 We have seen that one of the relatively easy way of showing that a function is integrable either Riemann integrable or Riemann Stilder integrable is the one criteria which we discussed last time and that is the following that let me write that is we have shown that f is integrable that is r a b alpha if and only if for every epsilon bigger than 0 there exists a partition p in this script p such that for this particular partition u p f alpha minus l p f alpha is less than epsilon and using this we have shown that every continuous function is Riemann integrable and also with a slight modification of same proof we have observed that every continuous whenever alpha is any monotonically increasing function every continuous function is also Riemann Stilder integrable and the main tool used there was if you remember that every continuous function defined on such a closed and border interval is uniformly continuous and using that we subdivided the partition in a particular way for continuous functions we can also notice one more thing in fact we have noticed that whatever be the partition this part is always true that is l p f alpha this is always less than or equal to u p f alpha then we have also observed let me keep this is less than or equal to any Riemann Stilder sum and this is less than or equal to upper sum and this upper sum is less than or equal to big m into alpha b minus alpha a and this lower sum is always bigger than or equal to small m into alpha b minus alpha a where as usual the small m and big m have the same meaning which we have discussed earlier namely that big m is the supremum of f x over the interval and small m is the infimum. What we also know is that whenever the function is integrable the upper and lower integrals coincide and upper integral is the infimum of the upper sums and so it will be always less than or equal to any upper sum similarly lower integral is the supremum of the lower sums so it is bigger than or equal to any lower sum so both these numbers lower integral as well as the upper integral they lie between any of the lower sum and the upper sum and in particular when the function is integrable the number integral is nothing but the common value because we say that the function is integral is same as saying that the upper integral is same as the lower integral so that common number will also lie between these two always common no not these two this and this this and so what we can observe here is that okay let me write it once again here if f is integrable that is if f is this or a b alpha then then we can say this that is this is less than or equal to integral a to b f d alpha and this is less than or equal to integral no this is less than or equal to up f alpha and we already seen that this is less than or equal to big m into alpha b minus alpha a and this is bigger than or equal to small m into alpha b minus alpha a okay and usually this function alpha will be a strictly monotonically increasing function so alpha b minus alpha a will be strictly bigger than 0 in most cases unless in some very trivial examples so what one can say further is the following okay that is I can say that this small m is less than or equal to 1 by alpha b minus alpha a into integral a to b f d alpha and this is less than or equal to big m okay this much is true for any integrable function right this if the function is continuous we can say something more okay and what is that this big m is a supremum okay and we know that if the function is so continuous on this interval then this value is attained that means there is some point let us say the point x 1 such that f at x 1 is equal to m right similarly there is some point let us say x 2 such that f at x 2 is m okay that means these values small m and big m are taken actually taken by the function and if the function is continuous we know that if it takes any two values it will take all the values which lie in between those two the famous intermediate value theorem so that means what that means there should be some point at which the function takes this value also right that will happen if a function is continuous okay so let us let us just see what is the consequence of that so if f is also continuous on a b then there exists a point c in this interval a b such that such that what should happen that is 1 by alpha b minus alpha a into this this is same as c that is 1 by alpha b minus alpha a into integral a to b f d alpha this is equal to c not c this is equal to f at c because there is some point c at which the function f takes the value given by this that is f of c is equal to this number or what is same as same that suppose we rewrite the same thing in a different manner that is integral a to b f d alpha that is same as f at c multiplied by alpha b minus alpha a and this statement that whenever f is continuous there exists c in a b satisfying this last okay that is called the first mean value theorem of integral calculus that is called the first mean value theorem of integral what is the theorem that if f is continuous see once it is continuous it is automatically integrable that is what we have already already shown then there exists c such that integral a to b f d alpha is equal to f c into alpha b minus alpha a that is called first mean value theorem okay we will use the standard short form for mean first mean value theorem of integral calculus by mean value theorem because it looks very similar to mean value theorems of differential calculus it talks about some point c inside the interval a b such that something happens at that point and why first obviously it means something is going to follow something some more theorems are going to follow that is why but anyway it will not happen immediately after some time alright so that is about this continuous functions there is one more class of functions in case of which we can prove in a straight forward manner that those functions are integrable and those are monotonic functions okay so that is another theorem that we want to say let us as usual just as we have done in case of continuous functions we shall first prove for Riemann integrals and then we will go to this arbitrary Riemann Stelios integral so if f is monotonic monotonic on a b then f is of course we are always assuming that this function alpha is okay for the time being I am not talking about alpha then f is Riemann integrable okay let me write in words then f is Riemann integrable okay now what happens if f is monotonic okay f is monotonic means what it is either monotonyically increasing or monotonyically decreasing let us prove in one of the cases prove in the other case will be similar okay so assume f is monotonyically increasing in order to know what is happening in this case see the first thing that we can observe here is that what we have defined let us recall the definitions of this big mi and small mi that is this big what was this big mi big mi was nothing but supremum of fx where x belongs to this ith sub interval xi minus 1 to xi okay but we know that f is monotonyically increasing so what can we say about this big mi for an increasing function on an interval what will be the maximum value that it takes at the right hand end point okay so this is nothing but f of xi right similarly small mi that is nothing but infimum of the same and that will be nothing but f at xi minus 1 right f at xi minus 1 right okay now for any such partition suppose we look at the difference between the upper sum and lower sum since we are talking about the Riemann sums let us just look at u pf minus l pf okay so this will be the thing but sigma i going from 1 to n sigma i going from 1 to n it will be nothing but f xi minus f xi minus 1 multiplied by delta xi okay right and delta xi is nothing but xi minus xi minus 1 okay alright now suppose I choose a partition in such a way that this delta xi is constant okay suppose I choose a partition in such a way that this delta xi is constant if the delta xi is constant then there is only one possible if there are n sub intervals each must be of length 1 by n okay so choose a partition in such a way okay that n we shall choose later okay but first we will say that if there are so choose p such that choose p such that delta xi is see there are n sub intervals and if each interval is has to be of the same length and the total length is b minus a so each of the delta xi must be b minus a by n okay so each of the delta xi must be b minus a by n okay and so what will be the partition that is also easy to say we know that a is x naught then x 1 will be nothing but a plus a plus b minus a by n that is x 1 x 2 will be a plus 2 times b minus a by n etcetera etcetera okay so in general one can say that xi is equal to a plus i times b minus a by n choose this partition okay so for this partition what will happen to this up f minus l p f this delta xi is b minus a by n so that will come outside the summation sign okay that will come outside the summation sign so this is nothing but b minus a by n and multiplied by this sigma i going from 1 to n f xi minus f xi minus 1 okay so once that once that the delta xi has come outside what will remain suppose you take this sum for example first element will be f x 1 minus f x 0 second will be f x 2 minus f x 1 etcetera okay so it will be simply f of x n minus nothing but f b minus f a nothing but f b minus f a okay so you can put that here f b minus f a so suppose you do that simplification so what we get this this up f minus l p f this is nothing but this b minus a divided by n and multiplied by f b minus okay you can say then okay now come come back this criteria what we said that f is integrable if and only for every epsilon there exists a partition p such that up in our case this alpha is just x okay so up f minus l p f is less than epsilon okay all right now in our case what is up f minus l p f it is b minus a f b etcetera okay there is only one thing that depends on the partition and that is this number n everything else is fixed okay so the only question is this given epsilon can we choose n such that this whole thing becomes less than epsilon that is obvious right because we can make n sufficiently large so that this b minus a into this number is suppose this number is fixed suppose you call that number k okay this whole thing is k by n okay then choose n such that n is bigger than epsilon by k okay m is bigger than epsilon by k then this difference will be less than epsilon okay and that means that that means that f is Riemann integrable right f is Riemann integrable okay all right is this clear now coming back to Riemann's Tildes integrable Riemann's Tildes integrable see one can do all these calculations up to this okay this will be replaced by up f alpha minus l p f alpha okay let me let me carry on here okay suppose instead of Riemann integrability suppose I look at Riemann's Tildes integrable then this will not change okay b mi and small mi those numbers depend only on f okay so if I look at this difference p f alpha minus l p f alpha this will be sigma i going from 1 to n then multiplied by f sorry f xi minus f xi minus 1 and this is multiplied by instead of delta xi it will be delta alpha i okay it will be delta alpha i okay and what is delta alpha i recall delta alpha is a thing but alpha xi minus alpha xi minus 1 okay so instead of delta xi there we have delta alpha i here we have delta alpha i here and what was our next argument we have said that we can choose a partition in such a way such that that delta xi is constant the delta xi and that the delta xi turned out to be if it has to be constant it has to be b minus a by n so what will be the change here we will want is delta alpha i to be alpha b minus alpha a divided by n okay delta alpha delta alpha i should be alpha b minus alpha a divided by n now the whole question is whether okay that is if we want to imitate that we want this okay we want delta alpha i is equal to alpha b minus alpha a divided by n okay for example suppose I suppose I want let us say suppose I look at say delta alpha 1 what is delta alpha 1 that is alpha 1 is nothing but alpha at x 1 minus alpha at x naught this should be same as delta alpha b minus alpha a alpha b minus alpha a divided by n okay or what is the meaning it means alpha x 1 should be same as alpha x naught plus alpha b minus alpha a by n okay now if you compare this situation here here it was easy to say that you just choose x 1 such that x 1 is a plus b minus a by n okay choose choose x 1 x 1 is same but a plus b minus a by n but here what is the case we want to choose x 1 such that alpha x 1 is equal to alpha x naught plus alpha b minus alpha a by n okay so the question is whether such an x 1 exists or not okay whether the function alpha takes this value or not okay right that is the question okay and in general we only thing that we know about the function alpha is that it is monotonically increasing okay only thing that we know is it is monotonically increasing right but we know nothing further but if we impose some additional condition see is it is it clear to you that alpha x 1 that is anyway bigger than or equal to alpha see this is this alpha is monotone this is a non negative number okay so so this alpha x naught is less than this alpha x 1 and this is obviously less than or equal to alpha b minus alpha a okay or not alpha b minus a this is I can say less than or equal to alpha b okay that is clear all right because alpha I can say that alpha b alpha x naught is nothing but alpha a alpha x naught is nothing but this is nothing but alpha a alpha a plus alpha b minus alpha a divided by n okay and one can say that for any positive n this will be less than or equal to alpha b minus alpha a and then hence this whole thing will be less than or equal to alpha b okay right so if we knew that alpha is continuous if we knew that alpha is continuous then we can say that it takes this value it takes this value so it will take all the intermediate values so I can find x 1 such that alpha at x 1 is equal to this otherwise we cannot say this otherwise so if we add the hypothesis of continuity to alpha then I can say I can choose alpha x 1 in this fashion having chosen alpha x 1 use the same technique I can choose alpha x 2 also in the same fashion so if alpha is continuous then we can choose the partition x naught x 1 in such a way that each delta alpha satisfies this condition okay otherwise you may or may not be able to do this okay so what it means is that this will work if you assume that f is monotonic that alpha should be not only monotonic but also continuous and that is remember that is true for rebar integration this because in in case of rebar integrals alpha x is a thing but x that is a continuous function okay so that is automatically satisfied here okay so we can say that if f is monotonic on a b then f is rebar integrable also if alpha is continuous then that means all this is assumed f is monotonic alpha is also monotonic increasing and continue continuous then f belongs to this r a b alpha now we go to the another task just so just as in case of differentiability we have done if we know that a function is differentiable then what can be for example suppose we know that f1 and f2 are differentiable then f1 plus f2 is differentiable etc a similar thing we want to do in case of integration okay that is suppose we know that some functions are integrable then constructing functions using those functions and whether those new functions are also integrable or not that is the question that we are going to deal with now right so let me start with this okay so suppose this is a theorem okay suppose f and g are that is this means inverse f and g are rebar integrable with respect to some monotonic increasing function alpha okay then okay there are several statements we want to make okay I shall write one by one and then we shall discuss the corresponding proofs also all these proofs depend basically on the definition of integrability okay so first thing that we want to say is this f and g are also sorry f plus g f plus g is also integrable f plus g is also integrable and we can say something more integral a to b f plus g d alpha this is same as integral a to b f d alpha plus integral a to b g d second thing is okay f or g it is suppose we take any k for every real number k in r okay for every real number k k times f is also rebar still integrable integral a to b k f d alpha this is same as k times integral a to b f d alpha okay what what does it mean in words that if you take two functions in this if f and g are rebar integrable rebar still just integrable then there are some as well as product with a constant is also integrable okay in other words since you are also doing a course in linear algebra simultaneously it is basically same as saying that this is a real vector space set of all ribbons still just integrable suppose alpha is kept fixed then set of all ribbons still just integrable functions is a vector space over real numbers if you take all functions on a b that is also vector space you can say this is a subspace of that and what can we say about can we also describe this part in terms of linear algebra using terminology of linear algebra have you studied what is meant by a linear functional yes have you studied what is meant by a linear functional so what does this say that integral a to b f plus g d alpha is same as integral f d alpha plus integral g d alpha and second thing says that integral k f d alpha is same as k times integral a to b f d alpha right so in other words if you take this map f going to integral a to b f d alpha or suppose you define phi f is equal to integral a to b f d alpha then phi is a linear functional on this vector space so basically proving these properties of this integral is basically showing that integration is a linear functional on this vector space of all integrable functions now since there are three functions involved here f g and f plus g and whenever we want to talk about the integration obviously upper sums lower sums are going to come into picture and so we have to discuss the upper sums lower sums of three different functions so we need some notations for this big m i small m i etcetera so let us start with this let us say that suppose we i will call this function h let h be equal to f plus g then we will have to talk about the upper sums corresponding to the function h corresponding to the function f and corresponding to the function g all right suppose let us say that we will have suppose I call this m i as the supremum of let us say hx hx from x in hx for x in xi minus 1 to xi I will get similar numbers for corresponding to f and corresponding to g right so let us say suppose I denote this by m i star as supremum of f x same thing x is in xi minus 1 to xi and m i double star as supremum of f x in not f x or g x for same thing x varying with the interval xi minus 1 to xi okay can we see obvious relationship between this m i m i star and m i double star what is that m i must be less than or equal to m i star plus m i double star okay that is so we can say that this m i must be less than or equal to m i star plus m i double star what does it say about the upper sums we can say that from this it follows that upper sum of h must be less than or equal to upper sum of f because all that you are doing is that multiplying each of this number by delta alpha and taking the sum so we can say that this implies that up h alpha is less than or equal to up f alpha plus up g remember we are going to use that same method to prove that h is integrable what is the method that given epsilon will produce a partition p such that up h alpha minus l p h alpha is less than epsilon that is the idea okay so what we can what we can do now is that okay now now this is true for every partition whatever be the partition that upper sum of h is less than or equal to upper sum of f plus upper sum of g that is true for every partition okay but we know that f and g are integrable okay we know that f and g are integrable so what we can say is that given any epsilon given any epsilon i can find a partition suppose i call it partition p okay let us let us start with epsilon okay let us let us start with an epsilon bigger than 0 what is the m our aim is to produce a partition p okay such that for that particular partition u of h up h alpha minus l p h alpha that should be less than epsilon okay that is the that is the aim okay but we know that such a thing can be done for f and g such partitions can be found for f and g so we shall use we shall use that fact first okay let us first stick to the upper sums one can say that there exists partition suppose i call it partition p1 okay there exist partition okay see here we are going to have the several partitions and we are going to use the thing that we have shown earlier that whenever you take a refinement the upper sums decrease and lower sums increase okay so we will find a partition corresponding to f corresponding to g and then take something which is common refinement of both okay so we can say there exist a partition p1 such that okay we know that f and g are integrable so integral is nothing but the infimum of the upper sums okay in fact that is what we had called upper integral but when the function is integrable those two things are same so i can say that i can find some partition such that the difference between the integral and the upper sum is less than epsilon or epsilon by 2 or epsilon by 4 whatever we want okay that is that is what we can do okay so i can say there exist a partition p1 such that so see u p p1 f alpha u p1 f alpha minus integral a to b f d alpha this is less than okay i do not remember okay let us start with epsilon by 4 okay if we require we shall change that epsilon by 4 to epsilon by 3 or 6 or whatever we require okay this can be done okay then one can say that similarly there exists a partition p2 such that upper sum u p1 g alpha is less than integral i can i can rewrite this inequality i can bring this integral to the other side okay that is convenient okay instead of writing like this i shall say that this this upper sum is less than this integral plus epsilon by 4 okay and similarly this upper sum is less than integral g d alpha plus epsilon by 4 okay now this was p2 right now consider the partition p1 union p2 okay consider the partition p1 union p2 for that partition both of this is inequality should satisfy should be satisfied okay right so okay suppose i call that partition now p3 okay let p3 be equal to p1 union p2 okay then what what we want to say is that then u p3 f alpha this is less than or equal to u p1 f alpha and that is less than integral a to b f d alpha plus epsilon by 4 similarly also u p3 g alpha this is less than or equal to since p3 is also a refinement of p2 okay this is less than or equal to u p3 g alpha and that is less than integral a to b g d alpha plus epsilon by 4 okay all right now you look at u p3 h alpha we have seen earlier that for any partition u p h alpha is less than or equal to u p f alpha plus u p g alpha okay so u p3 h alpha that should be less than or equal to u p3 f alpha plus u p3 g alpha that is this should be less than or equal to u p3 f alpha plus u p3 g alpha okay and this is less than integral a to b f d alpha plus epsilon by 4 and this is less than integral a to b g d alpha plus epsilon by 4. So, there are some should be less than this plus this plus epsilon by 2. So, this should be less than integral a to b f d alpha plus integral a to b g d alpha plus epsilon by 2. So, do you agree up to this? Now, whatever we have done for the upper sums is it clear that we can do a similar thing for the lower sums. What is the obvious changes? Here you will take small m i small m i star and small m i double star and inequalities will be reversed. Small m i will be bigger than or equal to small m i star plus that is in other words lower sum of h alpha will be bigger than or equal to lower sum of f alpha plus lower sum of g alpha. So, everything that you have done in a similar way you can do everywhere inequalities will be reversed. For example, this l p 1 f r y will be bigger than integral a to b f d alpha minus epsilon by 4. So, one can say that following a similar method you can find a partition. You can find a partition such that for that partition lower sum is bigger than this number minus epsilon by 2. So, I shall skip those details do that on your own to convince yourself. I will say that in a similar way we can find a partition. We can find a partition. Now, suppose I call that partition p 4 p 4 such that such that lower sum of with respect to this p 4 h alpha lower sum with respect to this p 4 h alpha is bigger than integral a to b f d alpha plus integral a to b g d alpha minus epsilon by 2. So, again we go is this clear. So, then again we go back to the same degree. Here we have partition p 3 for which u p 3 is less than this number plus epsilon by 2 and here we have partition p 4 such that this is bigger than this number minus epsilon. If this were the same partition we could immediately say that this minus this is less than epsilon, but that is not the case because. So, but we know that how to deal with such things. Just take the union of the two that will be a common partition for both common refinement and so that both the inequalities are satisfied. So, I will just this statement I require. So, I will go back here. So, let p be equal to whatever partition we have p 3 and p 4 p 3 union p 4 and we want to show that this p is the required partition p is a refinement of both p 3 and p 4 and we know that for the refinements upper sums decrease and lower sums increase. So, let me just write it here. So, I can say that this u p h alpha u p h alpha is less than or equal to this and hence less than or equal to less than this plus epsilon by 2. Sorry, this should have been everywhere this should have been that is correct. Yes sir third line this line third line here p 2 common sir second in the middle u p 2 g alpha is same. That is right because we have used the fact that p 3 is a refinement of p then where we are using that p is p 3 union p 4. So, u p h alpha is less than or equal to u p 3 h alpha and similarly lower sums increase. So, lower sum with respect to p should be bigger than or equal to lower sum with respect to p 4. So, we can also say that l p h alpha is bigger than or equal to l p 4 h alpha and so that is bigger than all these things. So, u p h alpha minus l p h alpha this is less than this less than what this minus this right this less than. So, when you subtract these two numbers will get cancelled. So, this will be epsilon by 2 minus. So, this is less than epsilon. So, what we have. So, this shows that this implies that h is Riemann Stilder's integrable. It shows this first thing and though it does not show immediately this you can observe that we have done whatever is required for showing this we have done whatever is required for showing this. Once we show that h is Riemann Stilder's integrable it means that integral h d alpha exists. It means that integral h d alpha exists and that is less than or equal to any upper sum that is less than or equal to any upper sum and also bigger than or equal to any lower sum. So, but we already have one partition for which we we not only proved this. In fact, we have proved many much more than this. We are in fact proved that u p h alpha is less than or equal to this number plus epsilon by 2. So, let us just recall that. So, once we know that h is Riemann Stilder's integrable it means that this number exists integral a to b h d alpha exists. That is something we already proved and moreover this number must be less than or equal to any upper sum. This number must be less than or equal to any upper sum. So, that means in particular this is less than or equal to this u p h alpha. This must be less than or equal to u p h alpha. But we already know that this p has the property that u p h alpha is less than or equal to integral a to b f d alpha plus integral a to b g d alpha plus epsilon by 2. That is something that we know. So, that means this is less than integral a to b f d alpha plus integral a to b g d alpha plus epsilon by 2. What is more? We also know that this is the integral must be bigger than or equal to any lower sum must be bigger than or equal to any lower sum. So, this must be in particular bigger than or equal to l p h alpha and we know that l p h alpha is bigger than or equal to again let us write this is bigger than or equal to l continue here. It is again the same 2 number f d alpha plus integral a to b g d alpha minus epsilon by 2. What does it mean? That is this number integral a to b h d alpha this lies between whatever is the sum of this 2 number plus epsilon by 2 and this minus epsilon by 2. So, what can we say about the difference between this and the sum of these 2? Suppose I want to look at this integral a to b h d alpha minus integral a to b f d alpha plus integral a to b g d alpha. Suppose I want to look at difference between this difference. In fact, we have shown that this difference is less than epsilon by 2 because this shows that this minus this number is less than epsilon by 2 and bigger than minus epsilon by 2. So, this difference is less than epsilon by 2, but epsilon was arbitrary remember that we started with any arbitrary epsilon and for that epsilon we have shown that the if you take this number and whatever is there on that is difference between left hand side and right hand side is less than epsilon by 2 for any arbitrary epsilon. So, the only way in which this can happen is that those numbers must coincide those numbers must coincide and that gives this result. That is if the 2 functions are remastered as integrable their sum is also remastered as integrable and integral of f plus g same as integral of f plus integral of g. Then what I would say is that the proof of this is also exactly similar. There are no new ideas involved here. In fact it is even simpler because see here what was the whole idea relating to the upper sum of this f plus g with upper sum of f and upper sum of g and similarly for the lower sums. Here you have to do only for the function k times f and that is very easy because you can say that upper sum of that is you can easily show this that for any partition p u p f alpha is say we want k times f we want k times f of course you will have to make a slight slight you have to be little bit careful here. You have to make the cases when k bigger than or equal to 0 k less than or equal to 0 etcetera. If k is equal to 0 there is nothing to be proved k is equal to 0 there is nothing to be proved because it is a constant function 0 and 0 that is integral. If k is bigger than 0 then we can say this if k is bigger than 0 then then upper sum of k times this is same as k times u p f alpha. And similarly one can say about the lower sums and so you can prove this at least when k is positive you can prove this at least when k is positive. If k is negative what will happen is that this will be k times the lower sum of this if k is negative upper sum of p k f alpha is nothing but k times lower sum with this and similarly for the lower sum and again you can basically use the same technique. So, no new ideas are involved to prove this so that is why I will skip this proof basically all that you need to do is given epsilon you have to produce a partition p such that for that partition difference between upper sum and lower sum is less than epsilon. And then this part you can prove in the similar way using this we will stop with that. There are many other properties of this type and those we shall see in the next class.