 Welcome to lecture number 33 on measure and integration. From today onwards, we will be looking at some special topics in measure and integration. We will start with looking at how to define integral for complex valued functions defined on measure spaces. So, topic for today's discussion is going to be integrating complex valued functions. So, for the coming one or two lectures, we will be fixing a measure space X as mu, which is sigma finite and complete. So, all the discussion will be on a fixed sigma finite complete measure space X as mu. We will denote by this letter c with a line in between that is called script c to be the field of complex numbers. For a complex valued function f defined on x taking values in c, we say that we define its real part and imaginary part. So, for any point x f of x is going to be a complex number because its values are in the complex field. So, that complex number f of x has got a real part and imaginary part. So, we define real f at the point x to be the real part of the value f of x. Similarly, imaginary f at x to be the imaginary part of the value f of x. .. So, let us observe that x going to real f x and x going to imaginary f of x are real valued functions. The first one is called the real part of the function f and imf is called the imaginary part of the function f. So, every function f which is complex valued has got a real part function and the imaginary part function and both are real valued functions. So, let us define what is called the measurability of a complex valued function. So, a complex valued function f from x to c is said to be measurable if both the real part of f and the imaginary part of f are measurable functions. So, note that f is equal to the real part f plus imaginary part of f. So, if both real part of f and imaginary part of f which are real valued functions are measurable on x with respect to the sigma algebra s, then we say that the function f itself is a measurable function. We will say f is integrable with respect to the measure mu if both the real part f and imaginary part f are integrable functions. So, real part f is a real valued function and imaginary f is also a real valued function. So, if both of them are integrable as real valued functions, then we say that the function f which is complex valued is integrable and we define the integral of f to be denoted by the symbol integral f d mu to be equal to integral of the real part of f plus i times the integral of the imaginary part of f, where this i is the square root of minus 1 which is normally used to write complex numbers. So, for a complex valued function f, we define it to be measurable if both real part and imaginary part are measurable and similarly we define f to be we say f to be integrable if both the real part and the imaginary part of f are integrable and that case we define the integral of f to be equal to integral of the real part plus i times the integral of the imaginary part of the function f. So, that is the definition of the integral of a complex valued function. .. We will study properties of this integral. Let us also denote the set of all complex valued integrable functions on x by l 1 x s mu. That is the symbol we are used to denote the integrable functions real valued integrable functions. We will denote the space of complex valued integrable functions on x s mu by the same symbol. In case we are referring to specifically the real valued functions, we will put a suffix r l lower 1 upper r x s mu. So, that will denote the space of all real valued integrable functions on x s mu. So, whenever need be and we want to specify that we are in the space of real valued functions integrable we will use this symbol. Otherwise the space of all complex valued integrable functions will be denoted by the symbol l 1 x s mu. So, next we will like to study properties of integral namely we will like to study the space l 1 of x s mu the space of integrable functions. So, let us start with the very basic property. So, let us take a function f which is a measurable function f is a measure complex valued measurable function on s. Then the first property as for the case of real valued functions is that f is integrable if and only if mod f which is a real valued function is integrable with respect to is integrable real valued integrable function. And further in that case we want to claim that the integral of mod f modulus of the integral is less than or equal to integral of the absolute value. So, let us look at how does one prove these properties. So, we have got a function f which is l 1 of x s mu. So, that means what is the definition of this? That says that the real part of f and the imaginary part of f are both integrable functions l 1 of x. Now, let us consider. So, look at the function absolute value of f. The absolute value of f is the real part of f plus the imaginary part of f square root. So, that is the definition of because it is a complex function complex valued function. So, absolute value of f can be written this way. And from here it is easy to see that this is always less than or equal to 2 times square root 2 times the absolute value of the real part of f plus absolute value of imaginary part of f. This is very easy in equality about complex numbers. So, it is basically saying that if you take a complex number then it is always less than or equal to square root. The absolute value of the complex number is always less than or equal to square root of 2 times the real part plus the imaginary part. One way of looking at this would be if you look at this definition. So, this is real part of f square plus imaginary part of f square. So, this will be less than or equal to 2 times the maximum value of real part or the imaginary part. And that is less than or equal to the absolute value of the real part plus the absolute value of the imaginary part. So, this term in the under the square root you can easily see it is less than or equal to 2 times the absolute value of the real part of f plus imaginary part of f. So, square root is less than or equal to this quantity. So, this is very easy in equality to prove for complex numbers. So, let me just write it as an exercise that you verify it yourself. Now, because f is integrable, so real part of f is in l 1, imaginary part of f is in l 1. So, these are both integrable functions. So, mod f is a non-negative real valued function less than or equal to 2 times an integrable function. So, this will imply that mod f belongs to is a real valued integrable function. So, this inequality implies that this is integrable with respect to mod f is. So, if f is integrable, so that implies mod f is a real valued integrable function. Let us look at the converse part. So, let us suppose that so conversely let mod f be integrable. So, is l 1 r of x. Then the real part of f absolute value is less than or equal to absolute value of f. That is a very simple state forward in equality that for any complex number the real part is less than or equal to absolute value of the. So, and similarly the imaginary part of f is also less than or equal to absolute value of f. So, these are real valued functions and they are less than or equal to a function f mod f which is integrable. So, that implies that the imaginary part of f and real part of f both are l 1 integrable functions, real valued integrable functions and hence this implies that f is integrable. So, we have proved. So, this proves so hence f belonging to l 1 of x if and only if mod f belongs to l 1 r of x. So, that proves the first part of the statement. So, f is integrable if and only if mod f is integrable. Let us look at the second part. So, we want to show now that mod of integral f d mu is less than or equal to integral of mod f d mu. So, this is what we want to show. So, let us write. So, let us denote by alpha the number on the left hand side. So, it is absolute value of f d mu. Note that f d mu is a complex number and its absolute value is denoted by alpha. So, we can write. So, let us write, let integral f d mu which is a complex number is absolute value is alpha. So, you can write as alpha times a raise to the power i theta for some theta between 0 and 2 pi. So, every complex number can be written as its absolute value at times e raise to the power i theta for some theta between. So, that implies that alpha is equal to e raise to the power minus i theta times integral f d mu times integral of f d mu. And this we will just show in the next property that integral of f a scalar multiple is same as I can write this as e raise to the power minus i theta times f d mu. We can write this as this. We will just in a minute we will prove this property that for any scalar complex scalar times the integral of f is same as integral of scalar times f whenever f is integrable. So, using that property we will have. So, we will just put it that we are going to prove this property soon. So, this is equal to this. So, let us write e raise to the power minus i theta f. It is a complex value function. So, it will have the real part. So, f 1 plus i times the imaginary part f 2. So, let this be equal to this. Then alpha which is equal to e raise to the power minus i theta f d mu its integral I can write it as integral can be written as integral of f 1 d mu plus i times integral of f 2 d mu. Because this function e raise to the power i theta f has got real part f 1 imaginary part f 2. So, its integral must be equal to integral of the real part plus i times integral of the imaginary part. But note alpha is real because what was alpha? Alpha was nothing but the absolute value of the integral of f d mu. So, it is a non-negative real number. So, because it is a non-negative real number and here we are writing it as an integral of f 1 d mu plus i times integral f 2 d mu. So, imaginary part must be 0. So, that implies that integral f 2 d mu is equal to 0 and alpha must be equal to integral of f 1 d mu. Alpha is non-negative. So, that implies I can write this also equal to absolute value of integral f d mu because alpha is a non-negative real number. So, that is equal to f 1 d mu and that for real valued functions we know this is less than or equal to integral of mod f 1 d mu. So, here we are using the property that for a real valued integrable function integral of f 1 d mu absolute value is less than or equal to absolute value of the integral. So, that is this. As a next step, let us observe. So, f 1 was real part of e raise over i theta times f. That was a real part of this function. So, that means what? That means that absolute value of f 1 is less than or equal to absolute value of e raise power minus i theta of f. So, real part of any complex number is less than the absolute value of that complex number. But e raise power i theta is a complex number of absolute value 1. So, this is same as mod of f. So, absolute value of f 1 is less than or equal to absolute value of f. So, that implies that integral of mod f 1 d mu is less than or equal to integral of mod f d mu. So, now let us combine these two facts that alpha was equal to less than or equal to integral of f 1 d mu and integral f 1 d mu is less than or equal to integral of mod f d mu. So, these two facts together imply that integral of f d mu which was equal to alpha is less than or equal to integral of mod f d mu. So, that proves the fact that f is integrable if and only if mod f is integrable and integral of absolute value of the integral f d mu is less than or equal to integral of the absolute value of the function. So, this proves the first property about integrals that for a integrable function f is integrable a complex valued function f is integrable if and only if absolute value of this complex valued function which is a real valued function is integrable and integral of absolute value of the integral f d mu is less than or equal to integral of f d mu. So, that proves the first property. Next we want to prove the linearity property part of which we have used already in the previous proof namely that if f and g are integrable functions complex valued integrable functions and alpha and beta are complex numbers then alpha f plus beta g the linear combination is also integrable and integral of alpha f plus beta g d mu is equal to alpha times integral f plus beta times integral g. So, to prove this inequality we will split the proof into two parts. So, as a first part let us prove for that if f belongs to l 1 of x and alpha is a complex number then alpha times f belongs to l 1 of x. Let us prove this part first because alpha times let us look at proof of this. So, let us write alpha as a plus i b it is a complex number. So, let us write it as a plus i b and so alpha f is equal to a plus i b times real part of f plus i times imaginary part of f. So, which on expansion I can write as a times real part of f minus when you multiply this with this minus b times imaginary part of f plus i times from here I will get b times real part of f and from the other one we will get a times imaginary part of f. So, the complex valued function alpha f is written as its real part is a real f minus b imaginary f and its imaginary part is b real f plus a imaginary f. Now, since alpha is integrable, alpha is sorry f is integrable implies that real f and imaginary f are both real valued integrable functions. So, they are real valued integrable functions. So, real f imaginary part f are integrable. So, a times real f is integrable, b times imaginary f is integrable. So, the difference is integrable. So, that means the real part of the function alpha f is also a real valued integrable function and similarly the imaginary part of alpha f which is b times real f plus a times imaginary f is also integrable because real f and imaginary f are real valued integrable functions. So, implies that alpha f is a l 1 function and what is the integral of it? .. So, I can write now the integral of this function. So, integral of alpha f d mu by definition it is integral of the real part plus i times integral of the imaginary part. So, it is integral of a real f minus b times imaginary f d mu plus i times integral of the imaginary part of the function which is b times real f plus a times imaginary f d mu. So, that is integral of alpha f. So, let us now use the properties that integrals is a linear operation for real valued function. So, the first integral I can write it as a times integral of real part of f minus b times integral of imaginary part of f d mu d mu. So, that is the first one plus i times b real part of f integral plus i times a integral of imaginary part of f d mu. So, using linearity we have split it into four parts, but now it is easy to check that this is nothing but a plus i b times integral of real part of f d mu plus i times integral of imaginary part of f d mu. Just open up the bracket and that is same as this. That means this is equal to alpha times integral of f d mu. So, we have shown that integral of alpha f d mu is equal to alpha times integral of f d mu. So, that is the first part of the linearity property that if I take a function f which is l 1 and multiply it by scalar alpha, then that is also integrable and integral of alpha f is equal to alpha times integral of f. Now, let us look at the second part of a requirement. Namely, if f and g belong to l 1 of x, then that implies that their sum is also in l 1 of x and the integral of f plus g is equal to integral of f plus integral of g. So, that is easy to verify because mod of f plus g, we want to show that is integrable. So, let us look at note. So, note as in real case is less than or equal to absolute value of f plus absolute value of g. So, by triangle inequality property, absolute value of f plus g is less than or equal to absolute value of f plus absolute value of g. So, that implies these are all real valued functions non-negative that will imply integral of f plus g d mu is less than or equal to integral mod f d mu plus integral mod g d mu and both of them are integrable. So, that is finite. So, that implies that f plus g is integrable is an integrable function. And to compute the integral of f plus g, that is simple thing. So, let us observe to compute the integral. So, we note that the real part of f plus g is nothing but real part of f plus real part of g. That is easy to verify and the imaginary part of f plus g is equal to imaginary part of f plus imaginary part of g. So, because f is integrable, the first one will give you that real part f plus real part g is a real valued integrable function. So, real part of f plus g will be integrable. Similarly, imaginary part of f plus g is also integrable and further we can write down the integrals. So, the integral of f plus g d mu by definition is equal to integral of the real part of f plus g plus integral of the imaginary part of f plus g by definition. But real part of f plus g is real part f plus real part g. So, this is the first one is real part of f plus real part of g d mu plus integral imaginary part of f plus imaginary part of g d mu. So, that follows from the simple result that we have just now shown. Now, these are all real valued functions. So, integration is linear. So, that splits into four integrals. So, integral of real part of f d mu plus integral of real part of g d mu. So, there is i here because real and imaginary plus i times integral of the imaginary part of f plus i times integral of imaginary part of g d mu. Now, the first and the third term combined together will give you that is equal to integral of f d mu plus integral of g d mu. So, integral of f plus g is equal to integral of f plus integral of g. So, that proves property three completely namely integral of f plus g is equal to integral of f plus integral of g. Let us look at some more properties. So, let us look at the property third property namely if f is an integrable function and e is any measurable set then the indicator function of e times f is also a integrable function and we write this as integral of the indicator function of e times f d mu is written as integral of f over e. So, that is one property and we want to show something more namely that if e and f are disjoint measurable sets then integral of f over e union e 2 is same as integral over e 1 plus integral over e 2. So, let us approve this property. Let us take a function f which is l 1 and e is a set which is measurable. Now, we want to look at e times indicator function of e times f. So, that is same as indicator function of e times real part of f plus i times imaginary part of f which I can write as indicator function of e times real part of f plus i times indicator function of e multiplied with the imaginary part of f. So, what we are saying is that for the function indicator function of e times f the real part is indicator function of e times v f part of f and this imaginary part is indicator function of e times the imaginary part of f. integral of. Because, real part of f is integrable, imaginary part of f is integrable. So, multiplying with the indicator function of E also gives them integrable. So, both real part of the indicator function of E times f is integrable, and the imaginary part of indicator function of E times f which is indicator function of E times imaginary part of f, that are real valued integrable functions. And, hence we can write the indicator function of E f, which we are denoting by integral over E of f d mu to be equal to integral of the indicator function of E of the real part of f plus i times integral over E of the imaginary part of f. And that if you recall, we had denoted it by integral over E of the real part of f plus at i or sometime also called data integral over E imaginary part of f d mu. So, what we are saying is the following, namely for an integrable function for f integrable and E belonging to a measurable set, integral of over E of f d mu is well defined and is nothing but integral over E of the real part of f plus i times the integral of imaginary part of f over E d mu and also d mu. Now, let us come to the second part. Now, suppose E 1 and E 2 are two sets, which are measurable and they are disjoint. E 1 intersection E 2 is empty. In that case, by the above claim integral over E 1 union E 2 of f d mu will be equal to integral over E 1 union E 2 of real part of f d mu plus i times integral over E 1 union E 2 of imaginary part of f d mu. Now, let us recall that for real valued integrable functions, integral over E 1 union E 2 whenever E 1 and E 2 are disjoint is nothing but the integral over E 1 plus the integral over E 2. So, this right hand side, I can use that property and write it further as, so that is equal to, I can write the right hand side as equal to integral over E 1 of real part of f d mu plus integral over E 2 of real part of f d mu. So, that is the first one plus i times the second integral gives me integral over E 1 of imaginary part of f d mu plus i times integral over E 2 of the imaginary part of f d mu. So, that is using the properties that for real valued functions, integral over E 1 union E 2 whenever they are disjoint spreads into 2 into the sum of that. So, now, I can combine real f 1 and imaginary f over E 1 together. So, I can write that as integral over E 1 of f d mu and the second combination will give me integral over f d mu over E 2. So, what we have shown is that integral over E 1 union E 2 whenever they are disjoint of a function f is integral over E 1 plus integral over E 2. So, that proves the required property 3 completely. We will extend this property a bit further. So, the next property is an extension of this property which says let f be a integrable function and let us take a sequence E n be a sequence of pairwise disjoint measurable sets and let us write E as the union of the sets E n. Then the claim is then the series look at the complex series summation 1 to infinity integral over E n of f d mu. This series is absolutely convergent and the sum of the series and implies as a consequence that integral of chi E times f is a integrable function and the integral of f over E is nothing but the sum of this series complex series sigma 1 to infinity integral of E n f d mu. So, to prove this property, we will be using the corresponding property for the real valued integrable function. So, let us write E is equal to union disjoint sets E i 1 to infinity where E i is belong to S and f is a integrable function. So, let us look at the series. So, the series we are concerned with is sigma i equal to 1 to infinity integral over E i of f d mu. So, we want to show that is absolutely convergent. So, that means what? That means we want to show that the series i equal to 1 to infinity absolute value of integral over E i of f d mu is convergent. So, this is what we want to show. But this note that just now we proved that the absolute value of the integral is less than or equal to integral of the absolute value. So, to prove it is convergent it is enough if we prove that the series. So, we will prove enough to show that the series i equal to 1 to infinity integral of absolute value of f d mu over E i is finite. But for that let us just observe that absolute value of f is a real valued integrable function and that we have already observed while discussing the integral of non-negative measurable functions that this is finite if f is mod f is integrable. So, this property is true by the property of real valued integrable functions because f in L 1 that is same as saying that mod f is a real valued integrable function and E i is being disjoint and we know this series is convergent. So, as a result we will have that the series summation over i integral over i E i's of f d mu is an absolutely convergent series. And now we only want to prove the claim namely, so claim is that integral over E of f d mu is equal to summation i equal to 1 to infinity of integrals over E i of f d mu. So, this is what we want to check. So, let us look at the partial sums of this series. So, let us look at integral over E of f d mu minus summation i equal to 1 to n integral over E i of f d mu. So, let us look at this absolute value of this. So, this is same as saying that this is this is integral of absolute value of the integral minus this sums. So, this we can write it as equal to absolute value of summation i equal to 1 to n of f over integral of f over E d mu minus integral over E i of f d mu. So, that is equal to this. Now, using the triangle inequality for the absolute value, I can write this as we can write this as less than or equal to summation i equal to 1 to n absolute value of integral over E f d mu minus integral over E i of f d mu. So, but that using the fact that absolute value of the integral is less than or equal to integral of the absolute value, I can write this is less than or equal to summation i equal to 1 to n and this is nothing but integral of chi E minus chi of summation. I think we may have to slightly modify the proof, but let us see this may also work out. So, E i absolute value times absolute value of f d mu. This I think will not work out. So, let us modify the proof a bit, because we want to say that as n goes to infinity, this goes to 0. This may not exactly happen in this case. So, let us modify the proof slightly. Let me go back to this step. So, this is what we want to analyze. So, to analyze this, we proceed as follows. Let us first write this is equal to, this is the finite sum i equal to 1 to n integral over E i and that we have already observed is equal to integral over f d mu minus. So, this I can write as integral over union of E i's i equal to 1 to n of f d mu. I can write it that. So, let us write this as absolute value of integral over E f d mu minus this sum. Let me write the integral over the indicator function of union E i i equal to 1 to n of f d mu. This I can write as the indicator function of E. So, this is less than or equal to integral over chi E of f minus chi union of E i i equal to 1 to n f d mu by using the fact that absolute value of the integral is less than or equal to. So, this is minus the union and the union I can write as the indicator function of E times f and now absolute value of the integral is less than or equal to integral of the absolute value. So, using that I come up to here and now let us observe that this where does this is a sequence of functions and where does it converge. So, let us observe this point that indicator function of union E i i equal to 1 to n f converges to the function indicator function of E of f. So, this converges to this function point wise. So, if I look at the difference. So, this difference. So, the chi E of f minus chi union of E i i equal to 1 to n f d mu f this goes to 0 as n goes to infinity. So, that is one observation. So, the integrant goes to 0 as n goes to infinity and this is less than or equal to. So, and this is absolute value of chi E f minus indicator function of union E i is 1 to n f d mu f is less than or equal to absolute value of the first one plus absolute value of the second function and both are less than or equal to f. So, 2 times mod f and mod f is a. So, this this function goes to 0 and this function is dominated by 2 times mod f which is a integrable function and mod f is a integrable real valued integrable function. So, by the dominated convergence theorem we will have that this goes to 0 as n goes to infinity. So, what we are saying is an application of dominated convergence theorem will tell me that. So, by dominated convergence theorem integral over E of f d mu minus summation i equal to 1 to n which we are trying to analyze absolute value of that. So, integral over E i of f d mu absolute value of that. So, this is what we wanted to show goes to 0 and this we got is less than or equal to. So, we got this is less than or equal to this integral and the integrant goes to 0 and is. So, integrant goes to 0 we showed that and is dominated by an integrable function. So, Lebesgue's dominated convergence theorem says this goes to 0 as n goes to infinity. So, that proves. So, hence integral over E of f d mu is going to be equal to summation i equal to 1 to infinity integral over E i of f d mu. So, this complex series converges and this is the sum. So, that proves the result for complex valued functions namely that if E is a disjoint union of sets measurable sets and f is integrable then integral of f over E is equal to summation of integrals over E n. So, these are the basic properties of complex valued integrable functions and one can actually also prove dominated convergence theorem for complex valued functions. Recall that for real valued functions we had three important theorems. One was the monotone convergence theorem and another one was for 2's lemma and the third one was dominated convergence theorem. The monotone convergence theorem was true for non-negative measurable functions and for 2's lemma also is valid for non-negative real valued measurable functions, but for complex valued functions no way of saying that a function is non-negative or something. So, you cannot expect monotone convergence theorem and for 2's lemma to hold for complex valued functions, but it is fortunately enough the dominated convergence theorem which is true for real valued integrable functions is also true for complex valued functions and let us look at a proof of that. So, the theorem says that if f is a sequence of actually measurable functions is enough complex valued measurable functions such that f converges to the limit f n converge to a limit f point wise almost everywhere and there is a function g which is integrable real valued integrable function such that mod of absolute values of f n's are dominated by g then the f is the limit function is integrable and its integral is equal to limit of the integrable. So, basically this is a straightforward application of. So, let us look at the proof of this. So, we are given that f n's is a sequence of measurable functions and mod f n's are dominated by a g which is in l 1 of r. So, that automatically implies is a non-negative function. So, this implies that f n is l 1 of x because integral of f n will be less than or equal to integral of g for real valued functions and that implies f n is l 1 also because f n x converges to f almost everywhere. So, we will imply that mod f is less than or equal to g almost everywhere and that also will imply that f is also an integrable function. Now, we want to prove that the integrals will converge. So, let us look at the integral of. So, look at the real part of f n real part of f n is dominated by mod f n is dominated by g. So, implies by dominated convergence theorem for real valued functions we get that the real part of f n integral d mu converges to integral of real part of f d mu. Similarly, the imaginary part. So, integral of the imaginary part of f n d mu converges to integral of imaginary part of f d mu and we already shown that for complex scalar multiplication the corresponding result told. So, i times that will converge to i times that and adding this to implies that integral of f n d mu converges to some of this limits. So, that is equal to integral of f d mu. So, what we have shown is that for complex valued functions monotone convergence theorem also holds. So, what we have done today is we have extended the notion of measureability and integrability from real valued functions to complex valued functions and we have shown that the integral for complex valued functions has same properties as that of real valued functions and monotone convergence theorem also gets extended. So, we will be using these results to in the next lecture to define what are called the p th power integrable functions and their properties. Thank you.