 So, let us try to answer these questions. So, let us note, supposing f is integrable. That means, that is, the least upper bound of L p f over p is equal to greatest lower bound of upper sums with respect to p, the partitions. So, given a partition p, what is the difference between upper and the lower sums? If f is integrable, then what is the difference? How big is the difference? If these two quantities are same, we should be able to say that the difference between the upper and the lower can be made as small as you want. So, let us write this as, so let it be given. So, this common value, we agree to denote it by integral f dx. That was a common value. Now, if this is the least upper bound, that means what? Look at a to b, then a to b fx dx. This is the least upper bound. So, least upper bound, it is the sort of the largest value. Nothing smaller will work. It is the upper bound and nothing smaller will work. Let us look at minus epsilon. This cannot be an upper bound for the set. That means what? Then there should exist a partition p such that this cannot be the upper bound. That means there has to be an element on the right side of it. So, it has to be less than l p f. There is a partition so that this happens. So, definition of least upper bound. Similarly, also with respect to the other one, there exists a partition q, a partition such that if I look at the integral, that is the greatest lower bound plus cannot be. That means there has to be an element on the left side of it. So, it has to be bigger than u q f. So, let us call it as 1, call this as 2. So, this must happen. Now, from here, I want to find out, this is a partition p, this is a partition q. These two may be different partitions, but I want to bring them to a common partition. So, how do I bring them to a common partition? Given two partitions p and q, I want to have a partition which is common to both, common to both in the sense that the new partition should be a refinement of both of the original ones. So, the simplest thing is put all the points of p and q together in one partition. So, let p union q denote the common refinement of p and q. Then what happens when you refine? When you refine, lower sums tend to increase. So, from one integral a to b, fx dx, which was less than this minus epsilon is less than L p f, which is less than L p union q f. Because, when you refine, you get something more for the lower sums. Similarly, two implies integral a to b, fx dx plus epsilon will be greater than the upper sum with respect to q, which is bigger because the upper sums tend to decrease with the refinement, upper sum p union q of f. Now, from these two, now partitions are same, p union q, p union q. I know that the lower sum is always less than or equal to the upper sum. For a given partition, the lower sum is always less than or equal to the upper sum. So, from these two, what do I get? Upper sum p union q f minus L p union q f. From this, what you will get? Can you say something is less than the same quantity minus epsilon, the same quantity minus plus epsilon. So, this is less than 2 epsilon. Is it okay? Add and subtract epsilon if you want. Or, if you want, you can calculate this plus epsilon, this minus epsilon less than or equal to 2 epsilon. So, that means what we have shown is, if f is integrable, if f is integrable, then there is a partition, so either the difference between the upper and the lower must be less than epsilon for any given epsilon. That is intuitively what we wanted actually. So, it says, thus f integrable implies for every epsilon bigger than 0, there is a partition p such that there is a partition. Do not confuse with that same p. There is a partition, what shall I call? p star such that upper sum with respect to p star minus the lower sum with respect to p star is less than f. So, this is one way of testing, whether a function is integrable or not. If integrable, then I should be able to do that. Can I say the converse is also true? Supposing f is a function, say that for every epsilon, there is a partition with this property. That means what? Lower is always less than or equal to upper and it says the difference between the upper and the lower is less than epsilon. So, what happens to that? It goes on increasing the lower ones. So, least upper bound, greatest lower bound, the difference I can make it small for any epsilon. That means they must be equal. So, conversely, so converse also holds. Conversely, if above holds, so if star holds, then f is integrable. Is that okay for everybody? If the difference between the upper can be made as small as you want, l p f is always less than u p f. So, the greatest least upper bound of l should be equal to the greatest lower bound of u. That means the function is integrable. So, this is one way of checking whether the function is integrable or not. So, the theorem says, we did not write it as a theorem. So, let us write the theorem says, f is a function from a, b to r. Then f, f is integrable if and only if for every epsilon bigger than 0, there is a partition p such that u p f is less than epsilon. See, keep in mind, we are not trying to find what is that common value. This only says, this is a way of checking whether something is integrable or not. Go back to something, a sequence is convergent if and only if Cauchy. So, limit exists if and only if it is Cauchy. It is something like that. Checking the limit will exist. For a Cauchy sequence, for a sequence, you can check, just check it is Cauchy. If you are not interested in finding the limit, same way to check whether the function is integrable or not, you have to just make the difference to be small, as small as you can. That is good enough. Let us look at another one. Now, supposing another way of looking at the same thing would be, supposing I can find some real number alpha, which lies between all upper sums and all lower sums. Whenever I give any partition and construct the lower sum, construct the upper sum, I know the lower sum is always less than or equal to the upper sum. But, somehow I am able to guess a number alpha, say that the lower sum is less than or equal to alpha, is less than or equal to the upper sum for every partition. Then, what that alpha should be? That should be the integral. That should be the common value, because I can make them small and small. That is the second part. Let us call it one. I would like you to write down, if f is integrable and alpha belonging to R is such that upper sum is bigger than or equal to alpha is a lower sum for every p, then that implies this alpha is equal to integral a to b f x d x. Actually, you can say, if and only if. Converse is also obvious, namely that f is integrable if and only if this happens, because if this such a thing alpha happens, then the difference between upper and lower will be small. So, this is the way you check whether a function is integrable or not. Let us look at some examples before we go to computing. Let us look at some examples. Let us look at a function f from a, b to r. f of x is a constant. It is a constant function. Then, what is the lower sum? c times b minus a. Upper sum is also c times b minus a. Whatever be the partition, upper sum is same as the lower sum. That is c times b minus a. So, that implies f is integrable and integral of f x d x is equal to, say, this sort of intuitively tells me that for rectilinear figure, the graph of the constant function is straight line. So, the graph below that, the area below that is a rectangle, actually. So, area of the rectangle with height as c, length as base as b minus a should be this. So, the integral is actually making sense that it is the area below the graph of the function in that sense. Let us look at f. Another thing. So, let us look at f of x. So, let us look at monotonically increasing. It is a monotonically increasing function. Then, I want to find out what is the upper sum and what is the lower sum. Can I make it small? So, how do I do that? It is monotonically increasing. So, what is the minimum value of the function in any interval? If it is monotonically increasing, the minimum value will be the value at the left end point of that interval. Maximum value will be the value at the right end point. So, let us write for any partition p, note for p, any partition, what is the lower sum? So, let us say p is the partition, which is a equal to x0 less than x1. So, that is equal to sigma. So, minimum value. So, at the interval xi minus xi minus 1, that is the length. So, f of xi, lower sum. So, left end point. So, I should be writing as lower end point. This is what the interval is, xi minus 1. What is the upper sum? So, that is sigma f of xi, xi minus xi minus 1. So, that will be i equal to 1 to n. So, what is the difference between the upper and the lower? This minus this. So, what is that equal to? So, that is, so from 1 and 2, I can compute. So, let us do one thing. Let us specialize the partition, so that I know what is xi. So, let us imply if the partition p. So, this is the partition. Let us say that what we are doing is, we are choosing the partition AB such that the length of each sub-interval is b minus a divided by n. That means what? I am dividing the interval a to b into n equal parts. So, that means p, if p is such that xi minus xi minus 1 is equal to b minus a divided by n for every. Then, what will be this difference equal to? What will be the difference equal to? Each one is b minus a divided by n. So, that comes out. The length of the sub-interval comes out. So, it is b minus a divided by n into sigma f of xi. So, that is, what is sigma of f i? Consequently, they will cancel or whatever it is. Let us keep it as it is. It does not matter. f of xi minus f of xi minus 1 i equal to 1 to n. Now, let me take it slightly further from here. So, that is same as saying upf minus lpf is equal to b minus a divided by n. Now, this I want to remove. The sum I want to remove, I want to make it a constant. f of xi, so what is the largest value of f? If we want to take the value like, it is the value at the end point b. So, I can say this quantity is less than or equal to f of b plus f of b twice. So, I am saying that this quantity is, if you like, two types, maximum value. What shall I write? Does it cancel out? Yeah, it cancels out. I do not have to do anything actually. So, what is the sum? f at the end point that is b and the last end point f of b minus f of a, consecutive terms will cancel out. So, I do not have to do anything. Now, why I am doing all this is, it says if I choose this particular partition p of dividing the interval into n equal parts, the difference between the upper and lower, 1 over n is coming here. So, I can make that this quantity as small as I want. So, that says, so choosing n large enough or we can just say that choosing n large enough, u p f minus l p f will be less than epsilon. For any given, I choose p such that this is possible. When p n is n equal parts, the upper minus the lower will be less than some constant divided by n. So, if I make the number of points, number of subdivisions large enough, then the upper minus the lower will become small. So, that says every monotonically increasing function is integrable. We do not know the integrable, but it says it is integrable. So, implies hence f monotonically increasing implies f integrable. So, let us look at second example for this. The same will be monotonically decreasing also. Instead of f b minus f of a, it will become f of a minus f of b. So, that is not a problem. Let us look at another simple example. Say, f is from 0, 1 to R, f is equal to the indicator function of rationals in 0, 1. So, remember we define what is called the indicator function of a set. So, that means this is equal to 1 if x is rational in 0, 1 is equal to 1 and it is 0 otherwise. That was the indicator function. So, for any partition for every p or for every p, a partition of 0, 1, what is upf? I want to calculate the upper sum. That means given any partition, it will have subintervals. I want to look at the maximum value of the function in any subinterval. So, what will be the maximum value of the function? Function takes only two values, 0 and 1. And any subinterval will have an irrational unit. So, for any subinterval, the maximum value will be 1. How small the interval may be? So, 1 times, so, upper sum will be capital M i that is 1 multiplied by x i minus x i minus 1 summation. So, consecutive terms will cancel out b minus a and that is equal to 1. Length of interval is 1. Is that okay for everybody? So, upper sum is 1 for every partition. What is the lower sum? So, let us calculate the lower sum. So, the lower sum with respect to any partition, the smallest value the function takes is 0 and that takes at every rational. Every interval has a rational inside it. So, the minimum value of the function in any subinterval is 0. Whatever be the length of interval does not matter. So, this is 0 for every partition. You are using the fact that rationals are dense and irrationals are also dense. So, that implies this f is not integrable. This function is not integrable.