 Okay, good afternoon, everyone. So I want to start by rewriting the precise statement of Birkoff's ergodic theorem, because in the last lecture there was just a little bit of confusion about the statement. So I tried to write a kind of cleaner statement, but let me write the precise statement here, because I think it's better to have it. So the statement works in abstract measure spaces, so that's why I think it's better to write it. So we have M is a measure space, F, M to M just needs to be a measurable map and a mere invariant probability measure, so then every integrable function, mu integrable function phi, the limit exists. So this is the first part of the statement, so I'm just rewriting again Birkoff's ergodic theorem, because I did not write it in its full generality and I think it would have been a good idea to have it. So in a completely abstract measurable setting, we have not assumed ergodicity yet, just the fact that F is invariant implies that this limit exists, which is in itself highly non-trivial, right? This just means that if you look at the average value of phi along an orbit, it does converge to an average, okay? Moreover, if mu is also ergodic, then this limit is exactly, sorry, I should say for then the limit, then for every phi, the limit exists for mu almost every x, of course. If mu is also ergodic, then the limit, and this should be 1 over n, sorry, limit 1 over n, sum i equals 0 to n minus 1, phi integral F i of x is equal precisely to the integral of phi d mu for mu. So just to spend a couple of minutes on this, what is the difference between these two and why do you need the additional ergodicity assumption to get the second statement? I'm not asking for a formal reason, but just intuitively, can you see the differences? So this is the limit exists, this is saying that the limit is equal to very specific value. Why can this not be true if mu is not ergodic? Why can you not hope to have this be true if mu is not ergodic, you have some idea? That's just for any invariant measure that's true, that's the definition of invariance. So for any invariant measure, F star mu equals mu, and this is true. Why can we not have this, why do we need the ergodicity for this, any ideas? So remember what does non-ergodicity mean? So we have our measure here, non-ergodicity means that mu can be, has at least two ergodic components, has several ergodic components, right? Mu is some kind of linear combination of various measures, okay? So there is a measure mu one and a measure mu two, and some combination of these is equal to mu. Now, what does the point do? For ergodic measures, what this theorem is really saying is that the distribution of the orbit of x inside this ergodic component is uniform with respect to this measure, right? Because the average value of phi, what is this? This is the average value of the function phi. Function phi can be, for example, the characteristic function of some set or it can be a more general function. But what this is doing is taking the orbit of x and is looking at the orbit of x and taking the average value of phi along the orbit. And what is this? You can think of the integral as the average value of phi also in a different way, right? This is the average value of phi on your whole space with respect to this measure. So these are two different ways of taking the average value of phi and this theorem says they coincide. And why do they coincide? Because this orbit is very uniformly distributed in your space. And therefore, if you take the average value of phi along the orbit, it will coincide with the average value of phi in your space because the orbit goes everywhere, it spends equal amounts of time everywhere with respect to this measure. Okay, this is what it's saying. So this cannot, of course, work for if the measure is not ergodic because if you choose a point that belongs to this ergodic component, it will never land in here, right? There is no reason that this should converge to the average value of phi over the whole space. Only with respect to the ergodic component to which it belongs. Just to give you some feeling for this. So how was it that I wrote it in the last lecture? I wrote it a little bit differently. So if you remember, we have the basin of the measure mu is equal to the set of x such that 1 over m, the sum i equals 0 to n minus 1, converges to the integral of phi d mu for all phi continuous. And what I said, which is really corollary of this corollary, is that if m is a metric space because we cannot even define continuous function if it's not a metric space, right? So if m is a metric space, then mu will be mu equals 1. And what is the difference between these two statements? It's a very subtle difference. I will not give a rigorous proof, but just to mention this, I think it's useful. So in this statement, this set of points for which this holds may depend on the function phi. So the statement is that for every phi, this limit exists mu almost everywhere. It means that the set of points where it does not converge has zero measure. But this actual set of zero measure might depend on the function phi. So for each function phi, there's a different set of zero measure for which this holds. And the same thing here. What we would like really is to find a set of full measure for which this convergence holds for all those points for every function. That's what we would like. And that is what we can get if we look at the slightly stronger situation in which we're looking at the convergence with respect to continuous functions, which we can do within a metric space. And in that case, what we're saying here is that there exists a set of full measure for which this convergence holds for every continuous functions. We cannot say that there exists a set of full measure such that this convergence, such that for that set, this convergence holds for all L1 functions for every point in the set. It's a very delicate difference, but it's a difference that I thought I should mention. Okay. Yes? So why would we need to assume it's a metric space? Well, in two ways. First of all, just to define continuous functions. And second of all is because the argument that you use to show that you can take this intersection. So basically, you're taking some kind of intersection of sets of full measure. So you're taking the set of points for which this convergence, for each function, you get a set of full measure. And then you take some kind of intersection of all these sets of full measures for all the functions. And so you use the fact that the continuous function have some dense countable subset of continuous functions. And you can take the intersection of this, of the full measure set for each of this countable subset. Yes? Do we need what? That is a good question, actually. I'm not 100% sure at this moment. I have to remind myself of the actual proof. Okay. Compact? Maybe. I don't think so. Yes, yes. I don't remember exactly. I will look it up and I will let you know. That's a good question. I wasn't planning on giving you an actual proof of this following from this. So I just wanted to emphasize this very subtle point, which is that in this case, the set depends on the function. And in this case, we are able to find a set for which this convergence holds for all functions. And with continuous functions, it's easier to do that. Okay. So what we want to do now is we want to start applying this theorem. As I said last time, this is a really powerful theorem because what it says is that if you know the measure, you know this limit. Okay. So really it's saying if you know that a system has an ergodic invariant measure, you pretty much know at least the statistics of the dynamics for almost every point. You know that almost every point is distributed in space exactly according to that measure. So it's a very powerful theorem. It's a theorem from the 20s, and it's still one of the fundamental theorems of dynamics. The difficulty, therefore, is to show that a measure is ergodic. And this is highly non-trivial. And as I said, in certain cases, you don't even have an invariant measure to begin with, so you need to first find an invariant measure and then show it's ergodic. Okay. So what we're going to start with in the next couple of lectures is some systems in which we already have an invariant measure and the problem is to show it's ergodic. And I will introduce a couple of interesting techniques to prove ergodicity of certain measures. So section ergodicity and circle rotations. So what is one of the simplest settings in which we know that we have an invariant measure? So if you have... So we write the circle, unit circle of length one, right, that we've used before. And we're going to prove the following theorem. This one. B and irrational circle. So there is one measure that we know already that is invariant here. What is this measure that is invariant here? Mod 1, yeah, in the circle. What is the measure that is invariant here? What? The what? I can't hear you. Yes, which measure is invariant? What measure is invariant for this map? Do you know one measure that is invariant? Which measure is invariant? What's delta zero? No, f of zero is not zero, f of zero is alpha. Unless alpha is also equal to zero. What is an invariant measure? If alpha is equal to zero, we have the identity, but alpha is irrational. We assume that it's an irrational rotation. So if alpha, as you remember, if alpha, okay, we need to review a little bit circular rotations, right? If alpha is rational, then every point is periodic, remember? And then we have the Dirac delta on the periodic points, okay? But the average of the Dirac delta on the periodic points. But if it's irrational, then we have no periodic points. But there is a measure that is invariant for all alpha, both rational and irrational. Another measure. A natural measure. An obvious measure, Lebesgue measure. Thank you. Okay? Lebesgue measure is invariant, because if you take an interval here and you rotate it, you get an interval here that is exactly the same length. Okay? It's invariant. f minus one of an interval is the same measure. So Lebesgue measure is invariant. Both in the rational case and in the irrational case, right? Just translation. So in the rational case, is it ergodic? In the rational case, it is ergodic? Why is that? Right? Right? So if alpha is rational, is Lebesgue measure ergodic? What if alpha is zero? Is Lebesgue measure ergodic? For example, let alpha equal to zero. f of x equals x, identity map. Lebesgue measure is invariant. Okay? Is it ergodic? It's not ergodic. You're sure it's not ergodic? You can take any interval here and it is invariant, fully invariant, and the complement is also invariant, and they have measures strictly between zero and one. Okay? What if x alpha is equal to one-half, for example, or one-third? Let's say one-third. Lebesgue measure is still invariant. Is it ergodic? Why is it not ergodic? Yes? Yes. That's because that's why it's invariant. Is it ergodic? Of course we can take an interval, but that's not the definition of ergodistic. It needs to be an interval that is invariant. Come on, guys. It's not ergodic. Why? What's ergodicity? Ergodicity is the definition that if f minus one of a equals a implies that mu of a equals zero or one. So to show that it's not ergodic, all you need to find is a set that satisfies this. Okay? Not just any set with measure between zero and one, but one that satisfies this. Where's the set? Yes? So you take this interval here, a1. You take the image here, a2. You take the image here, a3. Exactly. So if you take a equals a1 union a2 union a3, then it satisfies this condition, right? Because a1 maps to a2, maps to a3, maps to a1. The pre-image is exactly the same. a1 maps to a3, maps to a2, maps to a1. Because it's a rotation by one-third. Okay? So this is a set that satisfies this and does not satisfy that. You found a set that is fully invariant. It's not a connected set. It doesn't have to be connected set, of course. It's just a set of positive measure that's not fully invariant. You can see that you can do that for any rational number. You can easily do that for what we said is for every rational number, every point is periodic, and they all have exactly the same period, right? If the rational number is p over q, every point is periodic or period, q. Okay? So if every point is periodic period q, as long as you take a small enough interval and you iterate it q times, it will come back exactly on itself. If you take it sufficiently small, there will all be disjoint, all these images up to time, q and you will have exactly the same situation as you have here. Okay? Proof. This is a good exam question. So make sure you understand the argument. So if the map is rational, the big measure is not periodic. Okay? We've just proved that. I thought it would be easier to see, make a formal statement about this. So the question now is if it is an irrational rotation. So for an irrational rotation, we cannot use the same argument to prove that it is not ergodic. Because an irrational rotation, however small an interval you take, it will never come back exactly on itself because the points are not periodic. And indeed, we will show that then the big measure is ergodic. In fact, we will prove something more. We will prove, in fact, the big measure is the invariant ergodic. So what does it mean in terms of the distribution of points? The fact that it's ergodic. So we'll come back to this uniqueness point in a little while because it's very important. So by Birkhoff's theorem, corollary for Lebesgue, almost every x and every interval a, b in S1, we have that 1 over n times the sum. So we have that the frequency converges to the Lebesgue measure, right? So if we take an interval, A, B, take a point x, you iterate it and you wait and see what the frequency of visits to this interval is, then this frequency converges exactly to the Lebesgue measure, maybe. This is a simple consequence of Lebesgue measure, of Birkhoff's ergodic theorem. But in fact, we will prove in this particular case because it is the unique invariant measure, we will prove a stronger theorem. So this condition we say that x, orbit of x, orbit of x is uniformly distributed with respect to Lebesgue, okay? Just another way of talking about the convergence in Birkhoff's theorem with respect to characteristic functions. So we will prove the following theorem that says that if fS1 to S1 is an irrational circle rotation, then every orbit is... So this goes beyond Birkhoff's ergodic theorem, right? This says that this property here holds not just for almost every point, which is what you get from Birkhoff's ergodic theorem, but for every point. So the uniqueness is interesting here and we will study now some general properties to do with maps that have a unique invariant ergodic measure. So we're going to cheat a little bit. So I said that we're going to show that Lebesgue measure is ergodic, okay? So in this particular case, the way we're going to show that Lebesgue measure is ergodic is by showing that Lebesgue measure is the only invariant measure. If Lebesgue measure is the only invariant measure, why does that automatically imply that it's ergodic? Yes. Yes, also, yes. Exactly, it's only one point. So we proved before that the set of invariant measure is compact convex and non-empty. And if we prove that it's only one point and that the ergodic measures are the extremal points, they're always non-empty. Also, the set of ergodic measures is always non-empty in the case in which the space is compact and the map is continuous because they're the extremal points of the set of invariant measures. So if we show that Lebesgue measure is the only invariant measure, automatically it has to be ergodic, okay? So it is also possible to show that Lebesgue measure is ergodic by actually looking at the dynamics directly to show ergodistic. But we're going to do that for different class of maps. In this class of maps, we're going to take advantage of this property and we're going to prove that it's the only invariant measure. And that will imply that it's ergodic. And in fact, it implies additional properties such as this. So having a unique invariant measure is a very strong property of a dynamical system. So let me write this as a general definition. Definition f x to x is uniquely ergodic if it admits a unique ergodic f probability measure. So I put ergodic in parentheses because it follows from the uniqueness of the invariant measures, right? It's not part of the definition, really. It's just that if it's got a unique measure, this measure must be ergodic. So can you give me another example of uniquely ergodic system besides these circle rotations, which we're going to prove now, a simple example that we've seen many times before? What other system can you think of that has only one invariant measure? You're right, actually. The adding machine has a unique invariant measure. It's related to the circle rotation, but it's not completely obvious to see what this measure is. What about a very simple, much simpler example than that? So what kind of example? Okay, you can have a constant map that really maps everything to a single point. That's true. And why does that make it uniquely ergodic? Only the delta at that point. And do you have to map everything to that point? Or can you think of a little bit more general situation that's similar to this one? Sorry? Well, all you need is every point to converge to a unique fixed point. What maps do you know that have this property? Exactly, contractions. If you have a contraction mapping on a complete metric space and so on, everything converges to a unique fixed point. How do you know that there cannot be any other invariant measures except for the unique fixed point? So if you have some set here, you have your unique fixed point and you know that everything converges to this fixed point. So the Dirac delta in piece clearly an ergodic invariant measure because we've already said, shown that if you have a fixed point, the Dirac delta on that point is a fixed point. How do you know in this case that there's no other invariant measures? Why? It would have another fixed point, why? Yes? Right, if you had another set of positive measure or another set that was fully invariant and had positive measure it would have to be a contraction on that set. You'd have to have another fixed point. You're basically what you said is correct. Let's try just a slightly simpler way of saying this. Suppose you had another measure that was not the Dirac delta. Then there would be some other set that would have positive measure. It would have to be. So suppose you have some set that does not contain P that has positive measure for some other invariant measure. What do we know about this set? A, the fact that it has positive measure with respect to some invariant measure. Yes? Exactly. After submitted, the points would come back by Poincare recurrence theorem. Also by Birkhoff's ergodic theorem Birkhoff's ergodic theorem will show if this measure is ergodic and every measure can be decomposed into ergodic measures so you can assume it's ergodic and then this will come back infinitely many times. But even without Birkhoff's ergodic theorem that's like just by Poincare recurrence theorem you know that all points will come back. And is this true? Everything is converging to P and A does not contain P so they do not come back. So there's no recurrence. So invariant measures are absolutely connected to recurrence. As you can see from the Poincare recurrence theorem and the Birkhoff ergodic theorem that gives a much more precise quantitative statement by what we mean by recurrence, the frequency of the recurrence. So you can use this fact to understand situations like this and know that if all these points never come back there can be no invariant measure that is living in that place. And therefore in this case this map is uniquely ergodic because it has only one in a kind of trivial way. I mean it's uniquely ergodic in a simple way. But what we're going to show is that circular rotations are uniquely ergodic and that is much less trivial because circular rotations you have recurrence everywhere so it's certainly not immediately obvious that you have only one invariant measure. So the way we're going to prove it is by a general theorem. So suppose f x to x is a continuous map of a compact matrix space and suppose there exists a dense set phi of continuous functions such that so continuous functions phi such that phi in phi, there exists a constant phi bar which depends on phi such that 1 over n sum i equals 0 to n minus 1 of phi composed with f i converges to phi bar uniformly, uniquely ergodic. So there's a, I haven't finished writing the theorem there's a converse as well but let's first just think a little bit about what this says and make sure we understand the statement. So you take a continuous map of a compact matrix space and you take the space of all continuous functions on x and you suppose there exists a dense subset of continuous functions which converge uniformly to the constant function. So notice this, I could write x here but I'm not saying that for each, well I'm saying that this converges as functions. This is a family of functions and they converge uniformly to this constant function here. It's a bit like saying that for every x you converge to a constant to the same constant uniformly in x. We have uniform convergence of continuous functions. Then f is uniquely ergodic. At the moment there's no reason you should see any connection between uniform convergence and uniquely ergodistic but this is what we're going to prove. Let me state the converse also. There's a converse. Conversely suppose f is uniquely ergodic then for all phi x to r continuous there exists some constant phi bar that depends on phi such that, so let me call this condition here star such that star holds if it is the unique invariant measure then phi bar is equal to the integral of phi. So what's the difference between this and Bilkoff's ergodic theorem? So here this of course is obvious for the invariant measure. For the invariant measure we clearly have, well we don't have the uniform convergence. This is what Bilkoff's ergodic theorem says that this for every x this converges to the integral. So it's clear that if we have this convergence it must converge to exactly this for almost every point. And here we're ashamed that in fact it converges uniformly as a function which means that it converges for every point uniformly. So it's a much stronger form of convergence and this follows from the fact that it's uniquely ergodic. It turns out it's like a strengthening of Bilkoff's ergodic theorem for the specific uniquely ergodic case. Because there's nothing else it can converge to that. If there were other measures then these averages would have to converge to other things but because there's nothing else they can converge to in some sense they have to all converge to the same limit. Roughly speaking. Okay so this is the statement and then we're going to use the statement to prove that irrational circle rotations are uniquely ergodic. We're going to find a dense set of continuous functions and prove uniform convergence. And then we're going to apply this to show some strong kind of uniform distribution property for circle rotations. So first of all let me just make a small remark is that this holds the mark theorem also holds for complex valued functions observables phi x to c. Okay this is actually simple even in Bilkoff's ergodic theorem. I mean all these theorems it's simpler and it's more natural to study for real valued observables but if you look at the proofs you see there's no difference. Okay you can do everything for complex valued observables and we will use this a little bit later on. So maybe before I start the proof of this theorem let's just take a couple of minutes break and stretch your minds. Okay so let's prove this theorem now. So there's two parts the first part that says that if this happens it's uniquely ergodic and the second part which is the converse. So the first part we're going to split into two lemmas. First of all we're going to show that if you have this uniform convergence for a dense set of functions then you have the same uniform convergence for every function. And then we show that if you have uniform convergence for every function then f is uniquely ergodic. So two conceptually easy steps. And then we will prove the second part of the theorem. Okay so let's state the first lemma under the same assumptions I won't write them but under the same assumptions as before we have a continuous map of a compact metric space. Okay if star holds for all phi in phi then it holds for all. Okay so let's prove this it's basically just a kind of approximation theorem. So let's write to simplify the notation let me write this bn of x phi is equal to 1 over n sum i equals 0 to n minus 1 phi equals phi of x. Okay this is the averaging that we always work with and we can think of as a kind of bulk of averaging operator in some sense as we shall see. So if phi is in phi okay what is the assumption? Star says that there exists some phi bar which depends on phi. Okay such that this is exactly the average so this is a function such that bn of x phi converges to phi bar uniformly in the variable x uniformly in x. So now let psi, what did I use? x or m here, x I think for this space. X is continuous, arbitrary continuous function so we want to show that the same thing is true for x. So let epsilon greater than 0 and choose phi in phi such that so what does dense, what does it mean to say that this set is dense in the continuous functions? What topology? That's right. Uniform topology that's what. So if we don't specify talking about continuous functions you always talk about the C0 topology or the uniform topology. So by the density what does this mean? This means that we can approximate psi uniformly. So there exists phi such that the supremum overall x in x of psi of x minus phi of x is less than epsilon. This is what it means that we're epsilon close using the density. And what does that mean in terms of the bulk of averages of psi? So this means that at each point of the orbit of x phi and psi are epsilon close. So these errors, the difference between these two is accumulating a rate of epsilon every iteration but you're dividing by n. So the average error is still epsilon. So we have that bn x phi minus bn x psi is less than epsilon for all x and all n because it's uniform. Uniform topology, uniform convergence and so since we know that this is converging to phi bar, that means that this is also uniformly close to phi bar. So we have that the supremum overall xn of bn x psi minus phi bar is less than epsilon. And the infimum of xn of bn x psi minus phi bar is also less than epsilon. So does this mean that this bn x psi are converging to phi bar? Is this converging to phi bar? Does this mean that these are also converging to the same phi bar? Why not? Exactly. Very good. So just a remark here. This is not showing that these are converging to phi bar because the choice of phi depends on epsilon. But what it means is that this and this are close because they're both close to the same thing for a given epsilon. And so we have that the supremum of overall xn of bn x psi minus the inf of xn bn x psi is less than 2 epsilon. Since epsilon is arbitrary and this shows that they can be arbitrarily close and so they're also converging. So star holds. So a simple kind of approximation argument to show that that's why we take a dense set and this uniform convergence shows that we can extend it to every continuous function. And now we show that if it holds for every continuous function, for lemma, if star holds for all phi x to r continuous then f is uniquely ergodic. So why is this term? Why does it imply that there's only unique invariant measure? Unique ergodic invariant measure here. Well, let phi x to r continuous. So by Birkhoff's ergodic theorem we have that 1 over n integral i equals 0 to n minus 1 phi composed with fi of x converges to the integral of phi d mu for almost every... So how do we complete the argument here? What is our assumption? What does star say? Remember star? Light. And so that's light, that's light. So phi bar is equal to integral d mu and 1 over n sum of i equals 0 to n minus 1 phi composed with fi of x converges to the integral of phi d mu for all x in x. For every x it converges to this. Uniformly. So what was the final observation? If it converges for every phi, sorry, for every point, for every point of time I have it just converged to this. Can there be another invariant measure? We're trying to show that there exists only one ergodic invariant measure. So we pick one ergodic invariant measure and we get this. Now suppose I pick another ergodic invariant measure. Then what? Then I would get exactly the same thing and I would get that for every point I converge to the integral with respect to a different invariant measure. But this convergence is unique, right? The convergence is unique. You cannot converge to do different things. So that would mean that for those two invariant measures the integral with respect to every phi coincides and therefore they have to be the same invariant measure. Okay? If mu tilde is another invariant ergodic measure, probability measure, then we have that 1 over n i equals 0 n minus 1 of phi composed with fi x converges to the integral of phi d mu tilde. For all x and so mu is equal to mu tilde. So this completes the proof of the fact that if you have uniform convergence for a dense set of functions then the map is uniquely ergodic. Notice that we've used Bielkoff's ergodic theorem in the proof here. Okay, so now we want to prove the converse. Suppose that f is uniquely ergodic. Then we have the uniform convergence for all continuous functions. So now lemma suppose is uniquely. So then for all phi continuous takes this phi bar such that star holds. So again by Bielkoff, we know that we have this convergence almost everywhere, right? So we know that 1 over n sum of phi composed with fi of x converges to the integral of phi d mu from mu x. So let phi bar equals integral of phi d mu. Okay, and we want to show that we have actual uniform convergence for all x. Okay, and we show. So how are we going to show that star holds? How are we going to use the unique ergodicity assumption? What do you think? What's the strategy of proof? Yeah, we've used Bielkoff's ergodic theorem to choose our constant, right? We want to show that star holds. So we want to show that for every phi there's a phi bar. So now by Bielkoff's ergodic theorem we know what this phi bar needs to be. If this is going to be true the only possibility for phi bar is this. Because we already know that this converges for almost every point. What we need to show is that this converges uniformly to this. Okay, so how are we going to prove that it's going to be uniformly using the unique ergodicity? So basically we're going to look at the convergence for other points and we're going to show that this would be a contradiction. If they converge to anything else then you must have another invariant measure in there. If we suppose we have a point here for which this uniform convergence does not occur we can show that this must converge to some other constant that is gives rise to another invariant measure. So let's try to do that. So what do we need to show? What's the negation of uniform continuity? So suppose by contradiction that star does not hold. So what's the negation of star? Okay, this is not completely. So then the negation is that there exists a continuous function. There exists a continuous function. There exists epsilon. There exists sequences xk in x and nk going to infinity such that 1 over n sum i equals 0, 1 over nk nk minus 1 of f i composed with f i of xk minus phi bar is greater than or equal to epsilon. So how are we going to get a contradiction from this? Well, if you remember how we put the existence of invariant measures we took sequences of this kind and we showed that any limit point is invariant any accumulation point of the sequence. So we're going to do something very similar. Now define a sequence of measures nu k is equal to the measure 1 over n sum i equals 0 to n minus 1 so this should be nk nk of f i star delta in xk which is just the same thing by the definition of this is equal to 1 over nk the sum i equals 0 so this should be nk always nk minus 1 of the delta Dirac in f i of xk So then we have for every k what do we have here? We have the integral of phi with respect to this measure so what are we trying to do here? So we're trying to contradict we will show that in fact this cannot happen so any kind of sequences any choice of points and sequences and so on because of the uniform convergence this cannot happen this is actually converging uniformly to phi bar this is what we're trying to show so by contradiction we suppose we find these sequences so this is for every k this is greater than equal to epsilon so we're going to construct these measures which are specifically built by looking at these times where these averages are far away from the average that you're supposed to have and therefore we're going to get a limit of these measures which cannot be the original measure because it's far away from this limit and therefore that's going to prove that there exists another measure so let's look at the average of phi with respect to this nu k by definition this is equal to the integral of phi we respect to this measure here 1 over nk sum i equals 0 over nk minus 1 of delta fi of x this is this measure and here I can write this as I can take the sum and the average out as I've done before 1 over nk the sum i equals 0 to nk minus 1 of the integral of phi with respect to the Dirac delta measures in fi of x xk thank you and this is just equal to 1 over nk the sum i equals 0 to nk minus 1 of again remember when you integrate with respect to Dirac delta that's just the value of the function at this point so this is just phi composed with fi of xk okay so by defining the measures in this way I get that the integral of phi with respect to the nu k is exactly this and this is exactly the bulk of average that we had here so combining with this we just get okay and so we have that integral of phi d nu k minus phi bar is less than equal to epsilon for all k okay so remember that this phi bar is the integral of phi with respect to d mu right so what are we going to do here we have a sequence of probability measures here we can take a limit point okay so by the weak star compactness of the spatial probability measures there's a limit point right so m m is compact so let nu in m such that kj converges to nu some subsequence kj tending to infinity then because of this uniform bound we have then we have the integral of phi d nu minus phi bar which is the same as integral of phi d nu minus integral of phi d mu greater than equal to epsilon okay in particular so we have the nu is different from nu and how do you know there's invariant so arguing exactly as in the proof arguing proof of the Krilov Bogoljubov Bogoljubov okay Bogoljubov theorem we have the nu is invariant okay but ergodic averages is f invariant but for all ergodic components nu ergodic averages ergodic converge to converge to the integral of phi which is different from phi bar and this is a contradiction well maybe it's like more simple I could simply say that it's a contradiction because nu is f invariant and it's different from mu from this and therefore we have contradiction to the assumption of uniqueness okay let me say that actually it's a little bit simpler the contradiction is that nu is f invariant and nu is different from okay contradicting okay so today has been quite a technical quite a technical day I think we will finish here and the next lecture we will apply this result to show that the circle rotation is uniquely ergodic and we will also prove a nice little application to number theory in ergodic theory there's quite a few applications to number theory and I will give you some examples as we go along okay so I think we can finish here for today thank you