 This goes by the theorem Cauchy criteria for. So, this is a theorem. Let us write it as theorem Cauchy's criteria. f n converges to f uniformly if and only if f n is Cauchy Cauchy in B X R. So, that is. So, we have seen uniform convergence. We have seen when a sequence does not converge uniformly, a criteria for that. We have seen now a criteria which says when does a sequence converge uniformly, namely it should be converging in the B X R and that is same as saying it should be Cauchy there. So, we know we can test sequences being converging uniformly or not. So, let us finally look at whether this serves our purpose of saying that when point wise convergence limit was not continuous functions, converging point wise the limit was not continuous. Convergence uniform differentiability did not imply the derivatives converge and integral also we had an example that f n are integrable, converging point wise did not imply that limit is integrable. And these are important things as far as the convergence problems are concerned. These are important problems in analysis because whenever you want to go beyond algebra, algebra is 1, 2 and 3 only finite number of operations. When you can look at only polynomials to generate functions which are non polynomial, non rational functions, you have to go to analysis. That is why we have to do all that exponential logarithm trigonometric. All those functions were cannot be obtained using algebra alone. You have to go to limits. Let us write theorem 1. f n converges to f uniformly implies and each f n continuous at some point x naught implies f also continuous at the point x naught. So, let us write a proof that continuity is preserved in uniform convergence. So, let we want to analyze. So, for convergence what we have to analyze? f of x minus f of x naught this difference. We want to show f is continuous, we have to make this thing small whenever x is close to x naught. So, f n converges to f uniformly. So, let us somehow bring in f n inside it because something is given about f n. So, let us bring in f x minus f n of x plus I have subtracted f n of x minus f n of x naught plus f n of x naught minus f of x naught. I am adding and subtracting. What I am f of x? I subtracted f n of x, I add f n and then I subtract f n at x naught, add f n at x naught and the final term is there as it is. So, I have added and subtracted and used triangle inequality. Why I am doing that? I am forced because something is given to me about f n. To use that fact, I have to bring in f n, then only we can use it. Now, if I look at this quantity, the first one is less than or equal to norm of f minus f n. The first term, this term and this term, let us keep it as it is if you like. So, it is mod of f x f n x minus f n at x naught plus and this quantity again at the same point. So, less than or equal to f n minus f norm infinity. Now, I know that this quantity is small because f n is converging uniformly to f. Both these quantities are small, this and this. What is this quantity? Middle one, that is f n of x minus f n at x naught, same function at the point x naught, but f n is given to be continuous. So, this quantity is also small. So, we can just simply write using the fact that each f n is continuous and f n minus f goes to 0. So, this equation star implies f is continuous. I am not writing that epsilon every time kind of a thing. So, if you want to write all that thing, you can write given epsilon bigger than 0. I can find some stage n so that this quantity is small, this quantity is small less than epsilon and by continuity for f n after that stage for that stage, this will be less than epsilon by continuity. So, I can find a delta such that you want me to write everything? Let me write so that you feel happy. So, what does this mean? So, here is what I am saying. Let epsilon greater than 0 be given. So, this is how formal proof will be written. Choose n naught such that mod of f minus f n is less than epsilon for every n bigger than n naught. That is by using the fact that f n is converging to f uniformly also. So, let us take this f n, n bigger than n naught. Let us take n naught itself, f n naught being continuous. There exists a delta bigger than 0 such that mod x minus x naught less than delta implies f n naught x minus f n naught of x naught is less than epsilon. So, this is 1, this is 2, then in this equation star, then in star put these values. So, this is less than epsilon, this is less than epsilon, this is less than epsilon and this is less than epsilon when x is close to x naught by delta. So, you get it. So, in star, x minus x naught less than delta implies f x minus f x naught will be less than 3 epsilon. It does not matter. If you want it very nice, you could have cut down everything. So, that is the idea of the proof. So, you should understand what you are doing. This thing I can manage, I can make it small by using the fact that f n converges to f uniformly. I can make this thing small using the fact that f n is continuous at the point x naught and I can make this small again by uniform continuity. So, the middle term is small whenever x is close to x naught by continuity, this will be small. So, that is what we have written. Given epsilon, there is a delta and so on. So, that proves the thing. So, this is, that means, continuity is preserved in uniform convergence. So, let us prove that theorem 2 says, if f n belonging to B x r are Riemann integrable and f n converges to f uniformly, then I should Riemann integration. So, I should not write x now, because Riemann integration is only for some intervals. So, we are taking x is equal to interval a, b. Integrability of functions is defined Riemann integral only for close bounded intervals. So, here x is taken as, so we are looking at all bounded functions on a, b to r, which are Riemann integrable and f n converges to f uniformly. f is a function on a, b to r, then is also Riemann integrable and integral of f n a to b dx converges to integral f dx a to b. So, one has to put a slightly stronger condition. Point wise, we saw that even the function may not be integrable at all, limit may not be integrable. In fact, you can give examples of functions where Riemann integrable f n converges point wise and the derivative function is integrable, but the integrals do not converge. So, many such examples are possible. We are not going into all those things. What we are saying is uniform convergence at least is one criteria, which allows you to pass on under limits. Integrability is reserved. I think we will prove it next time, because the proof requires a bit of more partitions and so on. We will also analyze what happens to differentiability. We will show that differentiability also is not preserved under point wise convergence. In fact, one has to put very strong conditions for differentiability also, even for uniform convergence. So, these are slightly deeper theorems. This thing that when does integral of f n converges to integral of f for Riemann integrable functions, as I had mentioned, is the beginning of the story of Lebesgue integration. In general, it does not happen. It happens for uniform convergence. What other class can happen? So, we will stop here.