 So, let us just recall, we had started looking at alternate way of describing integration. So, we defined integral by upper sums and lower sums. So, here is an alternative way of doing that. So, given a function on a, b to r and a partition, one defines what is called a sum spf. So, in the sub-interval xi minus 1 to xi, choose any point ti. It is normally called a tag, but so look at the value of the function at that point into the length of the interval. So, that is the area of the rectangle with this height. So, that is an approximation to the area below the curve. So, this is called a Riemann sum of f with respect to the partition p. And one says f is Riemann integral, integrable if the limit of this as a norm of the partition becomes 0 exists. And that means that given epsilon becomes 0, there is a delta such that whenever the norm of the partition is small less than delta, the difference between this sum and that number l is less than epsilon. So, one writes this as that limit of the sums is equal to the number l and that l is called Riemann integral of the function. For the time being, we call it R integral of f. So, note that the function f is not assumed to be bounded in this to start with because we are just looking at the value of the function at a point in between in the interval xi minus 1 to xi. Whereas, in the upper and lower sums, we need to have the function to be bounded to start with. So, what we want to prove is that if a function is Riemann integrable or integrable in this sense, then it is also integrable where the upper sums and the lower sums and the two integrals are equal. So, to prove that, we need to first show that if a function is integrable in this sense, then it is also a bounded function. So, that was the theorem we were proving that if f is Riemann integrable, then f is bounded. So, we had almost completed the proof of the theorem except for the last step. So, let me just go through the proof again. So, since f is Riemann integrable by the definition of integrability, given any number say epsilon equal to 1, there is a partition such that the sum spf minus the number L is less than epsilon. The limit of spf is equal to L that exists. So, that means given any epsilon, there is a partition such that they are closed. So, we can take epsilon equal to 1. So, that means this sum is less than 1. So, let us take two points S i and T i and look at the sum with respect to choice of S i and sum with respect to the choice of T i. Then this, the spf with respect to the choice is less than, this is with respect to T i and there is a typo here, one should have a S i here. With respect to the choice S i and with respect to T i, each one of them is closer to L by 1. So, the difference between the two is less than, between the two of them is less than 2. So, in particular, if I choose S i to be equal to T i, then everything will be 0 except the first one. So, let us choose that that is what we had done last time. Choose S i equal to T i for all sub intervals except the first one. So, that all the remaining sums will be equal to 0 and you will get f of T 1 minus f of S 1 is less than 2. And from here, this is for every S 1 and T 1 in the first interval. So, you can fix T 1 and that means f of S 1 is less than 2 over this plus f of T 1. So, here T 1 is fixed. So, the right hand side is a constant. So, that proves that up to here we had done it. So, just fix now T 1 and this gives you that for every S 1 in the interval x 0 to x 1, this is bounded. So, it is bounded in the first interval and the same proof works for bounded in every other sub interval because you can choose S i equal to T i in the second interval and so on. So, that proves that if a function is Riemann integrable in the sense of what we have done just now, it is also bounded function. Now, since it is bounded, one can define what is the upper and the lower sum. So, we want to prove the statement that if f is R integrable, then also Riemann integrable and the two integrals are same. That means, the integral where the upper and lower sums is same as the integral where this method and both the integrals are same. So, let us write a proof of that. So, let us first suppose f is integrable. So, what does that mean? That means that via upper and lower sums. So, integrable, that means for every partition p of a, b, if I look at the lower sums, that is less than or equal to the integral a to b fx dx is less than the upper sum. And what do you want to show? So, we are given that it is integrable with respect to this to show that f is R integrable. Now, p going to 0 of Sbf is equal to this a to b. So, to show this, let us take epsilon greater than 0 be given. We want to show that this limit is equal to this. So, then by this integrability condition, so by star, there exists some delta bigger than 0 such that for every partition p with norm of p less than delta, see this integral is the least upper bound of Lpf and greatest lower bound of the lower sums. So, given epsilon, we can find a partition such that upper sum minus say epsilon, this cannot be that. So, there is a partition such that this happens and this is less than integral a to b fx dx. Is that okay? Because this integral is the upper bound of upf. So, this we should say, this is the lower bound. So, this minus, so we should say, we are looking at this as, I just, integral of this is the greatest lower bound of upf. So, upf minus epsilon cannot be, so this is okay. And similarly, Lpf plus epsilon is bigger than integral a to b fx dx. So, that is, now note one thing. So, let us note one thing that for every partition p, if I look at spf, what is spf? That is the value of that sum at, when we choose any point C i in that sum interval. So, this is always bigger than Lpf and always less than upf. Because this is the value at a point in the interval xi minus 1 to xi. Lpf is the lowest value you choose and that is the upper, you choose the largest value. So, now, let us combine this with this. So, it implies spf is bigger than or equal to Lpf and Lpf is bigger than integral minus epsilon from this, this star. So, this is spf is bigger than Lpf and Lpf is bigger than integral a to b fx dx plus epsilon. So, we will call it 2, call it as 3 and call it as 4. From 4 and 3, this happens and if I use second, then upper sum is less than this, sorry, this is minus epsilon. This is minus epsilon and upper sum will be less than this integral plus epsilon. So, upf, which is less than integral a to b fx dx plus epsilon. So, that means what? That means spf on the left, it is less than this. On the right, it is less than this. Same quantity minus epsilon and plus epsilon. So, that implies that this is for a given epsilon, we have found a partitions that means limit p going to 0 of spf is equal to integral a to b fx dx. Is that clear? So, what we have done is, given that the function is integrable, we have the integral exists a to b and that is the supremum of Lpf and the infimum of upf. We want to show that this value is also the limit of spf, the norm p goes to 0. So, to show that, we have to show that given epsilon, there is a delta such that whenever norm of p is less than delta, this quantity spf is close to this integral between at the most epsilon distance between them. So, now we start with using from arbitrary, we go to the upper and lower by looking at the definition. So, given epsilon, find a partition p such that the upper sum minus epsilon is less than this integral and similarly, the lower sum plus epsilon is bigger than that quantity. So, that is by the integrability by upper and lower sums. Now, we observe that Riemann sums, they are, you pick up points T i in a interval x i minus 1 to x i and Lpf is with respect to the lowest value of the function in that interval. This is the upper sum with respect to the maximum value. So, that is always true. So, now combine this to get the required thing. So, that says that if f is integrable, then it is also Riemann integrable and the two integrals are same. So, let us prove the converse. So, what will the converse be? Suppose f is Riemann integrable. So, that means limit of norm p going to 0 of S p f every Riemann sum exists and let us say, call it a number a. So, this limit exists and let us call it as a to show that f is integrable, f is integrable and the integral a to b fx dx is equal to this number a. So, when you want to show f is integrable, what does that mean? That means we should show that this number a lies between upper and the lower sums for every partition. So, to do that, so let us start with, because this is given to exist. So, limit norm p going to 0 S p f, that quantity we have called as a implies, given epsilon greater than 0, there is a delta such that for every partition p, norm of p less than delta implies the sum S p f is close to, this limit is equal to a. So, it is close to a. So, let us say it is less than a minus epsilon and a plus epsilon. So, that is the meaning of the limit is equal to a. Now, from this, so let us call this kind star, we have to go to upper and lower sums to show that the function is integrable in the other sense. So, let us write this partition. So, let p be any such partition, p is, let us give some name a equal to x 0 less than x 1, x n equal to b. As before, let us call capital m i equal to supremum of f x, x in the interval x i minus 1 to x i and small m i equal to the infimum in the same interval. We have to go to upper and lower sums. So, we have to write down all these quantities and these exist. Why do they exist? Because function is given to be Riemann integrable. So, this limit exists. So, by the earlier theorem, it is a bounded function. Because f Riemann integrable implies f is bounded. Just now we proved that. So, it is a bounded function. So, these quantities will exist. And f is a function on the interval a b to R. So, f is a b to R. This is a partition. Now, we are not given f is continuous. We do not know that. Otherwise, things would have been very easy, because this m i would have been attained at some point. But we are not given to be continuous. So, we can choose, select given that m i is the supremum, select some point x i dash belonging to x i minus 1 to x i, such that this capital M i minus epsilon cannot be the supremum. M i is the supremum. So, something smaller cannot be the supremum. That means there must exist some point x i dash such that this is less than f of x i dash. This cannot be. So, there must be a point on the right hand side. Similarly, select some point x i double dash belonging to the same interval x i minus 1 to x i, such that we have to go to upper and lower. So, that is why we are trying to maneuver all these things, such that small m i plus epsilon, that cannot be the supremum. So, m i is the supremum. So, this plus something cannot be the supremum. So, that means this must be bigger than f of x i double dash. Now, let us form the Riemann sums with respect to this choice x i dash and x i double dash. So, let us call this as 1, call this as 2. So, from 1 and 2, how do I go to upper sums? So, I want m i times the length of the interval. So, that is x i minus, x i minus 1 summation i equal to 1 to n. That will be the upper sum. So, here I multiply both sides by this number and try to add. So, what you will get? I will get minus epsilon into b minus a is less than, so, sigma f of x i dash into x i minus x i minus 1 i equal to 1 to n. I have multiplied equation 1, both sides by x i minus 1 x i and taken the summation. Here, when you multiply, things will cancel out and you will only get x epsilon times b minus a. This is the upper sum. So, u p f, this is the upper sum m i times minus epsilon b minus a is less than s p f. This is the Riemann sum with the selection of the point x i dash in that interval. Similarly, from 2, you will get, if I multiply this equation 2 by x i minus 1 to x i both sides and add, what will that equation give me? That will give me the lower sum with respect to the partition p plus epsilon times b minus a is bigger than s p f. Multiply by the length of the sub intervals in equation 2 and sum it up. Similarly, from 2, we will imply this thing. So, let us call it as 3 and call it as 4. Now, let us look. Upper sum minus sum quantity is less than s p f and lower sum is that property and here is s p f lies between this and this. So, let us combine these star 3 and 4. So, using star 3 and 4, what we have to do is, we have to multiply this by 3 and 4 and we will get. So, 4 says l p f plus epsilon times b minus a is bigger than s p f and what was s p f there? In star, s p f was bigger than a minus epsilon. So, we will star. So, that means l p f is bigger than a minus epsilon minus epsilon times b minus a. So, this term I take it on the other side is less than or if you like, you can just, if you want l p f on this side or l p f plus epsilon plus epsilon b minus a is less than a. So, rewrite this take everything on one other side that is a better one. So, this was using star and using 3. So, if I use 4, if I use 3, what I will get? So, using 3 now, so it will be similarly u p f. So, here is u p f minus epsilon. So, if I take it on, what is s p f? S p f is less than a plus epsilon. So, here is, this will be less than, this quantity will be less than a plus epsilon. So, that epsilon I can bring it on this side. So, that will be bigger than u p f minus. So, u p f minus epsilon minus epsilon times b minus a will be less than, this is less than a. So, this will be bigger than. So, what are we getting? S p f is it everything? This was l p f plus epsilon is bigger than s p f. It is bigger than a minus epsilon. So, everything on this side. So, it is bigger than, sorry, I should have written bigger. I wrote it wrong here. This was bigger here. And u p f, so that will be less than this quantity, less than a. Is that okay? This I just rewrote. So, as it was this. So, what does that mean? So, implies we have got the required inequality that l p f plus epsilon plus epsilon b minus a is between a is less than u p f minus epsilon minus epsilon times b minus a. So, how do I adjust all these quantities now? So, I have got a number a between l p f and u p f plus some small quantities. So, that small quantities are arbitrary. So, we can make it, how many of you are worried about this epsilon plus? I can re-adjust those things back. That is what we do in analysis. I can make it epsilon by 2 in the beginning, epsilon by 2 times b minus a. Or simply say that l, one simply says it is epsilon times 1 plus b minus a. Anyway, that does not matter. It is less than a is less than u p f plus epsilon minus epsilon times. One does not have to do all those cosmetic things. One can just write this as 1 plus b minus a. So, one simply says since epsilon greater than 0 is arbitrary, this implies that l p f is always less than or equal to a is less than u p f. So, arbitrary means I can let epsilon go to 0 if you want. Which sign? l p f is greater than a. l p f, did I make a mistake somewhere? l p f is this plus is greater than a. So, this was l p f is greater than a. So, this was l p a is bigger than. That seems a bit strange. It should not happen. That means either I have made a l p f. I want l p f to be, let us go through the steps. This step is okay because l p f is the limit. So, given epsilon, it is close to a. So, mod of l p f minus l p a is less than epsilon. So, a minus that is okay. So, this star is okay. Now, the m i is the supremum. So, m i minus epsilon cannot be the supremum. That is okay. So, there is a point with this property. So, 1 and 2 are also okay. So, u p f is less than p f. That is this. So, multiply this equation by x i minus 1 to x i plus b minus a. There is small m i. So, this is okay. So, in 1, if I multiply this is less than okay. So, u p f minus epsilon times b minus a is less than s p f. That seems okay. So, this part is okay. This part is okay. Third, so that means u p f. So, l p f is greater than this is okay. It seems okay. So, let us say from here what are we getting? l p f. This part gives me l p f is less than a minus okay. I think even that, I think I should not try it this way. Yeah, okay. I think probably the last line I should change. I think this only way should change. So, let us rewrite what is this. So, this says l p f is less than a minus epsilon 1 plus b minus a. This part l p f, this quantity I have taken it on the other side, b minus a. So, I think this is a place where we should say in this sense epsilon greater than okay. We can write this and similarly, the other one will be a plus epsilon times 1 plus b minus a will be less than u p f. This part will give me this. Is it okay? Now, just compare this quantity and this quantity. Can I say l p f is less than this quantity and that is less than or equal to this quantity because this is a minus something and this is a plus something, same quantity. And that is a non-negative quantity epsilon times 1 into b minus a. So, if I do that, I will get l p f is less than a times epsilon into this less than a plus the same quantity less than u p f. Now, here if I let epsilon go to 0, so letting epsilon converge to 0, I get l p f is less than or equal to a is less than or equal to u p f. That is okay. Which one? No, no, I am using only this part. Which part? Is this okay? l p f plus epsilon times b minus a is bigger than a minus epsilon. So, is this part okay? Yes. I think, okay. Let me probably, I think this is the time to show you probably the better way of writing this in case. Let us look at. See, s p f, given epsilon greater than 0, you can find the delta, so that this quantity happens. There is that number a is close to it, where p is a partition. So, it has sub intervals i i x i minus 1 to x i. Now, capital M i and small M i are the supremum and the infremum. So, that means what? I can find a point x i dash. So, that x i dash is bigger than M i minus that quantity. Because this quantity M i minus cannot be a supremum. So, there must be a point. And similarly for the other one, okay, for the infremum, small M i. So, now, if I multiply both sides by the length, so multiply both these equations by the length of the intervals. So, I will get these two, okay. And what is this quantity? So, what is this quantity? f f x i multiplied by lambda of x i, right? That is s p f, right? And that quantity, that instead of epsilon in the written proof, it is epsilon divided by 4 times v minus a. So, that is okay. So, you get upper sum is less than s p f plus epsilon by 4, right? Instead of that epsilon into 1 plus epsilon, what we wrote, okay? And now, this combined with this first one, s p f is less than a plus epsilon by 4. So, this and this combined together, you get upper sum is less than, so you get upper sum is less than a plus epsilon by 2. And similarly, you will get the lower sum is bigger than a minus epsilon by 2. And now, combine 1 and 2. What does 1 and 2 give you? The lower sum, okay? The difference between the two, upper and lower is less than epsilon, okay? So, go through this proof from the slides again. So, basically the idea is that Riemann integrability gives you tags that points t i arbitrarily. You can connect it with the upper and lower sums by looking at the definition of m i to be the supremum. So, that gives you that this number, upper and lower is less than epsilon. So, it is integrable and a must be that sum.