 Then, historically, Riemann said, I do not like this idea. He did not say that, but essentially he said, why we should assume that the function should be bounded? So, his idea was very nice one. He said, intuitively seems, so this is a and this is b. He said, this is the graph of the function f of x and what you want to do? You want to capture the area below the graph of the function. So, this is the area that you want to capture. So, to capture this area, what you are doing is, you are trying to take upper sums and lower sums and trying to capture them in between. But let me do this way. Let me take any point in this. Take this height and take a small near by point c plus delta x. So, that will give me this. So, look at the area of this strip and try to imagine the whole area being made up of these thin strips. So, what is the area of this strip? You can take f of c into delta x and try to sum it up. And the idea of summation is same that you take partitions and make partitions finer and finer. But his idea is, instead of taking the minimum and the maximum in the sub-interval, take any value in that sub-interval and take any rectangle of that height. So, let me define that. So, definition, f is a function on a, b to r. So, let p be a partition x1 less than xn equal to b. Take any partition and choose any point c i belonging to x i minus 1 to x i. Take any point in the closed interval x i minus 1 to x i. Look at the height f of c i into the length. So, that is at the strip. Take the summation that gives you an approximate size of the area below the graph of the function. So, let us call it as spf. Now, there are two things one should observe. To define this sum, we do not have to assume f is bounded. We do not have to assume because we are looking at the value of the function at some point. So, we are taking the height of the function at that point into. So, right hand side is meaningful. So, this sum is well defined. We do not have to assume. At the same time, note that the point c i is a arbitrarily chosen point in the interval x i minus 1 to x i. You can take the left end point, you can take the right end point, you can take the midpoint or you can take any point. It does not matter. So, look at this and look at the limit norm of p going to 0. Make the partition finer and finer spf. So, if x is, we say f is, now we will call it as Riemann integrable if this limit. So, what is the meaning of this limit exists? Concept of a limit again, but norm of p going to 0, what is the meaning of this saying that limit exists? So, let me explain that because that is for every epsilon bigger, sorry, limit exists. So, first of all, what is the limit? One should say that there exists some number a belonging to R such that for every epsilon bigger than 0, there is a delta such that whenever the norm of the partition is less than delta, that implies this sum spf minus a is less than. For every choice, when p is a partition spf depends upon the choice. So, this says for every choice of those point c i, whatever choice you choose, you construct as spf, that should be close to the number a. That is what the limit means. So, you say it is integrable and this number a is called the integral. We should say Riemann integral for the time being because we have notion of integral coming from upper sums and lower sums called the Riemann integral. Now, once we give another way of interpreting the area, we should say that this way of defining the area is same as the one we have done by upper and lower sums. Both are same. There is no difference between them. So, let us prove that. But before even proving that important thing comes, namely that if f is Riemann integrable and if we want to prove that it is this integral is same as the one which we obtained from upper and lower, but that was defined only for bounded functions. So, there should be a theorem saying that if f is Riemann integrable implies f is a bounded function. So, let us prove that first. If f is theorem, first f a, b to R, f Riemann integrable implies f is bounded. In total, it seems quite okay because if the function is unbounded, those steps, areas, you can keep on increasing. That is one way of looking. But let us look at another way of proving this. So, f Riemann integrable implies there is a number a such that for every epsilon bigger than 0, there is a delta such that for every p, norm p less than delta implies spf minus a less than epsilon. The boundedness means what? We should be able to show that for every point x, mod of f of x is less than or equal to some constant. That is what we want to show. So, let us say this p is a partition. So, something say a is equal to x is 0, x1 less than x equal to b. Then for every point s i and t i, the spf requires choice of a point in between the interval x i minus 1 to x i. So, let me choose some point x i t i between x i minus 1 and x i. Then for every such choice, we have, so let me write corresponding spf f of s i x i minus 2 minus x i minus 1 sigma i equal to 1 to n minus a less than epsilon. That is the sum corresponding to the choice of s i. But let me also choose the sum corresponding to some other choice f of t i x i minus 2 x i minus 1 minus a less than epsilon. For two different choices, let us, so that is what it says. But we only want at a point, so let me make a choice. So, let us choose s i equal to t i. Let us try to take from these two equations. Let us subtract them and see what we get. Sigma of f of s i minus f of t i. If I subtract, I want to look at this. So, that means if I, what will I get less than, oh sorry, into that length of that interval. So, let me minus f of t i into x i minus x i minus 1. I want to estimate this quantity. So, I can add and subtract a. See, this is summation, this, the first term minus mod a, this minus mod a. So, summation f of s i minus capital A times this plus minus of that. So, I am saying this is less than 2 epsilon. Is that okay? Because what is the first term? Summation f of s i times this length. So, I can add minus capital A. So, I will get 1 epsilon by using triangle inequality. And that other one, I will combine it with the other one. So, triangle inequality. So, add and subtract capital A in this thing and use triangle inequality and use this 1 and 2. Is that okay? Or shall I write that step? So, let me probably, in case you feel uncomfortable, let me write. So, the reason is because this quantity is this summation f of s i into x i minus x i minus 1 minus a plus the other term minus summation f of t i that thing minus a. Add and subtract. And now use triangle inequality. So, this will be less than or equal to absolute value of this plus absolute value of that and use 1 and 2 to get the required thing. So, this is less than 2 epsilon. So, let me call this as 3. So, in this 3, let us take s i equal to t i for every i bigger than or equal to 2. Then what will happen? That terms will be 0. Those terms will be 0. So, what is left is the first term f of s 1 minus. So, that will give me implies f of s 1 minus f of t 1 absolute value into the length x 1 minus x 0 is less than x 1 minus that is positive anyway less than 2 epsilon. Is that okay? In equation 3, I am putting s 2 equal to t 2, s 3 equal to t 3 and so on. All remaining terms will be 0 except the first one. So, first one will be f of s 1 minus f of t 1 into the length. So, that is x 1 minus x 0. So, that is the case. So, what is mod f of s 1 is less than or equal to? What is mod of f? So, mod of f t 1 into x 1 minus x 0 plus 2 epsilon. Is that okay? From here, this implies this. Is that okay? Mod of a minus b less than something. So, what is mod a less than? At triangle inequality again, you can add and subtract if you like. Now, look at this quantity. f of t 1 is the value at some point t 1, length of the interval. So, this is less than or equal to some number. f of t 1 is the value of the function at a point t 1 of your choice. You can take at f of t 1 to be a left end point a 1 a. So, f of a into the length. So, you can put maximum b minus a. This is the length of the sub interval. You can bind it by the total interval plus 2 epsilon. Is that okay? f of t 1, t 1 is a arbitrary point in the interval x i minus 1 to x i. So, let me take the arbitrary point as the left end point a. So, t 1 is equal to a x 1 minus x 0. That is the length of the sub interval. So, that is less than the length of the full interval. That is b minus a 2 epsilon. So, call this as some constant m. So, what is s 1? s 1 is an arbitrary point in the first part. So, what we have shown is the function is bounded in the interval x 0 to x 1. s is arbitrary. So, since s is arbitrary, s 1 is arbitrary in the first part. So, that is a 2 x 1. f is bounded in x 0. That is a 2 x 1. So, what we have shown is we have found a partition. Now, so how do I get the first one? By choosing s i equal to t i. I can do it for the other ones now. s i equal to t i everywhere for all i except 2. So, I will get boundedness in the second interval and so on. Similarly, other sub intervals. So, let me write. So, similarly f bounded in each one of x i minus 1 to x i for every i and that proves it. You can divide by that also. So, to be very precise, it should have been multiplied by this. So, 1 over of b minus a. Put that also if you like. Take it on the other side. This is a constant multiplying. That is what your question is. This multiplied by x 1 minus x 0. If I divide, it will be 1 over of this quantity. So, 1 over of, I should not say I write 1 b minus a. I should write it as something else. I think f of s 1 minus, that makes it smaller. There is a minor thing. I think from here, if you want to go, then you should, yes, that is a small h. I should, because this will not give you directly this. You have to divide by that. x 1 minus x 0. So, this is not correct. We should have x 1 minus x 0 multiplied by this is less than or equal to this quantity. So, we will have 1 over of x 1 minus x 0. Can I bind that with something? Yes, I think.