 Welcome to lecture number 36 on measure and integration. In today's lecture, we will be looking at the fundamental theorem of calculus for algebraic integrals. If you recall, we had started our lectures with a motivation saying that for Riemann integrable functions, the importance of fundamental theorem of calculus is there and that is why it is easier to integrate functions whose derivatives are known. However, after Riemann extended this notion of integral beyond continuous functions, the fundamental theorem of calculus no longer remains true if the integrand in the Riemann integral is not continuous. So that made the class of functions which were integrable was extended, the fundamental theorem of calculus no longer was true. So, efforts were made to extend the notion of integration and that is how we got the notion of integral, Lebesgue integral. And today, we will show how fundamental theorem of calculus holds for Lebesgue integrable functions. So, today's topic is going to be fundamental theorem of calculus for Lebesgue integral. Let us just recall fundamental theorem of calculus for Riemann integrable functions. It says, if you are given a pair of functions small f and capital F defined on a interval a b, then they satisfy the relation that f of t minus f of s is equal to the integral s to t of f x d x for every pair of points s less than t in a b, if and only if the function f is continuous and the function capital F is the derivative of capital F is small f. So, this equation that f of t minus f of s is equal to integral from s to t of f of x d x holds, if and only if the function f is continuous, the function capital F is differentiable and the derivative is equal to small f. So, in a sense, this relates differentiation with integration. So, it essentially says that for every continuous function, its indefinite integral is differentiable and if you integrate the derivative, you get back the function. So, we would like to analyze this theorem for Lebesgue integrable functions. So, we are going to look at a function f, which is integrable on the interval a b. So, let us fix a function f, which is integrable on a to b and look at the indefinite integral, namely capital F of x is equal to integral on the interval a to x of f t d lambda t, the integration being with respect to the Lebesgue measure lambda. So, what we want to do is, this function capital F is called the indefinite integral of the function small f. So, we would like to analyze the properties of this function capital F in detail. So, to analyze these properties, we have to look at some classes of functions. We will make the definition first and then come back to this function capital F. So, let us take a function f, which is defined on an interval a b, taking real values and let us take a partition p of the interval a b, partition being starting with a equal to x 0 and the point x 1, x 2, x n are equal to b. So, the p is a partition with points a and b and in between points are x 1, x 2 and x n. So, let us take a partition of this interval a b and define what is called the variation of the function f over this partition in the interval a to b. So, this is denoted by capital V lower a upper b indicating that we are over the interval a to b with respect to a partition p of the interval a b of the function f. So, what is this quantity? So, you look at the variation of f on each sub interval of the partition. So, look at f of x k minus f of x k minus 1 absolute value of this. So, this is how the function varies on the interval x k minus 1 to x k. So, look at this number absolute value of f of x k minus f of x k minus 1 and look at the summation of all these variations 1 to n. So, this is how the value of the function changes at the end points of the partition. So, this number is called the variation of f over the interval a b with respect to the partition p. It depends on the partition the points and on the interval of course, a to b. Then we define look at the supremum of the all these variations of the function f over the interval a b with respect to various partitions of the interval a b. So, this supremum is called the total variation of f over the interval a b and we denoted by V a b of f. So, V a b of f is nothing but the supremum of the variations with respect to various partitions V a b of f with respect to p. Now, obviously, this number is a non-negative number. It could be equal to plus infinity because it is a supremum of non-negative numbers. So, we say that the function has bounded variation if this number V a b f which is a supremum of the variations with respect to various partitions is a finite quantity. So, we say a function f defined on an interval a b is a bounded variation if the supremum of the variations of the function f with respect to all the partitions the supremum of this is a finite quantity. So, we will look at some examples of functions of bounded variation. Obviously, every monotonically increasing or decreasing function is a function of bounded variation. So, let us just check that. Let us take a function f defined on interval a b taking values in R and let us say f is monotonically increasing. Let us take a partition p of a b. So, let us a equal to x 0, x 1, x n equal to b. So, this is a partition of the interval a b. So, we want to look at what is f of x i minus f of x i minus 1, x i minus 1 absolute value of this. But this quantity because f is monotone. So, this is same as f of x i minus f of x i minus 1 because it is monotone. So, this value is bigger than this. So, absolute value is same. So, if we take summation on both sides 1 to n, so summation i equal to 1 to n. So, that will give us all consecutive terms will cancel. So, that will be f of x n minus f of x 0 which is equal to f b minus f of a. So, that means that this variation of the function with respect to any partition is equal to f b minus f of a. So, this implies that f is of bounded variation because the supremum also will be equal to this. So, this is every monotonically increasing function is of bounded variation. Let us check that if f and g are two functions on a b defined on the interval a b, then the function f is of bounded variation, then f itself is a bounded function. So, let us check that also that every function of bounded variation is a bounded function. So, f on a b to r, we are given that the variation of f from a to b is finite. So, let us take, so this is interval a, this is b, so here is a, let us take any point x. So, look at f of x. So, this f of x, absolute value of this is less than or equal to f of x minus f of a plus f of a by triangle inequality. Now, if I look at the partition of the interval a b by three points a, x and a, then what will be the variation with respect to that partition? That will be f of b minus f of x, absolute value plus the absolute value of f of x minus f of a. So, this first term will be less than or equal to variation of the function with respect to that particular partition plus mod f of a, where what is the partition p? So, p is the partition of three points a, x and a. So, this will be one of the terms in that variation, so it will be less than or equal to this and that is less than or equal to the variation over a b of f, so plus mod f of a. So, if f is a function of bounded variation, then mod f of x is less than or equal to the total variation plus this quantity, which is a finite quantity. So, let us call it as some number m. So, this is less than or equal to, so we can call this number as m. So, this implies f is bounded, so implies f bounded. So, every function of bounded variation is also a bounded function. Next, let us look at the class of functions of bounded variation and we want to check if f and g are of bounded variation, then so are the functions f plus g, f minus g, f into g and alpha f for every alpha belonging to R. That means the class of functions of bounded variation is a nicely behaved class and these properties are all easy to check. Let us just check, say one of the properties, say f g. Let me check only one of them, say f and g on a b to R, such that the variation of f is finite and g also has bounded variation. So, let us take any partition p of a to b, say x 1, x 2, x n minus 1 and x n and let us analyze f g of x i minus f g of x i minus 1. So, we want to look at this quantity and we are given f and g are functions of bounded variation. So, since f is a bounded variation, so that is also a bounded function. So, we can say this is less than or equal to, so let us assume that mod f x is less than or equal to m for every x. So, let us assume this quantity, because f is a bounded variation, so it is bounded. So, we can write this quantity as f of x i mod of x i less than or equal to or there is no need to write that. Let us just, because f g into f of x i is f of x i and g of x i and f of x i is less than or equal to m. So, I can write this is less than or equal to m times g of x i minus g of x i minus 1. So, that will be less than or equal to m times. So, when we take the summation on both sides, summation i equal to 1 to n, so this will be less than or equal to summation i equal to 1 to n. So, that will mean that the variation of a to b with respect to p of f g is less than or equal to m times the variation of the function g over this. So, this happens for every, so implies that f g is of bounded variation. So, basically these are all easy properties to check that if f and g are a bounded variation, then f plus g is of bounded variation and f minus g the product alpha times f all are functions of bounded variation. So, here are some facts about the functions of bounded variation which are of importance. So, we will prove 1 by 1. So, let us take a function f of bounded variation. Then for any two points x and y in a b with x less than or equal to y, let us denote the variation of f over the interval x to y, if x is less than y and if is equal then we denote it y equal to 0. So, v x y v lower x upper y denotes the variation of the function f in the interval closed interval x y, if x is strictly less than y, if x is equal to y obviously we write it as equal to 0. Then it has the following properties. The first property is that the variation of the function over an interval is additive. That means if we take the variation of the function over the interval a to b, then it is same as the variation over a to c plus the variation over c to b. So, let us prove this property. So, here is the interval a to b and here is the point c. So, we want to check that the variation over a to b of f is equal to variation over a to c of f plus variation c to b of f. So, to check that let us start, let us take a partition p of a b. So, take a partition p of a b. Then this gives us, so let us look at variation of a to b over p of f. So, that will be equal to summation. So, let us say the partition points are x i minus f of x i minus 1 i equal to 1 to n absolute value of this. Now, this partition may or may not have the point c inside it. So, in any case let us insert the point c inside this, even if it is not there, that means let us say it lies it in x i minus 1 and x i. So, introduce the point in the partition, new partition. So, then in that case p can be written as p 1 union p 2, where p 1 is a partition of a to c and p 2 is a partition of c to b. So, we look at the partition and divided into two parts, the partition p 1 in the points lying in a to c and we add the point c here and p 2 is here. And then this, if we introduce that point c in between, so that will tell us that this quantity is less than or equal to, so this is less than or equal to sigma i equal to 1 to that point, so 1 to that point let us call this point as c as something. So, that is, so or essentially it is same as the variation over a to b of p 1, a to c of p 1 of f plus the variation c to b of p 2 f f less than or equal to because when you introduce the point c here, so f of x i minus f of c plus f of x i minus 1 plus. So, this sum the i th absolute value sum will spread into two parts, which is less than or equal to one term will be accommodated here, the other will be accommodated here. So, as a consequence of this, we get that, so because of this, we will have that v a b of partition with respect to the partition p is less than or equal to, this is p 1 and p 2 are particular partitions, so is less than or equal to c of f plus v c b of f. So, we get that the variation with respect to any partition p over the interval a b is less than or equal to the variation, total variation over a to c plus the total variation over c to b and this happens for every partition, so that implies that the supremum, which is nothing but the variation of f over a b is also less than or equal to variation over c of f plus variation c to b of f because this happens for every partition, so we can take the supremum and that is also, so we get one way inequality, namely the variation of f over a to b is less than or equal to the variation a to c plus variation c to b. To prove the other way round inequality, let us fix epsilon greater than 0 and select partitions say p 1 of a to c and p 2 of c to b, such that what happens, so v a c of f is the supremum, so minus a small number epsilon by 2 cannot be the supremum, so there must exist a partition p 1 of a to c, so either this quantity is less than the variation a to c of p 1, p 1 of f and similarly for the second one, so while we are just using the fact that v a c or v c b of f is the supremum, so similarly we will get a partition p 2 of c to b such that this is less than or equal to the variation of the function with respect to the partition p 2, so that will give us, so adding these two equations we will get, so if we add these two equations, we will get v a c of f plus, so adding these two equations will give us this plus v c b of f minus epsilon is less than or equal to v a c p 1 f plus v c b p 2 of f and now we realize that the partition p 1 and p 2 put together, these two quantities are nothing but variation of a to b of p 1 union p 2 of f, so we have a partition p 1 on a to c, we have a partition p 2 on c to b, so these two variations put together give us the variation with respect to the partition over the whole interval which is the union of the two a b and that is less than or equal to v a b of f because it is the supremum. So, we get for every epsilon the v a c, the variation over a to c plus the variation over c to b minus epsilon is less than this, so epsilon is arbitrary, so that implies v a c of f plus v c variation over c to b is less than or equal to variation a to b of f, so that proves that the variation behaves nicely namely it is, so this proves the first property namely that the variation is additive over the interval which we are doing, so variation over a to c plus variation c to b is equal to the variation over a to b of f, so the next property we want to check is the following namely that if we take the variation of the function on the interval a to x, where x is varying in the interval a to b then it is an increasing function and that property is quite easy to verify, so let us just verify that, so we have got the interval a to b and here is the point x, so we want to check that v a x of f is increasing, so let us take a point, so let x be less than y, so here is a point y, so then for any partition for every partition p of a to x, if we look at p union the point y that gives us is a partition of that is a partition, so call it as call this as p 1 is a partition of a to y, so thus the variation over a to x of f with respect to this partition p any partition p of a to x, we are joined the point and get a partition, so that will be equal to the variation a to x of p plus, so we can write it, it is less than or equal to plus mod f y minus f of x, because this is variation of p with respect to f same thing plus I did, but this is nothing but the variation of a to y of p 1 of f and that is less than or equal to variation a to y of f, so every partition p of a to x, we look at the variation a to x of that is less than or equal to the variation a to y of x, so we can take the supremum, so that implies, so that will imply that variation a to x of f is less than or equal to variation a to y of f, so the variation is an increasing function, we could have also looked at in another way by the previous property, the variation over a to y of p is equal to variation a to x of p plus variation x to y of p, by identity property we could have done that and now noting that this quantity is always bigger than or equal to 0, so it is bigger than or equal to a to x of p, so that is another way of proving that v a x of f is increasing, so v a of f, not p of f, so variation being additive, so that will give us this, this is an increasing function, so that proves property 2 and let us look at the third property which says that for any point x and y, mod of f x minus f y is less than or equal to the variation between a to y minus the variation from a to x, so that property also is quite easy to verify, so let us we want to, so we have got the interval a to b and we have got the points x and y, so let us look at, we note that f y minus f of x is less than or equal to mod of f y minus f of x and that is less than or equal to v a y of f minus v a x of f, so that is just now we verified that, so this is less than or equal to, so that says that the mod of f y minus f of x is less than or equal to this quantity, so that is just now we have verified that, so that is easy to verify also. Next let us look at the property that the function v a x f minus f of x is an increasing function, so that also follows from the inequality just now proved, namely just now we approved the fact that f y minus f of x is less than or equal to v a y of f minus v a x of f, so that says that v a y of f minus f of y is bigger than or equal to, this is bigger than or equal to v a x of f minus f of x, so that says that this function is bigger than or equal to this function as a function of x, so this is an increasing function, so these are the properties of the variation of the function over the interval and as a consequence of all these properties, we get an important theorem namely that the characterization of functions of bounded variation, we saw in the beginning that any monotone function is a function of bounded variation and we also saw that the difference of and sum of two functions of bounded variation is a function of bounded variation, so sum and differences of two monotone functions will again be function of bounded variation and there is a characterization by Jordan which says the following, namely that if f is a function, then f is a bounded variation if and only if it is a difference of two monotonically increasing functions, so one way of this property we have already checked, because if f is a difference of two monotone functions, then each monotone function being of bounded variation, the difference will be again of bounded variation. So, let us and the converse is also easy to prove, so we want to prove the converse namely that if f is of bounded variation, then it is a difference of two monotone functions, so that is easy because we can define g of x to be equal to the variation of the function on the interval a to x and just now we observed that the function v a x, the variation over a to x minus f of x is a increasing function, so and if we take the difference of these functions g and h that is precisely f. So, every function f can be written as a difference of two monotonically increasing functions, so that is gives an important theorem of characterization of functions of bounded variation, namely a function is a bounded variation if and only if it is a difference of two monotone functions. Here is an important theorem due to Lebesgue that every monotone function is a different function which is differentiable almost everywhere. Let us just recall that in your courses in analysis, you must have seen this fact that every monotone function is a function which is continuous everywhere except at countable number of points, meaning that a monotone function can have discontinuities which are at the most countably many and you know that a countable set is a set of measure 0 is a null set. So, one can say that elementary property about monotone functions that every monotone function is continuous almost everywhere and this is a far reaching generalization of this fact that not only every monotone function is a function which is continuous almost everywhere. In fact, one proves that every monotone function is differentiable almost everywhere and this theorem, the proof of this theorem is quite technical. So, we will not be proving this theorem. We will assume this theorem for our discussion today. Those who are interested in looking at the proof of this can look up the proof in the textbook. It is given in all details there. So, the theorem says that every monotone function is a function which is differentiable almost everywhere. So, to analyze the properties of that indefinite integral, let us define another class of functions which are called absolutely continuous functions. So, we say a function g is absolutely continuous if for every epsilon bigger than 0, there is a delta bigger than 0 such that whenever any finite number of collections are mutually disjoint sub intervals a i b i are given in a b such that the total length of this intervals is less than delta then the total variation of g over the end points of this intervals. So, namely summation of g b i minus g of a i is less than epsilon. This definition looks very much like continuity. It says or actually is very much similar to absolute continuity. So, if we take when i is equal to 1, so that says whenever two points x and y a and b a i b i are closed then little by f of g of b i minus g of a i is closed. So, it is a generalization of the notion of uniform continuity. So, it says let us look at once again. So, it says that for any given epsilon there must exist a delta such that whenever there are disjoint intervals a i b i of total length less than delta then the variation g b i minus g a i summation 1 to n is less than epsilon. So, this property of a function is said to be absolute continuity of the function. Clearly, every function absolutely continuous function is also uniformly continuous. Let us look at the indefinite integral of an integrable, Lebesgue integrable function on a to b f x is equal to a to x f t d lambda t. We want to prove that this function is absolutely continuous. So, let us prove the fact that if f is l 1 of a b and we define capital F of x a to x f t d lambda t then this function capital F is absolutely continuous. So, to prove this we will do it in two steps. Let us first assume the case 1 namely f is bounded f is a bounded function. So, let us say mod of f of x is less than or equal to m for every x belonging to a b and let us take a partition p of a b. So, a equal to x 0 less than x 1 less than x n equal to b. Then let us observe what is f of x i minus f of x i minus 1 absolute value of this. So, that is equal to absolute value of integral a to x i f d lambda minus integral a to integral a to x i minus 1 of d lambda absolute value of this. And now let us note that this right hand side is integral a to x i minus a to x i minus 1. So, this is x i minus 1 and this is x i and this is b. So, integral a to x i minus 1 is nothing but the integral over this portion. So, this we can write f of x i minus f of x i minus 1 absolute value. We can write this is equal to absolute value of integral x i minus 1 to x i f d lambda. And now using the fact that absolute value of the integral is less than or equal to integral of the absolute value, we can write this as integral over x i minus 1 to x i mod of f d lambda. So, for every i we have this. So, that implies that summation i equal to 1 to n of f x i minus f of x i minus 1 absolute value will be less than or equal to summation i equal to 1 to n integral over x i minus 1 to x i mod f d lambda. So, this is for any points x i y i we will have this property. And now because f is bounded, so we can say this is also less than or equal to mod f is less than or equal to m. So, m times sigma i equal to 1 to n mod x i minus x i minus 1. So, for any set of points we get this property. So, as a consequence, so let us choose delta bigger than 0 such that m times delta is less than epsilon. So, for a given epsilon we will select delta such that this is true. Then sigma of b i minus a i, so if less than delta will imply that just now what we showed that i equal to 1 to n mod of f of b i minus f of a i will be less than or equal to m times summation i equal to 1 to n b i minus a i which is less than m times delta less than epsilon. So, we will get whenever we have intervals a i b i say that sigma of the length of the intervals is less than delta the variation at the end points of this intervals f b i minus f a i is less than epsilon. So, that will show that f is absolutely continuous. So, f is absolutely continuous if small f is bounded. So, that is case 1. Let us look at the second general case. In general, when f is in l 1 of a b, we can approximate it by bounded function. So, let us define f n to be equal to let us define f n as mod of f the maximum value of the minimum value of mod f and the number n. So, whenever the graph of mod f goes above n, we take the value as n. So, cutting the graph at the line equal to n. Then, this is for every n bigger than or equal to 1, then each f n belongs to l 1. Each f n is less than or equal to mod f and f n converges to mod f. These are all easy to verify because n increases. So, f n will converge to mod f. So, implies by dominated convergence theorem that integral of mod f n d lambda a to b converges to integral a to b of mod f d lambda. Not only that, we can even write this happens not only over the whole of it. We can say this also. In fact, this is also true that integral over e of mod f n d lambda converges to integral over e of mod f d lambda for every set e, borel set, Lebesgue measurable set inside the interval a to b. So, we will use this fact and note that each f n we have defined is a bounded measurable function. So, using this fact, we will have, so let us fix. So, for e contained in a, b fix, we can find n naught such that integral of mod f over e d lambda minus integral over e of f n naught d lambda is less than epsilon. So, for a given epsilon, we can find n naught say that this is true. So, then that is same as saying that integral over e mod f minus f n naught d lambda is less than epsilon. Now, let us analyze integral over e of mod f d lambda. So, that is less than or equal to, we can write as integral of mod f minus f n naught d lambda over e actually equal to plus integral over e of f n naught d lambda. Now, let us observe that the first integral is less than epsilon. So, it is less than epsilon plus integral over e of f n naught d lambda and f n naught is a bounded function. Actually, it is bounded by f n naught is bounded by the number n naught. So, that says that this is less than or equal to epsilon plus n naught times lambda of e. So, what does it mean? So, this means that if lambda, if n naught lambda of e is less than epsilon, then integral over e mod f d lambda will be less than 2 epsilon. So, this is for any set e. So, now we will apply it to our case. So, let us consider. So, let us take let epsilon be given. So, select delta bigger than 0 such that lambda of that summation b i minus a i equal to 1 to n is less than delta. Let us select delta bigger than 0 so that this lambda naught of e. So, let us such that this is less than n naught of this is less than epsilon. So, what we are saying is in a sense that take the set e to be the finite disjoint union of those intervals a i b i to be less than epsilon. Then what will happen? Then whenever these intervals are disjoint and the total length is less than less than delta. So, this is less than delta. Then we will have that integral of mod f will be less than epsilon. So, this will give us the following fact namely. So, thus for every epsilon bigger than 0, we can find delta bigger than 0 such that. So, in fact, delta bigger than 0 such that for intervals a i b i i equal to 1 to up to n disjoint in a b, if we take e to be that set union of a i b i over i equal to 1 to n, then that will give us that such that the disjoint intervals say that for disjoint intervals, if the sigma b i minus a i n naught right. So, that was n naught. So, this is n naught. So, that is equal to n naught is less than delta is less than epsilon that will imply. So, this is less than delta. So, is that this is less than delta say that n naught delta is less than epsilon. So, this is equal to n naught delta less than epsilon, that will imply integral of mod f over e d lambda is less than epsilon. And this will give us for the unbounded case that, so that will imply that if you look at f of V i minus f of A i summation i equal to 1 to n, that we know is less than or equal to summation i equal to 1 to n integral over mod f d lambda. And that is nothing but, so which is nothing but integral over the set E mod f d lambda, where E is the disjoint union of A i B i and that is less than epsilon. So, that will prove that the function is absolutely continuous. So, this will prove in both cases that the function capital F is absolutely continuous. So, this is an important example of absolutely continuous functions and if a function f is in L 1, then f of x, the indefinite integral is absolutely continuous. And every absolutely continuous function is a function of bounded variation. This is another obvious fact that every absolutely continuous function is a function of bounded variation. So, let us prove that that if f is A B to R and f is absolutely continuous that implies f is of bounded variation. So, let epsilon equal to 1 be given, select delta bigger than 0 such that whenever A i B i's are disjoint i equal to 1 to n disjoint in A B with sigma B i minus A i 1 to n less than delta. That will imply that the variation f of B i minus f of A i i equal to 1 to n is less than 1. So, that is less than 1. So, essentially that means that here is the interval A to B. So, whenever we have disjoint intervals in a length at the most delta, then the variation is less than 1. So, let us divide this interval. So, let P equal to A equal to x 1, x n equal to B be a partition of A B such that x i minus x i minus 1 difference is less than or strictly less than delta. Then by absolute continuity the variation of the function f in x i minus 1 to x i is less than epsilon, because wherever we take any partition of this any sub interval, then with the variation inside that by absolute continuity is less than is less than 1 epsilon equal to 1. So, that implies that the variation over A to B of f, which is summation i equal to 1 to n variation x i minus 1 to x i of f, which will be less than or equal to, if these intervals are n. So, that will be less than n, which is finite. So, that will prove that every function, which is absolutely continuous is also of bounded variation. And every function of bounded variation is a difference of two monodon functions and every monodon function is differentiable almost everywhere. So, we will get that this function f, which is an indefinite integral, is also differentiable almost. So, this is the beginning of the fundamental theorem of calculus for Lebesgue integrals. So, today we have looked at the property, namely if f is an integrable function and look at its indefinite integral capital f of x, then we have shown this function is absolutely continuous function and hence it is a function of bounded variation and as a consequence of bounded variation, there is a difference of two monodon functions and hence differentiable almost everywhere. So, this is the first, essentially the first part of the fundamental theorem of calculus saying that the indefinite integral is differentiable almost everywhere. We will continue this in the next lecture and we will show, in fact, that not only the indefinite integral is differentiable almost everywhere, the derivative is actually equal to the function, the integrand, namely the function f, almost everywhere. So, that will be the first part of the fundamental theorem of calculus. Thank you.