 OK, so good afternoon, everyone. So after a couple of lectures of quite intense background and motivation, I think now we are ready to get into the technical part of the course and start proving some results. So throughout, we will always have some space or some set, metric, let's say compact metric space. We will always have the Borel sigma algebra. And we let m equals the set of all mu Borel probability measures. Then we also have some map, which in several situations will be continuous or differentiable. If this is actually also linear, has some additional, is a manifold, a differentiable structure, certainly at least we require it to be measurable. That just means that the pre-image of every element in the Borel sigma algebra is an element of the Borel sigma algebra. So that's just a kind of minimum level of regularity. So this gives us a dynamical system in the usual way. We iterate it. So the purpose of this course is to understand how to use the notation and the formalism of probability measures to understand the dynamics of this map. So first of all, we need to make two definitions. So the first one is we say that mu in m is f invariant if mu of the inverse pre-image of a equals mu of a in the sigma algebra. So in a moment, I will give several examples and we will discuss this. Notice that the fact that as in the previous course, the fact that I'm putting f to the minus 1 here does not imply necessarily that f is invertible. What I mean is the pre-image of the points in a. So f minus 1 of a, by definition, equals the set of x such that f of x belongs to a. This is the pre-image of the set a. So easy exercise you can check. It might be strange to see it written in this way. If f is invertible, then this is the same thing. You can just apply f to both sides. And this is the same thing saying that mu of f of a is equal to mu of a. But as we shall see in some examples, when f is not invertible, this is the most natural definition. And this definition says somehow that mu and f are related in some way. So that mu is a particular probability measure that is related to the dynamics of f. We will see in some examples what we mean by this. The second definition is we say mu in m is algorithmic if for all a measurable sets a such that f minus 1 of a equals a, then either mu of a equals 0 or mu of a equals 1. Yes, the second definition. It depends on f because this is a condition on the set a that depends on f. So this condition is true only for sets that satisfy this condition here. So it's saying that if you have a set that satisfies this condition here, it cannot have measure a half or anything between 0 and 1. In general, every set, this is a probability measure. So any measurable set has a well-defined measure that is between 0 and 1. It can be 0, it can be 1, or it can be anything in between. What we're saying here is that a set that is fully invariant, sometimes we call it fully invariant, has to have either 0 measure or full measure. So it depends on f in the choice of the class of sets to which this dichotomy applies. So these two definitions are fundamental to the whole course, okay? And we'll start by just giving some simple examples and then for the rest of the course, we will develop some more sophisticated examples. And we will show how these properties remarkably can help us to study the systems in quite a certain level of detail. So some examples of invariant and ergodic measures, so example one, suppose that f of p equals p. So p is a fixed point. And let measure mu equals the Dirac delta measure on p. So here we have fixed point p that is mapped to itself. Here we have our space, m, our metric space. So I claim that delta p is invariant, f invariant. If once f is fixed sometimes we just say invariant, right? Here this should be f ergodic, ergodic with respect to f as well. It's invariant but if f is fixed sometimes we don't state it again. It's invariant and ergodic. So both of them are fairly easy but let me prove the invariant just so you start getting a feeling of what it's about, right? So to see that it's invariant, to see delta of p is f invariant. Choose a set, let A be a subset of M, okay? So we take a set A. So we need to show this. So what are the possibilities for the measure of the set A? There's only two possibilities. You can have measures 0 or 1, okay? So we just have to check those two possibilities. So suppose first that p belongs to A. Then the measure of A is equal to 1. What about the measure of f minus 1 of A? Exactly, p is a fixed point. So the pre-image of the set A must contain p also. Because if the pre-image of the set A did not contain p, then A could not contain p either, okay? So the measure of the pre-image is also equal to 1 because f minus 1 of A also contains p, excuse me? If there exists only one fixed point, so you see, so I'm not assuming anything about the dynamical system here. I'm just choosing. I'm assuming that there exists one fixed point, but maybe all the points are fixed. Yes, for this example, I'm assuming that there exists one fixed point. But this could even be the identity map, okay? But after I choose one fixed point, I also choose a measure. I choose a specific measure. And I choose that measure to be the Dirac delta on this fixed point. I choose it like that. I'm not proving that there exists, there's nothing to prove. I choose that measure. So once I choose that measure, it's like the rest of the dynamical system becomes invisible to me. Because that's the measure I've chosen. This measure is what I'm looking at in the dynamical system. And this measure gives positive measure only to that fixed point. I don't know what happens outside this, okay? I choose that measure. If there are other points, I will also be able to choose other measures. We will see lots of examples of that. But for the moment, what I'm doing, I'm choosing this measure on this fixed point. And I want to show that this measure is invariant and ergodic, right? So these two definitions, as was correctly pointed out before, depend on two things. Both of these definitions depend on the measure and depend on the map. So you say a measure is invariant with respect to this map if this holds. And a measure is ergodic with respect to this map if this holds, okay? So I choose a measure. In this case, it's the Dirac delta on the point P. I want to show that it's invariant. So I need to show this for every, for every measurable set, right? Often this is not trivial how to do that. But in this case, it's not a problem because we are interested in the measure of these sets and there's only two possibilities. Either A contains the fixed point in which case it's got measure one, or A does not contain the fixed point in which case it's got measure zero, right? In the first case, if A contains the fixed point, we've shown that this holds. Because any set A that contains the fixed point, its pre-image also contains the fixed point. So it has measure one, which is the same as the measure of A. And therefore this is satisfied, right? If P does not contain the fixed point, then mu of A equals zero. But the pre-image also cannot contain the fixed point. Because if the pre-image did contain the fixed point, then its image A would also contain the fixed point, which it doesn't. So mu A equals zero and mu of F minus one A equals zero. So we have checked that this is, okay, so mu is invariant. What about ergodicity? Ergodicity is trivial, because this is in fact two for every set. It's measure either zero or one. So in this particular case, it's trivially satisfied, okay? So ergodicity is trivial. So let's look now at just a very simple generalization of this. Let's suppose that we have periodic orbit, right? So suppose we have Pn equals periodic orbit of period n, okay? So P1, P2, Pn goes back to P1. Doesn't have to have such a nice structure. It can be period one million, like the ones that we have seen in the shift map. For example, 2x mod one, we have periodic orbits of period a million, a billion, whatever, okay? Period n. So what measures shall we put on this periodic orbit? Can I put the measure on P1? Is this measure, so this is the Dirac delta on the point P1. Is this invariant? Good, okay? You can take A equals P1, okay? If you take A containing P1 or A even equal to P1, okay? Then this has measure one and f minus one of A is Pn, that's right. Is contains Pn, yes. And does not contain P1. Well, actually in this case it contains P1, but no, it doesn't contain P1 because the image of P1 is P2, that's right. Contains Pn. Or let's just put it this way, does not contain P1. So the measure of A is one and the measure of f minus one of A is zero, not invariant. So how do we put an invariant measure on here? Can we have an invariant measure on this periodic orbit? We need to have some measure on P1 and some measure on Pn, right? And some measure on each of the points along the orbit. Take the average, very good. So we can take the measure mu equals to one over n delta P1 plus delta Pn. So we leave it as an exercise, same exercise as this to show that this is invariant. What about ergodicity? Is it ergodic? Why is it not ergodic? So let's remember that ergodicity, let me write down the definition. So recall, mu is ergodic if f minus one of A equals A implies mu A equals zero or mu A equals one. It is this implication, is ergodicity, right? The fact that this implies this, this is the definition of ergodicity. So it looks like this measure is now distributed in various different pieces. So maybe it's not ergodic. But let's check. We need to check. So to show it's not ergodic, all we need to do is find a counter example. So find a set that satisfies this that has measure between zero and one. So proving non-ergodicity is often easier than proving ergodicity. So let's prove, either non-ergodicity or ergodicity. What set are we going to use? Shall we take a set like this that contains only P1, set A. This set has measure between zero and one. What's the measure of each point here? One over N. No, exactly. PN or shall we have both? Like this, is this good? Ah, we need to contain all the points. So now what's the measure of the set? There's a problem. What's the problem? So the only way, if you have any set, so if you have a set that does not contain any of these points, then its pre-image will not contain any of the points. So it will have measure zero, okay? Whether it satisfies this or not. But if it contains any of these points, then either it contains all of them, in which case it may satisfy this or not satisfy this anyway, it's got measure one. But if it does not contain all of them, then it cannot satisfy this. So it is ergodic, because every time it satisfies this, it must satisfy this. So this shows also how ergodicity depends on the map and not just on the measure. So the measure is indeed, in some sense, intuitively divided into some separate bits. But the map means that these bits are connected in a way that you cannot separate them. They are intrinsically connected to each other. So mu is f invariant and ergodic. Okay, so let's look at some more examples. So example three, fm2m is the identity map. Always check the identity map as an example or counter example of everything. So what are the invariant measures for the identity map? So the identity map, this is x, okay? x goes to x. Every point is a fixed point. So what are the invariant measures? Remember, so this is ergodicity. What's the definition of invariance? So mu is f invariant, means that mu of f minus 1 of a equals mu of a plus a in b. So if you have the identity map, what's f minus 1 of a? It's a. So if f is the identity, f minus 1 is also the identity. So f minus 1 of a is equal to a. So the actual sets are the same for any set, which means that any measure you take, this will always be satisfied. So for the identity map, every mu in m is f invariant, every measure, okay? That's a lot of measures. The measures you have for every point is fixed. So you have a direct delta on every measure. You have all kinds of combinations. You have many measures in general for this. What about ergodicity? That's correct. And why is that? So you said that it's ergodic. So mu is ergodic, if and only if mu is the direct delta of some p. Because the measure, one way we just did it before, right? If it's a direct, every point is fixed for the identity. So if you take the direct delta on the fixed point, then it's ergodic. What about non-ergodicity? Why, if it's not like this, why is it not ergodic? If somehow the measure lives on more than one point, why is it not ergodic? Well, for example, let's just try first a simple conversely. So if let mu equals 1 half of delta p plus delta q, where these are two points, why is this measure not ergodic? That's right. Or you can just take any set containing p. It satisfies this condition because it's the identity map. And its measure is equal to 1 half. So it's not ergodic. So in general, because the identity map is so, every set satisfies this for the identity map, then clearly any measure that gives positive measure to more than one point, you can split that set into two parts where both parts have positive measure. Because you can just take any set that contains more than one point, you can divide into two sets. One that contains one point and one that contains all the other points, for example, or somehow you can divide into two sets. So in this case, if mu gives positive measure to more if, let me say it a little bit differently. So if mu is not of the form delta p for some p in M, then there exists sets A and B in M with measure of A positive, measure of B positive, and A intersection B is empty. The only way that this cannot happen is if mu lives completely only on just one point. And in this case, you just need two sets with positive measure because every set satisfies this is automatic. This argument will not work in general. This statement is true in general. But in this case, since f is the identity, f minus A of A equals A. And so, and this gives a contradiction to a good density. Another couple of examples. Example four, a favorite example, f, 0, 1, 2, 0, 1, x goes to 2x, 2x, so 0, 1. So what invariant measures do we have for this map? Excuse me? Excuse me? So the only measures that are organic for the identity map are the Dirac deltas on a fixed point. So every measure is invariant. So in general, if you take the identity map, for example, on the interval, the interval has lots of measures. Remember, we did the examples last time. It has Lebesgue measure. It has all the measures that you get by putting some density, integrating some continuous function with respect to the Lebesgue measure. It's got the delta Dirac on the fixed points. It's got, well, on the fixed points, on the periodic points. It's got lots of measures. But in this particular case, every measure is invariant because every set, the pre-measure of every set is itself. So every measure is invariant. Now we ask, which of these measures are organic? If you take the Dirac delta on one fixed point like this, then it's organic because we showed it before. Remember, if you have a fixed point, the Dirac delta on that fixed point is organic. The question is, if you take another measure, for example, if you take a measure like this that is leaving half of it on one fixed point, half of it on the other fixed point, then it's not organic. You can see that. This measure is not organic because you can just take a set here that contains only p. This has measured 1 half. And it's fully invariant because f minus 1 of a equals a, in this case. But you can generalize this fact to say, OK, any measure that is not of this form must necessarily, because it's not fully concentrated on a single point, you must be able to find two sets to disjoint sets, both a positive measure. If you could not find two disjoints on a positive measure, then it would exist only on one fixed point. If it somehow exists on more than one fixed point, then you can do that. And because each of these sets always satisfies this, then that means you found two sets that satisfy this and the measures between 0 and 1. So it's not organic. So no measure is organic except the Dirac delta on the fixed points for the identity map. Yes? Any measure what? I don't understand your question. No, no. So, or you take measure only in 0, 1. Any measure where the only possible measure of any subset is 0 or 1 I think so, yes. Yes? Well, but how can a measure take just values in 0, 1? Because if, yes, it has to be the Dirac delta measure, yes. Otherwise, yes, otherwise it needs to take, there must be some set that has measure between 0 and 1. That's right, yes, I guess that's what, yes. OK, what about this map here? It has some fixed points, right? So it has the fixed point at 0. So we can have the Dirac delta on 0. It has some periodic points. Remember that it has periodic points? It has many periodic points. So each of these periodic points admits the average of the Dirac deltas along the periodic orbit that we, the example we gave before. Are there any other measures that might be invariant? There's really only one that we have studied specifically, that we know specifically, which is a candidate, which is, for example, the Bayek measure. These are the measures that we know so far, OK? So is the Bayek measure invariant? Is the Bayek measure invariant? Well, let's look. Let's take a set A. Let's look at an open interval. If the set A has a certain measure, the Bayek measure, what is the pre-image of the set A? Exactly. So the pre-image of the set A will have two components. What is the Bayek measure of this pre-image? Each one is measure half. Half of the measure of A, of the length of A. So the union of the two has the same length as the interval A, the union of the two, OK? So at least for these intervals, the Bayek measure of F minus 1 of A equals the Bayek measure. So strictly speaking, to check invariance, we need to check it for every Borel set, OK? But if, again, using a kind of argument that I won't go into but using the fact, the Kolmogorov extension theorem, this is one of those examples where it is enough to check this property on the algebra of unions of intervals, OK? And then if you check a property of this measure on this, then it automatically, by the fact that this measure extends uniquely to the Bayek measure, then this property also extends to the sigma algebra generated by this algebra, OK? A very fancy way of saying that in this case, or certainly at least as far as this course is concerned, it will always be sufficient to check certain properties for open sets or for intervals in this case. And these imply the same properties for all the sets in the Borel sigma algebra, OK? It's a technical detail that I don't want to go into, let's say, OK? But it follows by the uniqueness of the extension of the measures. So in this case, you can easily see that for intervals, it's clear that we have this property and therefore it satisfied for all sets in the Borel sigma algebra, and therefore this implies that the Bayek measure is F invariant. What about ergodicity? So in this case, ergodicity turns out to be true, but it is not something that you can just see. This we will prove later on in the course that the Bayek measure is ergodic, but it's a non-trivial proof. And we need some dynamics, and we need some interesting argument. And yes, then will it work? If we have x equals kx of 10x, 1, 2, 3, 4, 5, 6, x goes to 6x mod 1. Is this what you meant? Yes. So is the Bayek measure invariant for this case? Yes, by the same argument. So this is, in fact, an interesting observation. So just like one map can have many invariant measures, the same measure can be invariant for many different maps. It just happens to be the Bayek on the interval. It is a special measure, but there are many different maps for which the Bayek measures in Bayek. So ergodicity is true, but non-trivial. We will prove it later. And let me now give you one final example in which we prove. So as I said, sometimes non-ergodicity is easier to prove. And I will try this example to show you why ergodicity. The first reason why ergodicity is actually an important property, number 5. So again, let's take f, 0, 1. And let me draw the graph. So is the Bayek measure invariant in this case? What is the slope here of these graphs? What's the slope of this? 4? Look, this takes an interval of size 1 quarter. And it maps it to an interval of size 1 half here. What is the slope of this? Minus 2, because this interval of size 1 half goes to an interval of size 1. And this is 2. Plus 2, minus 2, plus 2. It's always 2. So let's take an interval here, a. What is the pre-image of this interval here? It will have two components, one here and one here. So what is the size of the union of these two components? It's the same as the size of a. What if we take it somewhere different? If we take it here, b. Then it will have two components here with the same topity, also here. You can check if you take it in the middle. Also you would get the same thing. So you always get invariants. So in this case, the Bayek is f invariant. What about ergodicity? Come? Why not? You take a small interval here, 0 epsilon. OK? Yes, but you can always find a set that is measured different from 1. That's not, be careful not to make the mistake, right? In the identity, I emphasize, you can also here. I can do the same thing. Also here I can find an interval that has positive measure. If my invariant measure is Le Bayek, of course I can always find two sets of positive measure, these joint sets of positive measure, OK? But what property do these sets need to have? Remember, what's the definition of ergodicity? Yes, but not just 0 or 1. There's only specific sets with certain properties should be constrained. The definition of ergodicity, if you take a set with a certain property, then it is not allowed to have measure between 0 and 1. It has to be 0 or 1. But what is the property that the set has to have? Right. So is that true for this? What is the pre-image of 0 epsilon? The pre-image of 0 epsilon is here. So it's not equal to the set itself, right? So to check non-ergodicity, if we want non-ergodicity, we need to find some set A such that F minus 1 of A equals A. And we need to find such a set that has this property whose measure is strictly between 0 and 1. Or we can try to find it here. But to save you some time, I'm telling you that here you cannot do that, OK? Where's here you can? What's set? You need a set. Yes, but you've bought these sets that you've, both of these sets satisfy this, but they both have said it measures 0 or 1. So it's not good enough as a count, for example. So far you've just confirmed ergodicity. You have to look around here, because this maps, no, because you see this maps over itself. It stretches. You need to think more global. You know what? I drew the wrong graph. Sorry about that. Sorry about that. No wonder you could not find it. Sorry. This is the graph it needed to be. If you can tell the difference between this and the other one. Yes. Yes, sorry about that. But it was useful, maybe, also to see the difference between the two. So what's the right set in this example? 0, 1, 2, 1 half. 0, 1, 2, 1 half. The interval 0, 1, 2, 1 half. Very good. So if we take the interval a equals 0 to 1 half, then you look at the pre-image of this, 0 to 1 half. The pre-image is exactly 0 to 1 half. No, I don't understand what the problem is. 0 to 1 half. 0 maps to 0. 1 half maps to 1 half. Ah, here. You're right. You're right, you're right. Yes, yes, yes. Yes, yes, you're right. There is a little problem here. So let me define this on the open interval. Let me define this on the open interval. Yes, yes, well done. So this set satisfies f minus 1 of a equals a. And the big measure of a equals 1 half. Yes, yes. No, that's right. Because exactly the only way, this lack, because actually it turns out, we shall see again that if you just now restrict yourself to this part, then this is ergodic restricted to this, and this is ergodic restricted to this. So the lack of ergodicity means that really for that measure, you have two dynamical systems, separate dynamical systems. That's what this lack of ergodicity means. Is that whatever's happening dynamically here is completely independent of whatever's happening dynamically here. The two systems are really two. If you want, you can change this. You can make a perturbation here. And this system will not know about it. So the lack of ergodicity means that the measure has two different components. In this case, the ergodic components, as we shall see of the measure, one ergodic component is Lebesgue measure restricted to 0.5. And the other components is Lebesgue measure restricted to 1.5.1. And each of these components is ergodic for the map. So really what we have is the non-ergodic measure split into two ergodic components. And as we shall see, that is a more general feature. Whereas in this case, because the interval 0.5 is mapped to this interval here, and this one is mapped to here, and the dynamics on each of these is similar to this one here. In fact, in this case, we do have ergodicity, and we cannot decompose the system into two different parts. OK? So after these examples, what are the theorems? So these are just some examples. In principle, we don't know whether any invariant measures exist or whether any ergodic measures exist. So let's state the first theorem. We want some general theorem about the existence of invariant and ergodic measures. Let's define Mf is equal to the set of probability measures, such that mu is f invariant. And E of f equals mu in Mf, such that mu is, so just a brief remark here is that in the definition of ergodicity, I do not assume that f is invariant. So the two definitions are independent of each other. A map measure can be invariant without being ergodic, but it can also be ergodic without being invariant, OK? Although in practice, at least in this course, we will really only be interested in ergodicity of invariant measures, OK? So generally, we're looking for invariant measures, and then once we know that a measure is invariant, we're interested if it's ergodic. So clearly, we have these are all subsets Ef. This is just a definition. So Mf is the set of probability measures on our space, which are f invariant, just the definition of a set of measures. It may be empty, OK? Ef is a set of invariant measures, which are also ergodic, invariant and ergodic, because I'm assuming that I'm in mu f. Yeah, maybe I should say just mu f ergodic for a given dynamical system. So I'm always assuming here that our set is given with this Borel sigma algebra, and the map and the measure is given. So the theorem, let M compact and f, M to M continuous. Then Mf is non-empty, convex, and compact. Mu, moreover, mu is ergodic if only if mu is an extremal point. So I will recall some of the definitions here. So this is several statements, OK? So the first statement is that mu f is non-empty. What does this mean? Not that there exists a delta measure, just that there exists a measure, a probability measure that is invariant. For example, there are systems with no periodic orbits, no fixed and no periodic orbits, right? What examples have we seen of systems with no fixed or periodic orbits? Which have no fixed orbits and no periodic orbits? Sorry? Yeah, the adding machine, very good. Symbolic system, adding machine where every orbit is dense, and also circle rotations with no fixed or periodic points. Circle rotations have Lebesgue measure that is invariant, because they're just translations. So if you take an interval and you translate it, the big measure is invariant. But just to emphasize that this non-empty is non-trivial, because if you have a fixed or periodic point, then it's trivial, because you can always take at least that delta. But in general, this might not be the case, OK? So the fact that if f is continuous, this is non-empty is non-trivial. In fact, the proof of it is very interesting. We will prove this, OK? Convex. What does convex mean? Let me just remind you. I'm sure you all know. But just for let me recall, mf is convex. So what does a convex that mean? If for all mu 0, mu 1 in mf, and for all t in 0, 1, the measure mu equals t mu 0 plus 1 minus t mu 1 is also in mf. So every linear convex combination belongs to the set, OK? That's what it means for the set to be convex. This is easy to see that if these two are invariant measures, then this measure is invariant. It's easy to just check from the definition, actually. So compact, what do we mean by compact? Compact, we need a topology. What's a topology that we mean by compact? Week start topology, OK? So I don't need to recall the definition of compact. I just want to remind you that we mean compact in the week start topology. Because there are other topologies. But this is what we mean by compact. Finally, what is an extremal point? So what this is saying is that within the space of invariant measures, the measures which are gothic are exactly the extremal points of this set. So a point x is extremal. So mu is extremal whenever we write if mu equals t of mu 0 plus 1 minus t of mu 1 implies t equals 0 or 1. So an extremal point is one that you cannot write as a linear combination of two other measures, of two other points. That's the definition of an extremal point because it's like imagine that it is on the edge in some sense, OK? These are just standard definitions. So in the next lecture, we will prove this theorem. It's the fundamental theorem that says that this sets a non-empty under very general conditions. Today, I just want to give a corollary of this. So corollary. Again, m compact continues. Then e of f is non-empty. And there exists unique probability measure mu hat on n such that 1 mu over e of f equals 1 and 2 and integral e f integral. Sorry. Yes. T. Let's see a little bit what we have just said. So yes. So this mf has a topology, the weak star topology. Therefore, it has a sigma, the Borel sigma algebra associated generated by this topology, OK? Therefore, we can define a measure on this set. So I will prove it in the sense that this is a trivial consequence of some very deep results in functional analysis, which I will state in a second. We will not state those results, OK? But I want to explain a little bit what this result means. So first of all, what does the statement mean? So yes. So the first statement is that e f is non-empty. And the second statement is that there is a probability measure on this set of invariant measures. And this probability measure is supported just on the ergodic measures, OK? And what this is saying is that for any invariant measure, in particular, a non-necessarily ergodic measure, if we want to integrate a function with respect to this measure, we can integrate it as a kind of a product integral. So here you're integrating with respect to this measure on the set of ergodic measures, because this measure lives on the set of ergodic measures. You're integrating the function phi with respect to all the possible ergodic measures, OK? This nu belongs to the set of ergodic measures. So for each ergodic measure, you take this value here. This is a function of the ergodic measure. And then you integrate with respect to this, which means you're taking a kind of average of the integral of this overall ergodic measures according to this mu hat. So what this is saying is basically that what we would like to say is that every measure is a linear combination of ergodic measures. Because you see, this says that the ergodic measures are the extremal points. And you would like to be able to say that every measure is just a linear combination so that every non-ergodic measure, like the examples we saw before, are just the average of two ergodic measures or many ergodic measures. Like remember before the example, we had two fixed points and we had half delta p plus delta q. Each of these is ergodic, but we take the average of them and we get a non-ergodic measure. So we'd like to be able to say that that's true, but this is not necessarily true because it could be morally a linear combination of an uncountable set of ergodic measures. For example, if you take the identity map on the circle and you take Lebesgue measure, remember for the identity map, the only ergodic measures are the fixed points, the Dirac deltas. But every measure is invariant. So if you take the identity map, say on the interval on the circle, then this is the set of all ergodic measures, sorry, all measures, all probability measures are invariant. The ergodic measures are just the Dirac delta. So we'd like to say that every measure, like for example Lebesgue measure, is a linear combination of ergodic measures. But it's not a linear combination of a finite number or even a countable number of such ergodic measures. And therefore this is the generalization of saying that the Lebesgue measure is the average of all the ergodic measures in some sense, weighted by this, okay? So of course this measure here depends on the map, F, obviously, right? So for a given map, it says that you can write any invariant measure as some kind of integral of this on the ergodic measures. So let me state a little bit what are the feelings that this follows from. So the corollary, so the truth of corollary is the Krine-Millman theorem. I'm not sure if you did this in the Gaussian functional analysis, but the Krine-Millman's theorem says exactly that if Mf is non-empty compact and convex, this implies that it is the closed convex hull and Mf is the closed convex hull of its extremal points. So in particular, if it's non-empty and it is the closed convex hull of its extremal points, in particular it must have extremal points. So in particular, Ef is non-empty, okay? Because by the previous theorem, this is the, these are the extremal points. And the second statement, so N1 and N2 are exactly showcase theorem for non-empty compacts, convex for non-empty. So these are really the formulations of general functional analytic theorems in the context of the specific set of probability measures that we have here. Okay, so I think we can leave it at this for today and the next lecture we will prove theorem one. Okay, thank you.