 Okay, so this afternoon we'll finally start proving some lemmas. It's taken a little while to get that. But let me just remind you of the theorem we're aiming to prove. So theorem we suppose M is a compact metric space. F, M to M is continuous. And the theorem is that the set of F invariant measures is non-empty, compact, and that EF mu is an EF, if and only if mu is an extremal point. So how are we going to prove? We start by proving that it's non-empty. So we start by proving that there exists an invariant measure. We start with this definition. We define a map of star from the space of probability measures to itself. This is induced by the map F from the space to itself. And this is defined in the following way. So mu maps to F star of mu, where F star of mu is defined in this way, that F star mu of some set A is given by mu of F minus 1 of A. So what does this mean? This is our space M. We have a probability measure mu. And now we want to use the dynamics to define a new probability measure, which we call F star of mu. How do we define it? Well, it's defined by the value it takes on various sets. So we want to know what the measure of this set A is. The mu measure of A is whatever it is, the mu measure of A. But now we're going to assign a new measure of A. How are we going to do it? Well, we will look at the pre-image of A. This is F minus 1 of A. We look at the mu measure of F minus 1 of A. And we define the measure of A by this new measure to be exactly the mu measure of this pre-image. That's a very good question. So what is F? So we need to check. I'll leave it as an exercise. So exercise F star mu is a probability measure, in particular 1. But this is easy. What is F star mu of M is equal to mu of F minus 1 of M, which is just equal to mu of M, which is equal to 1. Because the pre-image of the whole space is the whole space. The pre-image is all the set of points that map to some point in the whole space. Yes. That's right. No, the image is not necessarily the whole space, but the pre-image is. So this turns out to be a measure. This is, in some sense, the amount of measure that maps into A. Not in some sense, it's exactly that. So what this is saying is that with respect to the mu measure, this is the amount of mu measure that maps to A that comes from somewhere and maps into A. That comment in itself does not help you to understand why we're doing this, but we will see this in a second. So notice, first of all, that if the measure is invariant, what does the measure mean? So notice that mu is F invariant if and only if F star of mu equals mu. It's a fixed point, because from the definition, if F star of mu of A is equal to mu of F minus 1 of A, then it's a fixed point. So we will not really approach this, so we're looking for an invariant measure. What we want to show is that there exists a fixed point. But we're not going to be able to do it by just applying some fixed point theorem. We need a slightly different argument here. So similarly, we can also define, similarly, for all i greater than or equal to 1, we can define the iterates basically of this map. F i star of mu is by definition mu of F minus i. So this sometimes is called, so F star is called the pullback of mu. Or sometimes we say push forward. Pullback, push forward means more or less the same thing. So we want to find a fixed point for this map, so we're going to start by proving a few properties of this map of star. F minus i of A. What is F minus i of A? If you know what F minus 1 of A is, you should know what F minus i of A is. Shall I write it? F minus i of A equals, by definition, the set of points such that F i of x belongs to A. So first lemma. For all phi in L1 of mu, we have that the integral of phi with respect to this measure mu is equal to the integral of phi composed with F d mu. So this is a simple lemma that follows basically from the very definition of F star mu, but it helps you also a little bit to understand what this measure really means. So this is quite easy to understand what it means. So phi is your function, phi is an integrable function with respect to mu. So you've got your space, you've got your measure mu, you've got your function, and what this integral... So if you take the integral of phi d mu, maybe it's easier, it's better to represent it as a graph. So we have m and we have our function phi to R. So m does not need to be a one-dimensional space, but we can represent it like this as an abstract set. So remember, we talked about the fact that the integral really is just your intuitive notion of the area under the graph where you count this area as positive and this area as negative. So the integral can be zero, even though the function is not zero. But when you integrate with respect to a measure, then this weighting of the values of the graphs depends on the measure. If the measure lives only in this region, then this integral will only see the values of phi in this region. This is the key idea behind integration with respect to the measure. So what does it mean to integrate this function here with respect to this measure? It means that when you evaluate the function at a point x, you're not integrating the value of phi at the point x, but first you apply the dynamics. So you apply f, so you apply f of x, and then you evaluate the value of the graph there. So this integral here is somehow putting the dynamics in and saying we're taking the integral of the observable phi after one iteration of the dynamics. So it depends both on the way this measure mu is distributed and on what the dynamics does. For example, if the dynamics maps everything to one point, then this integral is going to evaluate just phi at that one point, even though you're integrating over the whole space. So this here is integrating the function after one application of the dynamics. And here this is integrating the original function, but with respect to a different measure. But this different measure from the definition is really kind of going backwards with the dynamics. So here you're going forwards with the dynamics. You're applying the dynamics and evaluating the map f and integrating with respect to the original measure. Here you're integrating the original function, but with respect to a measure that has been obtained by looking in some sense at the pre-images. So you can see intuitively that you should get an equivalence here. We're going to prove it in a second, just to get some feeling for what this is saying. So how do we prove this? Well, first of all, we do it for a characteristic function. So first, let phi equals the characteristic function of some set A. Then we just check this. Then this follows easily, because then we have the integral of the characteristic function of A with respect to this measure, f star mu. So what's the integral of a characteristic function with respect to a measure? It's just the measure of the set. So this is just equal to f star mu of A. Yes. Oh yes, everything, everything. All the measures, when I have a compact metric space, I always assume that we have everything is Borel sigma algebra Borel for some A Borel. Yes, every time, always whenever we talk about any sets, it's a Borel set. Okay, so by the definition of this measure, this is equal to mu of f minus 1 of A. And this we can write again as an integral of a characteristic function. So this is the integral of the characteristic function f minus 1 of A d mu. And now, key step, how can you write the characteristic function of f minus 1 of A? If you think about it, this is just exactly the integral of the characteristic function of A composed with f, because these are the same functions, right? Because if a point belongs to f minus 1 of A, then when you apply f, it will belong to A. So these two functions are either 0 or 1 exactly for the same set of points. So they're either A or 0 or 1, so it's the same function. So we have verified this property. So as long as phi is a characteristic function, we get that its integral with respect to this measure is exactly the same as the integral composed with f with respect to the original measure mu. So there's standard techniques actually that perhaps many of you have seen, is that in many cases in analysis, in real analysis, you can approximate, once you prove a certain kind of equality like this for characteristic function, it's kind of a standard argument to prove it for general measurable functions. I will let me write it out, let me sketch it out anyway, because I think it's useful just to remind of the argument. So first of all, from this, it follows also, so from this, it's also true for simple function. Okay, from this, it follows also for simple functions. Simple functions, if you remember, is a finite sum of characteristic functions. We talked about that in the first lecture. So if phi now is a non-negative integrable function, then by standard results, it is the point-wise, so then phi is a point-wise limit sequence phi n of an increasing sequence, increasing of simple function. So this is generally true that every measurable function is a point-wise limit of a sequence of simple functions. And then, so then the sequence composed with f is also increasing sequence of simple functions, converging point-wise to phi composed with f. Okay, so now, by definition of Lebesgue integral, that the integral, because this is an increasing sequence of functions converging to phi, then the integral of this converges to the integral of this and the integral of this converges to the integral of phi composed with f. So we have that the integral of phi n df star mu converges to the integral of phi df star mu and the integral of phi n df star mu, sorry, phi n composed with f. So the integral of phi n composed with f df star mu converges to the integral of phi composed with f df star mu. It's true in any case, but I just, this is what I want. Okay, so these are just two measures, mu and f star mu and because the phi n converges monotonically to phi, then the integrals converge, sorry? Exactly. So what do we want? So now for simple functions, we already have that this is equal to this because this is a simple function, this is a simple function so we know that this is equal to this and therefore the limits are the same and so we get that this is equal to this, which is what we wanted to do, okay? And then the same as for integration, this is for non-negative integrable functions. If otherwise you just take the positive part, the negative part and you do the same approximation argument and it's all exactly the same. Okay, so that's lemma one. Lemma two, f star m to m is continuous. So what do we mean by continuous here? Continuous with respect to the weak start topology. So what do we need to show? So proof, suppose mu n converges to mu. We need to show that f star of mu n converges to f star of mu. This is continuity. So by lemma one to phi continuous, we have integral star mu n is equal to the integral of phi composed with f d mu n and now this is a crucial point in which of course we use the continuity of the map of the dynamics, right? Because since this is continuous and this is continuous, the composition is continuous. So what do we know about the convergence of this integral here? What does this converge to? It converges to the integral with respect to f star mu, with respect to mu, right? Because the definition of this convergence is exactly that for any continuous function, this convergence with respect to this converges to the integrals with respect to this. So this converges to the integral of phi with respect to mu, phi composed with f with respect to mu and this is equal to, again by lemma one, this is just equal to integral of phi with respect to d f star mu. Yes, yes, yes. And phi continues in this case. In lemma one, you're right, lemma one we prove for any l1 phi and therefore in particular for phi continues and here we're just applying lemma one on the specific case with the specific case where phi continues. Yes, good observation there. Okay, and therefore we have that for any continuous function. So we have that for any, so we have that for all phi continues, integral of phi d f star mu converges to integral of phi, sorry, d f star mu n converges to integral of phi d mu, okay? And so, again, using the definition of weak star convergence, we have that f star of mu converges to f star of mu, which is what we needed to prove. So a simple corollary of this, one more corollary and then we can prove the first part of the main theorem. So a simple corollary is that mu is invariant if and only if integral of phi composed with f d mu is equal to integral of phi d mu for all phi continues. Questions? Sorry, what's the question? In the right-hand side. Ah, yes, thank you. Okay, so basically straightforward corollary of this is this implication, so proof. So one direction is just a consequence of lemma one, right? So if mu is invariant, if mu is invariant, result follows by lemma one, lemma one, right? Because you just, if mu is invariant, you get f star mu is equal to mu, so you immediately get this result. If otherwise, assuming conversely, we have that phi integral of phi d mu is equal to the integral of phi composed with f d mu, so we assume that this is true. And then again we use lemma one and this gives that this is exactly equal to the lemma to phi d f star mu, again by lemma one, and this is for all continuous functions. So what this means is that if you integrate with respect to mu or with respect to f star of mu, the integral is the same for all continuous functions. And this means that the two measures are the same. If you want to see it a bit more formally, you can see as the fact that the integral is linear functional on the space of continuous functions, and if two linear functionals coincide everywhere then they're the same. Which is exactly what the situation is here. So mu is equal to f star of mu, and so mu is invariant. So we're now ready to prove the existence of invariant measures. It's a nice proof. It's not so complicated. So proposition. So this is sometimes called the Krilov Bogulyobov theorem. Mf is non-empty. So the key part is the non-emptiness. The other two are fairly straightforward. How do we show that it's non-empty? Well, first of all, recall that if Bogulyobov, Krilov Bogulyobov, two Russians, Russian mathematicians, so if M is compact, then M is also compact in the weak start topology. Again, that is because M is in the dual of the continuous functions, that is the spatial measure, and it's the spatial probability measures. So what it is is the unit ball in the dual is compact in the weak start topology. This is the Bannach-Allaoglu theorem that some of you were talking about the other day. So how are we going to use this? Well, this is useful because that way we take any sequence, it's going to have some converging subsequence in M, and what we're going to do is we're going to do exactly that. We're going to take a sequence and take some converging subsequence and show that this converging subsequence is invariant. This is all probability measures, not necessarily invariant. So even though it's compact, you take an arbitrary subsequence or subsequence, it converges so what? But we're going to take a special sequence that's going to be defined using the dynamics and that's going to help us to show that the limit is invariant. So let Mu0 in M, by the way, how do we know that M is non-empty? It might be compact in a trivial sense in this case. But how do you know that the space of probability measures on the space? Exactly, because you can always take, if there's at least one point in your space, you can always take the Dirac delta, so there's a space. So for example, you could take the Dirac delta in one point here, you can take an arbitrary measure and then we define, for any n greater than or equal to 1, we define the first following family of measures, Mu n is 1 over n, the sum of fi star Mu0. So what is this average? Each of these, as we discussed before, is a probability measure, right? Each of these is the ith pullback of the measure and it's defined in the way we define f star of i. So each of these is a probability measure, so even though you don't know exactly what the properties of this is, you know that the sum of these measures divided by n is also a probability measure, right? Because you take any set and you measure it with all of these measures, you sum the measures you divide by n, so the total measure of the space according to this is 1, right? According to this. So this is a probability measure. Yes. Yes. Yes. If you take the Dirac measure on a fixed point, it's f invariant, yes. Absolutely, no. I'm just saying that the space of measures is non-empty, not the space of invariant measures. The space of invariant measures is what we're trying to prove that is non-empty. That's right. If there's a fixed point, then this is trivial. But in principle, there's no fixed point. So in principle, we don't know that there's a fixed point. That's right. So this is completely general. We just take an arbitrary measure which we know exists and we take these averages. So notice for future reference that this itself would also be a sequence of measures. Okay? But as it turns out, we're going to take now a subsequence that converges. Either way, either taking averages or not taking average, you get some sequence of probability measure. So some sequence converges. But we are going to want to show that this limit, any limit point is invariant. And for this, it turns out that it's crucial to take the average here. Okay? And now we will show that. So let mu and j converge to mu. Maybe some limit point. And we claim that this mu is an invariant measure. So how are we going to show that this is an invariant measure? We need to show. So we claim that mu is f invariant. And the way we're going to show that it's invariant is to show that f star of mu equals mu. This is that it's a fixed point of f star. Okay? So we already know. So by continuity of f star, we already have that f star mu and j converges to f star of mu. Right? Because by assumption, we assume that we have a converging subsequence. If we apply f star to both sides, then the limit converges also because f star is continuous. Okay? And so what we have is that mu and j converges to mu. f star mu and j converges to f star of mu. So we need to show that these converge to the same thing, that these are the same sequence, basically, right? And this will show that they have the same limits. Or we can just show, so it's sufficient, to show that f star mu and j converges to mu. Okay? Because f star mu and j converges to f star of mu. If it also converges to mu, then we will get the result. So it's really just direct computation here. So we have f star of mu and j equals, by definition, f star. What's the definition of mu and j? Is that 1 over nj? Sum i equals 0 nj minus 1 of fi star of mu 0. And now this f star, this is just an average, right? So when I take f star of this finite sum, it's just like taking the average of the pullback of each of these. This is just equal to 1 over nj. The sum i equals 0 nj minus 1 of f star i plus 1. Because I take one more iteration of f star on each measure of mu 0. And now this, I'm going to do the usual trick where you kind of separate the terms here. And this gives 1 over nj of the sum i equals 0. Sorry, this is a bit high. 1 over nj, the sum i equals 0 to nj minus 1 star mu 0 minus mu 0 plus f nj mu 0. So I just add and subtract these terms. And this I can write as equal to 1 over nj sum i equals 0 to nj minus 1 of f mu 0 minus mu 0 over nj plus f, which is equal to mu nj minus mu 0 over nj plus f star nj mu 0. So I've written, in conclusion, I'm calculating f star of mu nj and just by doing a little bit of algebraic manipulation on the definition, I write f star mu nj equal to mu nj plus some remainder terms. Some sense. So what I want to show is this, that f star mu nj converges to mu. And what I have is that f star mu nj is equal to mu nj plus something. So as nj goes to infinity, we will get that this limit converges to what? It converges to mu. Why does it converge to mu? What about these remainder terms? Abounded by what? Well, that would not be good enough. If these were bounded by 1, this is also bounded by 1. Exactly. They're bounded by 1 over nj. They're bounded by 1 over nj, not by 1, right? So if you look at the convergence in the weak-start topology, you've got to integrate continuous functions by these measures. This is a probability measure and you just get the integral with respect to this. But here you get integrals with respect to measures that have very little mass going to zero. So the contribution to the integral is getting smaller and smaller. And so this is converging to mu in the weak-start topology. And therefore this shows, so f star mu nj converges to mu. And so mu is equal to f star. So this simple calculation here shows that the every limit point of the sequence is f invariant, right? Because we've shown the invariance. And again, notice that we needed to use the fact that we had averages here because we played this little game, right? If we just took the limit of these sequences here, they might not, if we define the measures mu nj by just this f n star mu zero, we might not have been able to do this. Okay? So this proves that invariant, we just need to show convexity and compactness. So to show convexity is easy. Just now we have mu one, mu two, an invariant measures. Okay? Then T mu one plus one minus T mu two is also invariant. Okay? I will leave it as a simple exercise. You just use the definition of invariance and you get it immediately. Okay? For compactness it's also fairly easy. So suppose mu n, that mu n in mf, mu probability measure and mu n converging to mu because m is compact. So we know that we can assume that this converges to mu. We need to show that this implies that mu is in mf. So then for any continuous function, so again by lemma one, we have that for all phi continuous. We have that the integral of integral of phi composed with f d mu is equal to the limit of the integral of phi composed with f d mu n. So this is just the fact that mu n is converging to mu. Okay? And now this is the full equal to the limit as n tends to infinity of the integral of phi d f star. But, okay? But the limit of d f star is the f star mu n also converges to mu. Okay? So this is actually equal to... Sorry? Exactly. Exactly. So this is equal to... That's right. So mu n is invariant. That's right. Yes. So you have... This is equal to the limit n tends to infinity of the integral of phi d mu n. And so this is equal to the integral of phi d mu. Okay? And so we've shown... So the integral of phi composed with f d mu is equal to the integral of phi d mu for all phi continuous. And we've shown that this is if and only if mu is invariant. So mu. Okay? So I think this is... We're having a bit of a shorter lesson today so I think this is a good place to stop. The next time we will prove the fact that the ergodic measures are extremal elements of the ergodic invariant measures are extremal elements of the invariant measures. Okay?