 So, let us just recall, we were looking at point y's and uniform convergence of sequences of functions. So, let me just recall that we said that f n is a sequence of functions defined on. And we say f n converges to f point y's if f n of x, of course, f is also a function from x to r, converges to f of x for every x belonging to the domain. And that means, should not forget what that means, means for every epsilon bigger than 0, there is a stage n naught, which may depend upon, in general it will depend upon epsilon and the point x, such that mod of f n x minus f of x is less than epsilon for every n bigger than or equal to that natural number n naught, which may depend upon n and x. So, that stage may depend on epsilon of course, and also on x. So, then we defined what is called f n converges to f uniformly. If same thing, if for every epsilon bigger than 0, there is a stage n naught, which now will not depend on the point, but it will depend only upon epsilon. There is a stage, there is a natural number n naught, such that f n x minus f of x is less than epsilon for every n bigger than n naught and for every, so probably we should mention that for every x this happens. So, we gave a lot of examples of sequences, which converge point wise, which converge uniformly and we had started looking at properties of uniform convergence. We said that point wise convergence need not preserve various properties, namely if each f n is continuous, f may not be continuous, each f n is differentiable, then f may not be differentiable and if each f n is integrable, f may not be integrable. So, to analyze these properties for uniform convergence functions, let us recall, if f is a function defined on x to r and f is bounded, we defined what is called the infinity norm, that is supremum x belonging to x of mod f x. We observed that this is a norm and gives a metric r, so I think we called it b x r on the set of bounded functions with domain x and taking values in r. So, what is that metric? That metric is d infinity f g is equal to norm of f minus g. So, the relation of uniform convergence with this norm is, let us put that as a theorem, we proved it last time that if f n, if f n x to r are bounded and f x to r is also bounded, then f n converges to f uniformly, f n only if norm of f n minus f goes to 0. So, we had proved this theorem. Let me also point out that here, so let us observe a few things, some observations. One, if f n converges to f uniformly and each f n is bounded, then f is also bounded. So, if a sequence of functions converges uniformly to some function f and each f n is bounded, then the function f also is bounded. Hence, you can apply the earlier criteria because there we assumed f is also bounded. So, one way if f is uniformly convergent, each all f n's are bounded, then f is also bounded and hence you can write and hence norm of f n minus f converges to 0. That is the consequence of earlier theorem. So, let us prove this. So, f n converges to f uniformly. So, that means, what does that mean? That means for every epsilon bigger than 0, there is a stage n naught which depends only on epsilon such that absolute value of f n x minus f of x such that for every x belonging to x is less than epsilon for every n bigger than n naught. That is the definition of uniform convergence. So, let us specialize it. So, in particular, it is not necessary, but anyway let us take epsilon equal to 1. Then for every x belonging to x, we have f n x minus f of x will be less than 1 for every n bigger than that stage n naught 1. That means f n is close to f by distance 1 and f n's are bounded anyway. So, I can just apply triangle inequality hence for n greater than n naught epsilon fix. So, fix any one of the numbers. Mod of f x, I want to show it is bounded. It is bounded by some scalar for every x, then for every x belonging to x, I know f of x is close to f n x and f n's are bounded anyway. So, I can use the triangle inequality. Then for every x belonging to x, this is true. This is less than or equal to 1 plus supremum over x belonging to x of f n x. That exists. So, you can call this number as m if you like. So, implies mod f x is less than or equal to 1 plus m for every x. Hence f is bounded. So, if a bounded sequence of real value functions converges uniformly, then the limit also is a bounded function. So, that is 1. Let us look at second property that we looked at last time. So, let us do it again anyway. That is, if f n converges to f uniformly, each f n is continuous at x is equal to c, then f is also continuous at x is equal to c. So, continuity is also preserved under uniform convergence. So, let us prove that. So, to prove this, what we have to do? So, we want to make for, we want to analyze the distance between f f x and f of c. We want to say that this distance can be made small whenever x is close to c for continuity. And what we know is f n is converging uniformly to f. So, close to f, there is a f n at every point and f n is also continuous. So, both these facts. So, let us note this. I can add. So, let me write less than or equal to f of x minus f n of x for any n plus f n x minus f n at c. That will be small given because of continuity of f n. So, plus f n of c and the last is f of c. Why I am doing this? I want the left hand side to be small, but close to f, there is a f n, because f n is converging uniformly. So, I can estimate change f to f n and f n are given to be continuous. So, I can use continuity for the second term of f n. And finally, once again close to f n, there is a f. f n is close. So, all these three terms can be made small. So, how do I write it? So, note for every n for every x, this is true. So, by uniform continuity, sorry, by uniform convergence. So, this is how you think and now how you write for uniform convergence. Given epsilon bigger than 0, choose the stage n naught such that mod f n x minus f of x is less than epsilon for every n bigger than n naught and for every x. So, that is by uniform convergence. So, now let us fix some n bigger than this n naught fix by continuity of f at the point x is equal to c. Given epsilon as before that is already fixed. There is a delta such that mod of x minus c less than delta implies f of x minus f of c or f n. So, f n f n of c is less than epsilon. So, there is continuity of f n where given f n is continuous. So, we have fixed n bigger than n naught. Already epsilon is fixed. So, find a delta such that this happens. Then for every x with x minus c less than delta, f of x minus f of c will be less than or equal to 0. So, go back our starting point. Let us call that as star. When we said that f is close to f n, f n is continuous. So, that triangle inequality, the first inequality. So, that is what motivated us. So, in the star, use these things less than, at least, strictly less than 3 epsilon. And if you wanted nice thing, then you could have gone back and modified everything this by epsilon by 3, and this by epsilon 3, that could have been epsilon. Because epsilon is arbitrary. So, you can always make it as small as you want. So, that proves basically keep in mind what we want to show. We want to make this thing small and we are given that f n's are converging to f uniformly and each f n is continuous. So, bring in f n's. So, that is continuous continuity is preserved under uniform convergence. So, let us look at next something which is also preserved. So, third, fourth, I do not know what is it, property number 3. Let us look at integrability. Let f belong to R A B, f n's belong to R A B, n greater than equal to 1 or Riemann integrable functions on the interval A B and f n converges to f uniformly. Then f is also Riemann integrable. So, then f belongs to R A B and integral f n d mu dx converges to integral f dx A to B. By the way, I think, let me just make a point here that this theorem, when we said uniform limit of continuous function is continuous. Another way of writing that, let me put a note, f n converges to f uniformly, f n continuous implies. See, we will look at limit f n of x, n going to infinity, that is f of x and as x goes to c, that is continuity. So, limit x going to c is same as, you can write it as, you can take the limit inside, so it is limit x going to, n going to infinity, let me write, n going to infinity limit x going to c of f n of x. So, limit f n will be f of c and f is continuous. So, this conclusion of this, you can write it as like this. It is like interchanging two orders of taking limiting operations. Limit x going to c, limit n going to infinity is same as limit n going to infinity and limit x going to c. Interchange of limit operations are possible whenever the convergence is uniform. So, that is another way of writing this theorem, which is useful way of observing. There are two limit operations. So, you can interchange whenever there is uniform convergence, that is what it says. So, that is something coming here also. So, I can write this as, so it says a to b limit n going to infinity f n dx. What is the right hand side? f is the limit. In the right hand side, f is the limit. So, here this f is the limit. So, I can write it as limit. That is same as this converges, so that is limit n going to infinity integral a to b f n x dx. Again, there is an interchange of limit and integral here now. Integral of the limit is limit of the integrals. So, again, so it essentially says under uniform convergence, integration is a continuous operation kind of integration is continuous operation. Whenever there is a interchange, something implies continuity of something. So, here it is saying that the limit of the integrals is integral of the limit. So, the two operations are interchanged. So, let us prove this. So, we are given f n converges to f uniformly. So, this is given and each f n belongs to R a b. We want to show that f belongs to R a b, then only you can write it as integral. And integral f n d mu converges to integral f d mu. So, that is what is to be shown. Now, if you want to show that f is Riemann integrable, then you have to first show that it is a bounded function. f should be a bounded function. But that follows from the fact that just now we have shown. So, let us note each f n is Riemann integrable. So, that implies each f n is bounded. We are shown that every Riemann integrable function is also bounded. It implies that f is bounded because f n converges to f uniformly. So, all are bounded functions. The only thing that we do not know now is whether f is Riemann integrable or not. f is a bounded function. So, let us try to show that f is Riemann integrable. We have gotten it is bounded. Boundedness helps us to look at upper and lower sums for function. That way of defining integration by upper and lower sums is possible only when f is given to be bounded. In the Riemann definition, f is not assumed to be bounded. But anyway, we have gotten boundedness. For this, what do we do? For this, we want to show f is Riemann integrable. That means, for this, given epsilon bigger than 0 to find a partition P such that when is a function Riemann integrable, when given any epsilon upper and lower can be brought close to each other such that upper sum minus the lower sum is less than epsilon. So, this is what we want to show. So, given epsilon, if we can find a partition P such that upper minus the lower is difference is small, then we have to do. And keep in mind, how is the upper sum defined? Upper sum is, our partition is by looking at the maximum value of the function in the N e sum interval multiplied by the length of the interval summed up. And what is given to us? Given to us is f is, each f n is Riemann integrable and f n's are converging to f uniformly. So, close to f, there is a f n and f n is Riemann integrable. So, the idea would be the upper sums of f, we should try to approximate it by upper sums of f n. That is the root we should follow. So, let us do that. So, by uniform convergence, since f n converges to f uniformly given epsilon greater than 0, there is a stage n naught such that all are bounded. So, let us write mod f n minus f infinity is less than epsilon for every n bigger than n naught, because all f n's f are bounded. So, I can now write the supremum less than for every x and so supremum. So, that is same as saying, hence for every x, for every n bigger than n naught, let us write what is hidden here is that f n x minus f of x is less than epsilon. That is same as saying the supremum and hence this also is less than n. So, sometimes expanding things help. What we want to do is we want to relate f with f n, the maximum value of f with f n, the minimum value of f with f n. So, this gives me, this is same as saying, if I look at f x, it is less than, expand this inequality absolute value. So, this is f n x minus epsilon is less than f n x plus epsilon. The same as saying as this, the distance of f n from f is either this side epsilon or that side epsilon. That is same as saying this. Now, this will give us the required thing. So, let us choose the partition. So, given choose, so we have used uniform convergence. Now, we use the integrability. Choose a partition p. So, let us call it as a equal to x 0 less than x n equal to b of a b such that, so we are not fixed. So, let us, n bigger than n naught. So, let us take n naught itself, such that u p f n naught minus a n naught with l p f n naught is less than epsilon. What I have used? I have used the fact that f n naught is Riemann integrable because is Riemann integrable. So, there must be, given epsilon, there must be a partition so that things are closed. Now, let us, so this is, now that equation, let us keep that in mind. This was the equation when f n is close to star. So, let us use star n equal to n naught. So, what we have? f n naught x minus epsilon is less than f of x is less than f n naught of x. I am just specialized because this is true for all n bigger than n naught. So, let us fix a n naught so that we do not have any. See, this is what we are saying, f n naught x is close to f of x by distance epsilon. So, I can take the minimum values. So, implies, so let us look at the minimum value m i. Let me write f n naught. What is this m i of f n naught? That is the minimum value of the function f n naught in the interval x i minus 1. So, if, let me write, it is equal to minimum of f n naught x, x belonging to x i minus 1 and x i, then what does the upper one give me? What will this equation give me? This is f n naught x. So, it will be bigger than the minimum value. So, m i of f n naught minus epsilon will be less than f of x will be less than m i f n naught plus epsilon. Is that okay? This equation, this is true for all x. So, let us look at x in the interval x i minus 1 to x i. So, this will be bigger than the minimum value. It is less than f n naught x for every x. So, take the minimum over that. Is that okay? So, this equation gives me this. Now, from the minimum value, how do you go to the lower sums by multiplying by the length and any gap? So, let us do that. So, implies sigma m i of f n naught multiplied by the length. So, x i minus x i minus 1 i equal to 1 to n. When I multiply epsilon by those lengths, the length will add up, right? Summation. So, it will be just epsilon times b minus a. Is it okay? This part is less than f of x multiplied by the length is less than the corresponding thing. So, that is m i f n naught x i minus x i minus 1 plus epsilon times b minus a. What happened to the sum? Now, here is the sum. So, i equal to 1 to n. Is that okay? The previous equation, I have multiplied throughout by x i into x i minus 1 and summed up. So, the first term is m i. So, this first term is, is it okay? Let me just underline it. So, this thing, if you multiply by x i minus 1 to x i minus 1, the length of the interval and sum it up. When you multiply by epsilon, that should be giving you epsilon times b minus a. In between, I should put the sum also. I forgot to put the sum here. 1 to n. There is a sum everywhere. So, what does this give me now? Why f of x? No, it is okay. Actually, I can put here also the minimum of f of x. So, let me do that. Anyway, this I should do that. So, from here, when I multiply m i less than, I can put here also the minimum of f. Is it okay? See, from this equation, let me just simplify. This is true for every x. Now, in the first part here, take the minimum over f and not. So, this is the minimum for every x. So, minimum of the function f and not in the interval x i minus 1 to x i. So, that would be less than or equal to f of x for every x anyway. And there, you can take the minimum of f also. So, you can get here minimum. So, you will get this part. Take first the minimum over x. You get this quantity. This is less than or f of x. Take the minimum of f x over that interval. So, that gives you the minimum function of f i. So, m i. So, what is that? What is the notation I am writing? We are not writing that. Let me just write. What is that quantity? So, that quantity is m i of f. So, I am saying I can just write that. Is f of x? This quantity is less than f of x. So, it will be less than or equal to minimum also. And that minimum will also be less than or equal to minimum of that. So, I can take minimum everywhere in this inequality. That is what I am saying. One at a time. But you should do it one at a time to justify that. First, take the minimum over this part. This is for every x. So, minimum only in the left hand side. So, you will get this quantity. Less than or equal to f of x for every x. So, take the minimum in that interval. So, that is m i. This is less than or equal to f n not of x. So, take the minimum over that. So, that is m i. So, I should have… So, this f of x, I can replace it by… m i of f. Basically, the idea is f n is close to f by a margin of epsilon this side or that side. So, I can take the minimum over the interval. This is happening for every x. So, I can take the minimums. So, that means, what is the meaning of this? This means this is the lower sum. So, lower sum of small m i. So, lower sum of f n not respect to p minus epsilon times v minus a. This first part is less than or equal to the lower sum of f with respect to p. And the last term, this one is less than or equal… Why less than or equal to? It is less than. It is strictly less than. It does not matter actually. It is less than l f n not is the minimum over f n not. So, f n not of p plus epsilon times v minus a. Basically, what I am saying is that this equation, if I take the minimum over all x in the interval x i minus 1 to x i. So, equation holds for minimum also. Multiply by the length of the intervals and add up. That gives you the lower sums minus epsilon times this. So, similarly, the upper ones. So, that means what? The lower sum of f and the lower sum of this is less than. That means mod of l, we are writing f n first. f n not p minus the lower sum f p is less than epsilon. In the absolute value, I can write it this way. This minus v minus a epsilon times v minus a. I am writing that inequality back in terms of absolute value. Nothing more than that. So, similarly, you will have upper sum f n not p minus the lower sum of f p will be less than epsilon times v minus a. Instead of taking minimum, you can take the supremum and the same thing. So, hence, upper sum v f minus the lower sum. I wanted to estimate this quantity for f. So, that is why. So, now add and subtract. So, less than or equal to absolute value, triangle inequality, upper sum p f, but that is close to upper sum of p f n not. I mean, I should interchange because I am writing p first and then f should be consistent. So, that is. So, we wanted to estimate f. So, I take this one, f n not. No, that is u. So, sure. Thank you. That is upper sum. I said similarly, the lower and similarly, the upper. We are in that same thing here. Now, this quantity is less than or equal to f of x. So, maximum of that must be less than or equal to f of x, less than or equal to maximum of f of x. I am saying same inequality gives you both, lower as well as the upper. So, now, I wanted to estimate this quantity for. So, add and subtract. So, this is less than or equal to upper sum f p is close to upper sum of f n not. So, add and subtract p plus. So, I should add upper sum of f n not p minus lower sum of f n not p plus the last term now left would be lower sum of f n not p minus lower sum of f n p. Add and subtract. This technique by now we are very familiar. So, less than or equal to upper f and f n. So, this is this equation now. Call it 2, call it 3. Using 2 and 3, the first one is less than epsilon times u v minus a. The last also is less than same. So, 2 times this upper minus the lower is less than epsilon. So, that we have already seen. Where was that? Upper minus the lower here. So, we can call that as 1, if you like, where f n was, f n not was integrable. So, upper minus lower is small. That is what we started. And 2 and 3 are giving corresponding things between f n and f. So, plus. So, 1, 2 and 3. Using 1, 2 and 3, I get this. Is that ok? No problem. So, if you like this is epsilon times 2 of b minus a plus 1. Is it ok? Because epsilon I can always modify. Go back and change wherever required. So, implies f belongs to R a, b. So, the basic idea is because of uniform convergence f n is close to f for all x. That is an important thing. So, you can take the minimum over that sub-interval as well take the maximum over sub-interval and then multiply by the lengths and add up to get the corresponding. Now, f ns are given to be Riemann integrable. Yeah, but why in general, given that f n is bounded, why it should be Riemann integrable? What are you saying? So, here is a theorem. So, what modification you are suggesting? Small fs meaning what? Small f ns or what? Yeah, if a limit is Riemann integrable, why you should say all corresponding sequence is Riemann integrable? Why should it say that? For example, all f ns can come to a singleton and all f ns may not be Riemann integrable, but they converge to a constant function. When the limit is Riemann integrable, why should the function be Riemann integrable? Each f ns, that is, expecting too much, giving nothing and you are expecting too much. So, basic property we are saying is if f ns are Riemann integrable and f ns converge to f uniformly, then f also becomes Riemann integrable and integrals converge. We have not proved integrals converge yet. We have only proved f is Riemann integrable, but proving integrability is okay. So, also look at integral of f n d mu minus integral of f d mu. We want to say this converges, so we want to show that this quantity goes to 0. We want to show that this quantity goes to 0, but we know something about this is less than or equal to integral of f minus f n. Is it okay? Using the property at absolute value of the Riemann integrable is less than or equal to integral of the absolute value. Which is less than or equal to? Now, each function f n x is less than or equal to norm of into the length of the interval b minus a. The right hand side because the function f minus f n is bounded by the constant norm of f minus f n. So, I can take it out less than and this goes to 0 as n goes to infinity because of uniform convergence norm goes to 0. So, that is not much of a problem. Once you know that limit is integrable, convergence of integrable is not a big issue. But the important thing is that you need uniform convergence, f n converging uniformly to f to say that you can interchange limit and the integral sign. This is the beginning of a, I have already mentioned I think that the Riemann integral is not very well behaved with respect to limiting operations. Namely, point wise convergence need not imply integrability. Limit of integrable is equal to limit of integrable. So, that started the research for looking for an integral which has better properties compared to this. So, that is another point for the beginning of what is called Lebesgue integration, where this is a much better behave. You do not need uniform convergence. You need a slightly milder conditions to be true. This is now the question comes. See, what we are doing today is very important in mathematics to relate ideas. Continuity is an idea about how f and x converges to something f of x as x goes to c. If you take limiting operations, you know algebra of limits. If f plus g are continuous than f, f and g are continuous, then f plus g is continuous. Product is continuous. So, here it says if you go on taking limits. So, this is a very general thing to consider. You are looking at the class of real valued functions, say R to R. What are the properties that you would like to analyze? You would like to analyze. If I add, you can add functions. So, you would like to know under addition what is preserved. I can multiply under multiplication what is preserved. I can scalar multiply. Given a function f, I can multiply by c times f. What properties are preserved under that? Because on R, these are the structures available. Addition, scalar multiplication, product, you can take limits. Under all these, what are the properties which are preserved? What we have shown is, when we did continuity, differentiability and so on, if f is continuous, g is continuous, f plus g is continuous. We saw that f plus g is differentiable. If f and g are differentiable, product was differentiable. If f n's are differentiable, then can you say the limit is differentiable? That we saw is not necessarily true under point wise. So, that is why there is a need to put some stronger conditions which allow us to pass over limit and that operation interchange those two operations. Uniform convergence is one. Unfortunately, differentiability is not preserved under uniform convergence. One has to put slightly more stronger conditions. So, we will not prove that theorem. Let me just show you that theorem, so that you understand. So, this is the integrability. f n converges to f uniformly, then f is integrable. So, here it says, suppose not only f n converges uniformly, f n dash, each function is differentiable and the derivatives also converge uniformly. You need something more and also suppose that it converges at some point, then f n converges uniformly to a differentiable function and derivative. So, it is slightly more convoluted kind of a thing. You have to put conditions on the derivative itself to ensure that the limit function is differentiable. So, we will not go into that because it is not a very useful result in many situations. It just simply does not say f n's are differentiable converging uniformly, then limit is differentiable. So, we will not do that. So, let me just summarize what we have done till now. We had looked at sequences of functions and tried to analyze what are the limits of under what is the meaning of point wise convergence at every point. It converges and we observe that at point wise convergence is not very good behavior, not very well behaved. It does not have any one of the properties that we can think of continuity, differentiability and so on. So, for that one looks at what is called uniform convergence and it preserves continuity, it preserves integrability, it can preserves boundedness, but in some way it preserves differentiability, not exactly straightforward way. So, next time what we want to do is, I will stop here today. There is a reason for that. What we want to do is, start with sequences and see what is other use of sequences. For example, the algebraic operation of addition of numbers. Given two numbers, you can add them a plus b. You can three numbers a1, a2, a3, you can add a1 plus a. Any finite number of real numbers, you can add them, because there is operation of addition and inductively given any n, you can add them. Can you add infinite number of numbers? That is a question. How does one find a way of adding? So, what does it mean? Given a sequence a1, a2, a3 and so on, how do I say something like a1 plus a2 plus a3 plus dot, dot, dot? It makes sense or not? In what way I can give a sense to that operation, infinite addition of numbers. And that is very natural way of doing that. How do you add a1? You add a2, so a1 plus a2. You have got another one, add a3. So, up to an. So, you can go on doing it, but what eventually you want is, when you go on adding a1 plus an, whether they become stabilized somewhere. So, that is limit of a1 plus a2 plus an you will like to consider. So, that gives you a notion of what is called series of numbers. So, we will do it next time. So, we will stop here today.