is bigger than 1 over n naught and outside a n naught is 0. So, this is going to be bigger than so it is bigger than integral over a n naught of 1 over n naught d mu. So, that is what we are saying. So, once that is true and this is nothing but this integral and that is bigger than 0. So, in case mu of a n naught is bigger than 0 integral of f over a n naught is going to be bigger than 0 which is a contradiction because which is not true because we are given integral of f over every set e is equal to 0 which is not true. So, if this holds then this is a contradiction similarly if this holds one can approve it is a contradiction then the integral of f over b n naught will be less than strictly less than 0 naught equal to 0. So, in either case both of these are not possible. So, our assumption that mu of n naught is bigger than 0 must be wrong and hence mu of n. So, implies that the measure of the set n is equal to 0 and n was the set where so n was the set where f of x is bigger than 0. So, this is set as what measure 0. So, this is what we wanted to prove. So, we approved the property that if integral of a function over f is an integrable function and its integral over e is equal to 0 for every e belonging to s and f must be equal to 0 almost everywhere. So, this is a very nice property and useful property. Now, let us look at the additive property of the integral over the sets. So, this is as now we have proved then f. So, this is another property that if f is integrable then f must be a finite number for almost all x. So, a similar argument as before. So, let us prove that property also. It says if f is integrable then this implies that mod f of x is finite almost everywhere x. So, let us write once again the idea is let us write the set x n to be the set where mod f of x is equal to plus infinity either. So, f of x is equal to infinity or equal to minus infinity. So, together we have put them together in the set n. So, to show that the set mu of n is equal to 0. So, once again if not let us write n as x belonging to x such that such that mod f of x is bigger than say some quantity is not equal to 0. So, if this is not equal to 0 and f of x n is the set where f of x is equal to plus infinity. So, let us write f of x bigger than n. Let us write a n to be the set where f of x is bigger than n. Then each set a n is in the sigma algebra. So, we want to sorry this is not required. So, let us this is this is not required because we want to just want to show that f is finite. So, now observe that integral of mod f d mu. I can write it as integral over n mod f d mu plus integral over n complement mod f d mu. Because n and n complement together make up the whole space and just now we observed that integral of mod f over a set E is a measure. So, integral of mod f over the whole space can be written as integral over the set of mod f over n plus integral of mod f over n complement and on n the set is on n mu of n is equal to 0. So, the first integral is 0. So, it is equal to 0 plus and sorry yes let us observe this is so. So, if mu of n is bigger than 0 then what will happen? So, let us assume mu of n is bigger than 0 then this integral is equal to this integral plus integral over n plus integral over n complement. So, that means integral of mod f d mu is always bigger than integral over integral over n mod f d mu. Because I am just so what we are doing so integral of mod f is integral over n plus integral over n complement let us just drop the second term. So, integral of mod f over the whole space is going to be bigger than integral over n f d mu. So, once that is true and on n the function takes the value plus infinity. So, this is going to be equal to plus infinity multiplied with mu of n. So, if mu of n is bigger than 0 then this will be equal to plus infinity if mu of n is bigger than 0 and that is a contradiction that is not possible not true as f belongs to l 1. So, this integral must be a finite quantity and here we are saying in that case it will be equal to infinity. So, that proves that if a function is integrable then it must be finite almost everywhere. So, now let us come back to the question if f is integrable and e belongs to the set s then the indicator function of e times f is integrable that we have just now observed. So, let us write nu tilde of e as before the integral of f over e and we observed that this number may not be a non-negative number. However, it still has the property something similar to that of countable additive property for measures namely. So, let us state that property that if you take sets e n's in the sigma algebra s which are pair wise disjoint. So, and e is the union of the sets then the claim is that the series which is integral of e i's f d mu summation 1 to infinity this series is absolutely convergent and if we write e as the union then integral of f over e is equal to summation of integral of f over e i's. So, essentially we want to say that the integral of f of a integrable function over a set e can be written as summation integral of e i's where e i's are pair wise disjoint and this is we are saying is always possible for f to be integrable function and because why we are saying absolutely convergent. So, here is one should note that when e is equal to union e i it does not matter whether you write as e 1 union e 2 union e 3 and so on or any other order say e 2 union e 1. So, it does not the union does not depend on the order in which you write the sequence e i. So, that means in this series the summation should not depend upon the order of the terms and all are non-negative that means we should prove that these are absolutely convergent and that is what we want to prove that if e i's are pair wise disjoint then the series integral over e i of f d mu summation 1 to infinity is a absolutely convergent series and this integral of f over e is summation of integral over e i's. So, let us prove this property. So, let us we have got e i is a sequence of sets in the sigma algebra they are pair wise disjoint is equal to empty set for i not equal to j and e is equal to union of e i's 1 to infinity. So, first we want to show so first claim is that the series summation i equal to 1 to infinity of integral over e i of f d mu this is absolutely convergent. Let us observe what is absolute convergent means that means absolute values of this terms that is a series of non-negative terms that must converge. So, for that let us note that absolute value of integral over e i of f d mu is less than or equal to integral of mod f over e i d mu. So, that is the property of integrable functions that absolute value of the integral is less than or equal to integral of the absolute value and for non-negative function. So, mod f is a non-negative function and e is a disjoint union of sets. So, that implies that integral over e i of mod f d mu if I sum it up i equal to 1 to infinity that is same as integral over e of mod f d mu and f being integrable that is a finite number. So, here we have used two things one for non-negative measurable functions integral over a set is a measure. So, integral of mod f d mu over e i summation 1 to infinity is equal to integral of absolute value of f over e and f being integrable this is finite. So, that proves that the series integral f d mu over e i is absolutely convergent because this sum is less than or equal to sum. So, from these two it implies that this series is absolutely convergent. So, once the series is absolutely convergent it sum is equal to sum of the partial sums. So, now we can easily write so this implies that the claim holds. So, the series is absolutely convergent hence integral over e f d mu is equal to limit n going to infinity of the partial sums. So, i equal to 1 to n integral of f over e i d mu and that is nothing but partial sums and that is same as saying that sigma i equal to 1 to infinity integral over e i of f d mu. So, that proves that now the integral of a integrable function over a set need not be measured, but it has we can say this is a countable property of this integral that integral over e is equal to summation of integrals over e i is whenever e is a union of pair wise disjoint sets e i. So, this is the property that we have just now proved. So, these are some of the properties that we have proved about the integral of integrable functions and now we want to prove an important property we want to analyze this sequences of integrable functions. So, we want to analyze if the property that if f n is a sequence of integrable functions and it converges to a function f can we say that f is integrable and can we say integral of f n's will converge to integral of f. We have seen that this need not be true even for non-negative functions, but there under some suitable condition we can say that integral of f n's will converge to integral of f and that is an important theorem called Lebesgue's dominated convergence theorem. So, let us prove interchange of integral with the limits and look at the theorem called Lebesgue's dominated convergence theorem. So, it says let f n be a sequence of measurable functions such that there exists a function g which is integrable with the property that integral of mod f n's are less than or equal to g for almost all x in mu for all n. Then the claim is if f n's converge to f almost everywhere then the limit function is integrable one and secondly integral of the limit function is equal to limit of the integrable functions. So, let us observe once again what is given and what is true we are saying f n is a sequence of measurable functions and all the f n's are dominated by a single function g which is integrable and this dominance could be almost everywhere. So, all the f n's are dominated by a single function g and so the conclusion is if f n's converge to f then f is integrable and integral of f is equal to limit of integral of f n's. So, this is what is called dominated convergence theorem and it is an important theorem. So, let us prove this theorem. So, we are given that all the f n's are measurable for n bigger than or equal to 1 mod f n x is less than or equal to g of x for almost all x and for every n and we are given that f n x converges to f of x almost everywhere. So, to prove the required claim for the time being let us assume that this almost everywhere is everywhere that the proof is not going to change much. So, we will see that improvement. So, let us assume for the time being f n x is less than mod f n x is less than or equal to g of x where g is l 1 is a integrable function. So, that implies that integral of mod f n x d mu x is less than or equal to integral g x d mu x and g is integrable. So, that is finite. So, that implies that each f n is a integrable function. Also, mod f n converges to mod f because f n converges to f and each mod f n is less than or equal to g. So, that implies that mod f n x is mod f n x is less than or equal to g and that converges. So, that implies that mod f x is also less than or equal to g of x for every x. So, that once again implies integral mod f d mu is less than integral g d mu which is finite. So, this again implies that f is in l 1 of mu. So, under the given conditions we have shown that if f n are dominated by an integrable function and f n is converged to f then f is a integrable function. So, to look at the limits let us observe. So, note f n converges to f and mod f n is less than or equal to mod g. So, this implies that look at the sequence f n. So, look at the sequence f n minus g. So, look at the sequence of measurable functions of course they are mod f n is less than or equal to g. So, this will be negative. So, let us look at we want non-negative. So, let us look at g minus f n look at this sequence instead. So, this is a sequence of non-negative measurable functions because mod g is bigger than mod f n is bigger than or equal to g that is given. So, g minus f n so is a sequence of measurable functions and let us observe g minus f n is bigger than or equal to 0 because g is bigger than or equal to f n. So, this is a sequence of non-negative measurable functions and so let us write that g minus f n is a sequence of non-negative measurable functions and it converges to g minus f because f n converges to f. So, now we can apply Fatou's lemma. So, implies by Fatou's lemma that integral limit inferior of g minus f n d mu will be less than or equal to limit inferior of integral of g minus f n d mu. So, that is application of Fatou's lemma. So, recall we had Fatou's lemma which was applicable for functions which are not necessarily sequence of functions which is not necessarily increasing. So, look at this now let us compute both sides. So, what is the left hand side? So, this is equal to integral limit inferior of g minus f n is g minus limit inferior plus limit inferior of minus f n. So, that is the left hand side d mu and that is equal to integral of g. So, look at that is equal to integral of g d mu and what can we say about this? f n is all of integrable functions so everything is finite. So, this limit inferior of minus f n is equal to minus limit superior of f n d mu. So, this is a property of limit superior and limit inferior that limit inferior of minus f n is equal to minus of limit superior. So, this is equal to integral of g d mu minus integral limit superior of f n. So, limit f n is convergent so limit superior is same as f of x d mu x. So, that is the left hand side and let us see what is the right hand side once again so limit inferior of integral. So, this is equal to limit inferior of integral of g minus f n is integral g and that does not depend upon limit. So, it is integral g d mu and then limit inferior of minus so that will be minus limit superior of integral f n d mu. So, from these two so this is less than or equal to this so what does that imply? So, that implies that integral g d mu minus integral f d mu. So, I am just writing integral this is less than or equal to integral g d mu minus limit superior of integral f n d mu. So, everything is finite so I can cancel out this and negative sign gives you other inequality. So, the implies integral f d mu is bigger than or equal to limit superior of integral f n d mu. So, looking at the sequence g minus f n we got this is the g minus f n is non-negative converges to g minus f gives us this. Similarly, if I look at the sequence g plus f n that is again a sequence of non-negative measurable functions and application of Fatou's lemma. So, Fatou's lemma will give me that integral integral of f d mu is less than or equal to limit inferior of integral f n d mu. So, a similar application of Fatou's lemma to this sequence will give me this. So, 1 and 2 so 1 plus 2 together imply that integral f d mu is bigger than limit superior that is always bigger than limit inferior and that is bigger than integral f d mu implies that the sequence so limit integral f n d mu exists this limit exists and is equal to integral f d mu. So, that proves a dominated convergence theorem. So, the proof of dominated convergence theorem is essentially very simple it is just a set forward application of Fatou's lemma because mod f n is less than or equal to g implies g minus f n and g plus f n both are sequences of non-negative measurable functions. So, apply Fatou's lemma and you have the conclusion that integral of f is equal to limit of integral of f n. So, we have proved this under the conditions that f n converges to f n x converges to f of x and f n x is dominated by g of x for every x. The modification for this for almost everywhere things is simple and we will do it next time. So, thank you very much.