 Welcome back. We were discussing we were building up towards monotone convergence theorem. We said we defined point wise convergence and almost everywhere convergence of a sequence of functions f n of omega to f of omega. We said that even if f n of omega converges to f of omega for all omega or for almost all omega. It does not necessarily mean that integral f n d mu converges to integral f d mu. It is not true generally. We gave a counter example to show that this limit of limit an integration cannot be interchanged. We gave an explicit example where the integral of the limit and the limit of the integral were different. So, the monotone convergence theorem gives one sufficient condition under which you can in fact, interchanged limit an integration. So, the condition is not just convergence almost everywhere, but monotonic convergence almost everywhere. So, I will state the monotone convergence theorem. So, you have. So, what you have is omega f mu sum measure space and let g n be a sequence of measurable functions such that g n increases to g mu almost everywhere i.e. g n of omega is less than or equal to g n plus 1 of omega mu almost everywhere. So, limit and infinity g n of omega equal to g of omega also mu almost everywhere. Then integral g n d mu increases to integral g d mu. So, this is the statement of the monotone convergence theorem. So, you have. So, these are non negative. So, these are non negative functions. They are non negative measurable functions such that you have g n converging to g mu almost everywhere, but you also it is not just convergence. It is convergence monotonically. So, you have g n of omega. So, g n of omega is less than or equal to g n plus 1 of omega. So, this is true for all n is equal to 1 2 dot dot dot. So, this holds mu almost everywhere, which means that there may be a set of mu measure 0, where may be monotone density does not hold. So, g n of omega is less than or equal to g n plus 1 of omega. So, for every omega you have a sequence of increasing numbers. This will be a sequence of increasing numbers, except it may not hold in a mu measure 0 set. And similarly, you have convergence mu almost everywhere, g n of omega converges to g of omega for almost all omega mu almost everywhere. In that case, you are guaranteed that the limit and integral can be interchanged. So, here again I write an up arrow, because these the sequence of these integrals will be bigger and bigger, because these functions are themselves getting bigger and bigger. So, the limit of this integral. So, another way of writing this is limit intending to infinity integral is equal to integral of limit. So, this is saying that therefore, you can interchange limit and integration. So, long as you have monotonic convergence, not just convergence, but monotonic convergence almost everywhere. The statement of this theorem clear? Any questions on the statement? So, this theorem is perhaps the most important in integration theory. It is considered the cornerstone of integration theory and it has lots of applications. It also has some important corollaries, which we will do a little later. So, the proof of the monotone convergence theorem is little long, but not difficult. So, in fact, if you look at MIT open course where in lecture number 11, there is a full proof of the monotone convergence theorem and it runs to about 2 pages, which is not a very long proof come to think of, because it is a very fundamental theorem. It is a very important foundational result, but it proof is only 2 pages long. In fact, I suggest you look at it and how do you suspect that you will prove it? There is only one way to prove all results in integration. Start with simple functions. So, what you will assume? So, the way the proof builds up is that you assume that you have, you first assume that the limit function is a simple function. This g is some q, some limit function is simple and you have monotonic convergence to a simple limit function. Then you have, then you can explicitly prove that this theorem holds and the key ingredient in that is continuity of measures, continuity of any general measure, not just probability measures. You first prove it for simple functions and then you use the supremum definition for non-negative functions and you can use, you already proved MCT for simple functions. Then you use the supremum definition to prove it for arbitrary non-negative functions. So, there is all proofs in integration go like this. So, actually I had half a mind to do it in class, but it will probably take all of today's lecture and I may lose you here in between I guess. So, I decided against doing it in class, but it is not a difficult proof. I mean for a such a fundamental theorem in a measure theory, it is about 2 pages long and you can actually follow it and if you have difficulty following it, please do come to me. It is an interesting proof. So, I will not spend class time on it, because that is all I will be able to do today if I do it. So, in the example we gave, so this, so the example we gave. So, we said g n of last class, we said we took the functions, the sequence of functions as n 1 over n. Your sample space was 0 1 and you took I guess you put solid dots here, all the dots over there and in this case the integral of g n of omega was 1, but the limit function was 0. Limit of g n was 0 for every omega. So, it was not possible in this case to interchange limit an integration. In fact, you can see that monotonicity does not hold. After all if this is your g n of omega, g n plus 1 will look like. So, g n plus 1 would look like that, it will be at n plus 1 and 1 over n plus 1. So, while it is true, so g n has to get bigger. So, it is true that g n plus 1 is bigger than g n over here, but in this little interval g n plus 1 is 0, whereas g n is equal to n. So, in this set of positive measure, this set has positive measure namely the measure is 1 over n minus 1 over n plus 1. In that little set, you have failure of monotonicity. So, I mean which is 1 I mean if this were monotonic. So, if you this monotonic you will have by the MCT you will have convergence of the integrals as well. On the other hand, if you have to take g n of omega equal to 1 between 0 and 1 over n. So, for what I am saying is if you have if g n of omega were equal to 1 between 0 and 1 1 over n and. So, g n plus 1 will be 1 between 0 and 1 1 over n plus 1. In this case, the sequence of functions is. So, in this case monotonically decreasing. So, even so in this case you will have convergence of integrals. So, the reason it broke down here was it was bigger back, I mean the monotonicity did not quite hold. So, there is one property I had not proven. So, there was a property of linearity which I said this not easy to prove using the supremum definition. I said that we will get back to it after we do monotone convergence theorem. So, let us get back to linearity. So, this says that integral g plus h d mu is equal to integral g d mu plus integral h d mu. This is the property we never proved. As you might expect the proof goes in the usual sequence. You prove it for simple functions you take g n h s both simple and prove it. And then you have to generalize for this non-negative arbitrary non-negative functions as well. And then you can go to g plus g minus that is standard. So, now if first assume g and h are simple. So, you have a representation of g as sum over a i indicator of a i. i is equal to 1 through sum n finite and h as sum over. So, what I mean is g of omega is equal to sum over a i a i of omega. And similarly h is equal to j equals 1 through m b i indicator of some other b a some other sets. So, all this a i's and all this b i's are measurable sets f measurable sets. So, in this case what will g plus h be g plus h will also be simple. So, g plus h will have the representation sum over i equals 1 through n sum over j equals 1 through m a i I should write b j here. So, I will have a i plus b j indicator what a i intersection b j after all the function g plus h will take the value a i plus b j if your omega lies in a i intersection b j correct. So, now what you will assume. So, let us assume that these two are canonical representations. If I assume that these are canonical representations then I will have this a i's are disjoint and b j's are also disjoint. So, these guys will be disjoint for a different i i's and j's. So, these intersections will be disjoint correct. So, if I have. So, if I am looking at this integral g plus h d mu that will simply be sum i equals 1 through n sum j equals 1 through m a i plus b j measure of that guy right by definition measure a i intersection b j is it correct by definition. So, I can just rewrite this as sum over i equals 1 through n a i. So, this a i will come out of the j summation right and then I will write sum over j equals 1 through m mu a i intersection b j correct plus sum over j equals 1 through m. So, now I am looking at that summation. So, if I am looking at that summation I can bring the i summation inside right. So, that is just algebra you will get b j sum over i equals 1 through n mu same thing a i intersection b j good. Now, what is that equal to see this a i see this a i intersection b j if you just look at it as some sets index by j let us call this c j now. So, if you take. So, these are disjoint sets right because you assume that g n h are in canonical representation right. So, these sets are disjoint you are adding the measures of disjoint sets which means I should get the measure of yes exactly right. You will get the measure of the unions of all these guys over all these sets over j right. So, this will simply become. So, essentially. So, these are this is by countable additivity of measure finite additivity of measures this will become i equals 1 through n a i mu of a i right. I am just. So, you just have to establish that union over all that union of all these sets over j is simply a i right. And similarly the you flip the argument around if you take union over i this will give you the measure of the j b j right. So, you will get j equals 1 through m b j mu a well taking union over i. So, this should be b j right. So, that is exactly this is nothing but the integral of a g right this is integral g d mu plus integral h d mu. So, proving it for simple functions is quite elementary right. Now, you have to prove it for arbitrary functions arbitrary non-negative functions first right measurable functions measurable non-negative functions. So, there you use the monotone convergence theorem. Now, you say next let g and h be non-negative let g n and h n be a simple functions a sequence of simple functions right. There are sequences of simple functions such that g n increases to g and h n increases to h. So, what I am doing now is. So, I am given two non-negative functions right I am approximating these non-negative functions from below by a sequence of simple functions g n and h n. So, I have g n increasing to g and h n increasing to h. Now, the question is given two functions h and g can I always find simple functions that increase to say let us say g is some non-negative function. How am I guaranteed that I can always find simple functions g n that increase to g right is not clear right, but I will give you an explicit construction. So, this is possible. So, if you hold on a little bit I will show you that this is possible for now just take that this is possible. We will show that this is possible little later then we will invoke monotone convergence theorem. So, in this case. So, then you will see then you can show that g n plus h n will increase to g plus h right. If g n increases to a g and h n increases to h then g n plus h n will increase to g plus h fairly elementary actually all these can be almost everywhere it does not have to be for all omega right almost everywhere is enough. So, you have. So, this is true is not it. So, I have this. So, I have we have integral g plus h d mu is equal to limit n tending to infinity integral g n plus h n d mu right. Why is that true monotone convergence theorem right this is because of m c t right and this now I have h n and g n are simple functions. So, I have proven linearity. So, linearity among simple functions is already proven right. So, this will split right. So, this integral will. So, this you look at that will be limit n tending to infinity integral g n d mu plus. So, I can take the limit inside right. So, limit. So, so far I am right. So, I have just have skip one step. So, I have written this integral as integral g n d mu plus integral h n d mu and I am putting limit separately right that is that is always allowed. As long as the limits are well defined. So, now what happens. So, now g n is again a monotonic sequence of functions right. So, this will converge to. So, this will give you what you want this is again by m c monotone convergence theorem. So, is that proof clear. So, after the monotonic convergence theorem proving linearity of proving linearity of integral integral is a very simple matter except that you have to buy that this is possible right. You can always for any non-negative function g I can find a sequence of simple functions which increases to g right. It is possible to do this in many ways given any non-negative function g I can always find a sequence of simple functions that increase to g and I can do this in many ways. I will show you one explicit way in which this can always be done one valid construction I will show you. And this approximating a non-negative function from below using simple functions is anyway very useful to compute this abstract integrals. See remember that so far I have only given a supremum definition which I told you is conceptually fine it helps us prove a number of properties. But, it does not tell you how to evaluate the integral right. So, far if I give you some you go on integrate g of x is equal to x square over the 0 1 interval you actually do not know how to do it except treated as a lemma integral right. Now, I will tell you an explicit way of computing it. So, is this clear I am moving to the next topic. Approximating a non-negative measurable function from below using simple functions let g be measurable define. So, you are given some non-negative function g in some measurable space. Now, I am going to define a sequence g n an explicit definition of g n such that g n increases to g. Just one particular construction you may be able to do it in other ways as well. But, I will give you one construction which always works. We define g n of omega equal to i over 2 raise to n if i over 2 raise to n less than or equal to g of omega is less than or equal to well actually less than i plus 1 over 2 power n. And this is for i equals 0 1 2 dot dot till n times 2 power n minus 1. And if g of omega is bigger than or equal to n I define it as simply n. So, this is the sequence of functions I define. So, this is not I mean by no means the only way you can do it, but this works this is enough. So, let me just pictorially tell you what I am doing here. Of course, if I have to draw picture I have to assume let us say I have to assume that my space is r. Let me take omega equal to r. And so I have some non-negative measurable function. So, that is my g and this is my omega and my space is real r and I have Borel sigma algebra on it. Let us say Lebesgue measure on r just for the sake of concreteness. So, now what I am doing is I am I am at some level I am quantizing g that is really what I am doing. So, I am looking at all those omegas. So, let me draw may be little bit let me something like that. So, I am looking at some function. So, and I am considering let us say I fix some i let say I fix an i and whenever my g of omega is between i over 2 to the n and i plus 1 over 2 to the n. For those values of omega I define it as the i over 2 to the n the lower end of the interval. So, I am looking at some value i over 2 to the n let say that is my let say this is i over 2 power n. So, you fix an n I am defining the nth function now g n is what I am defining. So, you fix an n and you let say you fix an i and you look at i over 2 to the n and you also look at i plus 1 over 2 to the n. So, i plus 1 over 2 to the n will be somewhere here let say like that. So, that is i plus 1 over 2 to the n and you look at all those omegas for which your function is in this range in this is the function takes values in this range. So, in particular for us it will be if you look at that as well. So, it will be so that is it here there will be that little set and that little set that guy there and perhaps one more rate here that. So, you look at all those omegas where the function takes values between i over 2 to the n and i plus 1 over 2 to the n and in all these. So, you these are all the omegas where the function is in that range and in all these ranges you define the function to be equal to over there i over 2 power omega. So, here again the function will be here I do not have a different color of chalk, but you see what I mean. So, the function will be over here and similarly here similarly if you look at i plus 1 over 2 to the n i plus 2 over 2 to the n you will again approximated from below. You see what I am doing here. So, in essay I am really just quantizing the function, but I making sure that this g n of omegas less than or equal to g of omega correct and you can very trivially see the g n of omega is a simple function. Why is that true? So, I can in fact write. So, let me write it this way. So, can I not write g n of omega as sum over i equals 0 to n times 2 to the n minus 1. See you get n times 2 to the n of these subdivisions and if the function is greater than or equal to n you define the function to be equal to n. That is one way to do it. So, you have so many summations the functional value is equal to i over 2 power n indicator of omega for which i over 2 power n less than or equal to g of omega less than i plus 1 over 2 power n plus I will have just one more term that says n times the indicator of omega for which g of omega is greater than or equal to n. So, is that a correct representation of my function agree with that. Now, this is already in a simple form. So, only thing I have to verify is that these sets are measurable are these sets measurable indeed they are measurable because g is a measurable function and after all this set is simply the pre image of the interval this is just the pre image of the interval i over 2 to the n i plus 1 over 2 to the n pre image of that under the mapping g and this is an interval. So, pre images of Borel sets are f measurable. So, that is a valid f measurable set and similarly that is a valid f measurable set. So, you have already written g n in terms of a simple function explicitly a simple function. So, now I have the following three things are of quite easy to show. So, the first is that for each n g n is simple correct that we just argued number 2 we have that g n is less than or equal to g n plus 1 for all omega correct. Why is that true because if I take a finer quantization. So, if I increase n I am taking a finer quantization. So, in that case I will take. So, this will split into two further sub intervals and I will have to. So, if I split this further. So, I will lower approximate in that in that little interval and. So, the function I get the approximation I get will be larger correct for a finer quantization. So, that is true all right. So, this you can show and g n. So, g n increases to g right. So, when becomes bigger and bigger. So, my approximation will get better and better right because my g n of omega will be i over 2 to the n, but g of omega will be very close right has to lie between i over 2 to the n and i plus 1 over 2 to the n as n becomes very large g n of omega will converge to g of omega for every omega correct. So, I have a monotonic. So, I have a monotonic increasing simple function sequence of simple functions monotonic increasing to g right and this sequence can in particular be used in the previous proof of linearity correct with me. . Oh, this one let me draw a slightly bigger picture for. So, what I am trying to say is that no. So, what I am trying to say is that. So, let me draw a slightly bigger picture. So, you have a function like that. So, you approximated. So, earlier. So, you approximated like that right g was approximated like that. Now, you have one more level of. So, you are increasing n by 1 let us say and. So, you will have. So, that will split into right that is first split into 2. So, I may be different now, but that is ok, but the functional approximation will now be. So, here. So, just to zoom in right. So, I have let me zoom into that a little bit. So, I had that right initially I had that. So, I said. So, in that interval I approximated the function as that guy. Now, I am going to increase n right which means the width of each of these intervals used to be 1 over 2 to the n. Now, it is going to be 1 over 2 to the n plus 1 right. It means it is half is biggest the earlier width earlier. So, it is going to be like that right. And so, if you look at those omegas for which g is here I will approximate the function with that right. And here I will approximated as that right correct. So, now therefore, g n plus 1 of omega is bigger than or equal to g n of omega. Now, so the value of I may be different in the 2 cases that is ok, but you see the functional value is bigger bigger than or equal to right. So, g n plus 1 will be like that g n will be like that correct. So, you can just if you just draw big slightly bigger diagram like that you will figure out why this is true. And this is convergence which is also reasonably easy to show. So, I have now by. So, since I have all this I can definitely right. So, by m c t I have integral g d mu is equal to integral well limit of this correct. Why because I have monotonic convergence correct, but integral g n d mu I already know why it already have it in the simple form. So, I can explicitly write down an expression for integral g n d mu and by m c t. So, this limit is equal to the integral of g d mu. So, I am going to give you an explicit formula for integral g d mu in terms of a limit right. So, which is nice right because if I give you a function to integrate you can go ahead and evaluate the limit right which is an explicit rather than give you a supremum of a set of simple functions and. So, on which you will never be able to compute in practice right. So, this is a more practical way to compute the integral. So, integral therefore, integral g d mu is equal to limit n tending to infinity sum i equals 0 to n times 2 to the n minus 1 i over 2 to the n mu of. So, mu of omega for which i over 2 to the n less than or equal to g less than i plus 1 over 2 to the n mu of that plus there will be 1 more term is not it plus n times the measure of omegas for which g of omega is bigger than or equal to n correct. So, if you evaluate this limit you already have the integral of g d mu. So, this is an explicit formula. So, if your mu have to be Lebesgue measure you will simply take the. So, remember your this set right this set you will just simply take the Lebesgue measure which is the length of those sets if it is some other measure you have to take that measure. So, in effect what you are doing. So, let us for again continue assuming that mu is Lebesgue measure on R. So, in this case what you are essentially doing is you are chopping as up the vertical axis right for Riemann integral you used to chop the horizontal axis right. So, that is really the only essential difference. So, if Riemann integral you used to chop up the horizontal axis and look for the supremum infimum within those intervals. And you define the upper Riemann integral lower Riemann integral and compute the 2 see if the 2 are equal right. So, in this abstract integral this Lebesgue integral when mu is Lebesgue measure what you are doing is you are just chopping up the vertical axis the y axis instead of chopping up the x axis right. And you are looking at the measure of those sets for which the g lies in a certain range that is all you are doing. So, somehow just chopping up. So, chopping up the other axis instead of the. So, chopping up the vertical axis instead of chopping up the horizontal axis gives you a much more general notion of an integral. So, I will stop here.