 Ok, bravo povrza. Zelo tudi, da smo nekaj drobno zelo počaj, kako si se zelo, da smo imeli, izvediliko tako ljudi, ali je so pravdjev svoj. Vrzača to je, da ne bomo tudi, prikaj se zapravo, da se zelo, da se je v nekaj nekaj nekaj zelo, da nekaj se kaj, da se počaj jaz bomo počaj po vzetku x, y, z in nekaj intervju omega z f double prime of x over f prime of y times f prime of z. The supremum of all possible combinations, this is clearly less than, this is equal to actually equal to the supremum over all x, y in omega of f, w squared. Because you take the supremum over all possibilities, right? So this was in a conversation with Hoku and Salako, I think yesterday we realized this. Since this is the supremum, then of course it means all possible combinations of this. So of course this is always less than or equal to this one, because if one of the, you can always take here the biggest one of these two, right? And you can put it in here, or the smallest one, the smallest one of these, right? And so this is equal, less than or equal to k. So what we had is, we had this as our assumption, okay? And we had this that came out in the proof in the equation, and we just needed to show that this was less than or equal to k, so it's this simple observation. Okay, so let's get back to what we completed last time. So last time we completed the proof of the fact that if we have a piecewise C1 map that is uniformly expanding and satisfies a certain property, which is this property here, then Lebesgue measure is ergodic. So today the first thing I want to do, two things, depending on how much time exactly we have, the first is an application of this result to a specific map called the Gauss map. So the Gauss map is a very classical dynamical system that was studied by Gauss for reasons that are slightly different from the ones we are studying today, because it has some connection to number theory related to continued fractions. But this is the map. X goes to 1 over x, mod 1. So this is a very interesting map, because it turns out that it is indeed a full branch map. So what does the map look like? So for all n in n, we have that f maps the interval 1 over n plus 1, 1 over n. Let's just map 0 to 0. OK, let's map it on the open interval. OK. Well, when 0 goes to 1 over 0 is not really well defined. No, no, no. Ah, here it's OK. Yes, but we want to iterate it. OK. We can also define it just to be 0 in 0. f of 0 equals 0, and f of x equals 1 over x to the other points. So this is a full branch map, and you can easily see what the structure of this map is, so it maps the interval 1 half, from 1 half of 1, it maps to, let's see, so 3 quarters, so something like this. And then it maps the interval 1 third, 1 half, also to the whole thing, to everything. It's not piecewise affine, right? It's got a small, the derivative changes a little bit, and then 1 fourth, 1 third, like this, and so on. It's got countably many branches. It's a C1, piecewise C1 full branch map. Lebesgue measure is not invariant in this case, because if you take an interval here, AB, this is not piecewise affine, so the preimages, it has a countable number of preimages, and some of them does not correspond to this Lebesgue measure, but Gauss himself, he found an invariant measure for this map, a very explicit invariant measure for this map. So mu is called the Gauss measure, and is given by mu of A equals 1 over log 2, the integral of A, the integral of A of 1 over 1 plus x, the mu. Let's put Lebesgue measure here. So what we're going to prove is this theorem. As mu g mu is invariant, ergodic. So let's prove, first of all, that it's invariant. This we're going to prove directly. Lemma 1 mu is invariant. So let A equals AB be an interval. And then we have that mu of A is equal to the integral, is equal to 1 over log 2 integral AB dm, and this is just the integral over this interval, which is in fact 1 over log 2 times log of 1 plus B over 1 plus A. You can compute this integral explicitly in this case. So what do we need to do? We need to estimate. So this is the measure. This is what this measure gives to this interval AB. Of course, it depends on the positions of AB, and that's what you get. And now we look at the preimage of AB. So A equals AB has a countable number of preimages. So mu of the preimage AB is equal to mu of the union, the union n equals 1 to infinity of the intervals 1 over n plus B 1 over n plus A. Because that is exactly what the form of these preimages is. If you look at the explicit form of the map and you look at these preimages, so these are two points such that this 1 over x, 1 over this gives A, 1 over B gives this. And so all the images are exactly of this form for all the different values of n. You can calculate them explicitly. So this interval here is 1 over 2 plus B over 1 and 1 over 2 plus A. So this interval here, this is the interval 1 over 2 plus B and 1 over 2 plus A. When you take this and you take 1 over x, so this maps to 2 plus A mod 1, which is equal to A. So this point here, sorry, is the other way around. This is 2, yes, B and A. This is A maps to here. You can easily see that the image of this interval maps to AB, because you take 1 over A of this gives m plus A mod 1 gives A and this 1 over m plus B, 1 over this mod 1 gives B. So these are all the preimages of these intervals. So this is what we need to estimate. We want to show that the measure, mu measure defined like this of this union is exactly equal to the measure of AB. So this is just going to be a direct calculation, another one line calculation. So what is the measure of this union? These are all disjoint, so we can write this as the measure of the sum, so we can write this as the sum n equals 1 to infinity of 1 over log 2 and then for each interval, since that is an interval, we will use this formula here, so we will get log, sorry, I am getting a little bit clamped here, log of, let me write it like this, 1 plus B, which in this case is this the second endpoint, 1 over m plus A divided by 1 plus 1 over m plus B and this is equal to 1 over log 2, the sum n equals 1 to infinity of if you kind of multiply this out and actually I want to write this as a product, so I take the log out, let me take the log out and write this as a product, n equals 1 to infinity of, ok, if I kind of simplify this, what I get is n plus A plus 1 over m plus A times n plus B over n plus B plus 1 and if I write this out, this is 1 over log 2, log, so what is this product? So for n equals 1 I get 1 plus A plus 1 over 1 plus A times 1 plus B over 1 plus B plus 1 times, now I put in n equals 2 here and I get 2 plus A plus 1 over 2 plus A times 2 plus B times 2 plus B plus 1, ok, and I continue like this and you see that a lot of things cancel, so what cancels? We have here 2 plus A cancels this 2 plus A 2 plus B here cancels this and then these other ones are also going to cancel with what comes afterwards and what you're going to be left with is just log 1 over A so if you write the next time you will see that these cancel also and so we get that 1 over log 2 times log of 1 plus B over 1 plus A and this is exactly the definition of mu of A B, yes. In order to enter this sum inside the algorithm we have to know that this sum converts how do we prove that this sum converts? In order to here well you're just doing term by term I mean you're just writing the sum and I think it's fairly straightforward in fact to see that it converts so this is very unusual to have an explicit formula for the invariant measure like this it's very, very special case there's a couple of other examples but in general it's very unusual in this case however we have an explicit formula and so we can check directly by this calculation that this is indeed invariant so this way this proves that we have a measure that is invariant now what we're interested in is ergodicity so this is when we would like to apply our result from the previous lecture because this is exactly a piecewise C1 full branch map what did we prove in the previous lecture can we just apply our result here? that's right so there's two issues one is we need to check the conditions so we would need to show that it's uniformly expanding and it satisfies this condition on the second derivative over the first derivative squared we can show that in a second but what is the conclusion of the theorem that we proved last time? ergodicity of what? of Lebesgue measure but this is not Lebesgue measure but this measure is absolutely continuous with respect to Lebesgue measure is that enough? this equivalent means that they are both absolutely continuous with respect to each other which is not always the case but often is ok let's go step by step first of all let's show that Lebesgue measure is invariant so first we show sorry is ergodic first show that Lebesgue is ergodic so we need to show two lemmers lemmer 2 and the derivative is uniformly expanding is this true? why? ok I will leave that as an exercise because it's very simple ok so the derivative is actually everywhere bigger than or equal to 1 it's only equal to 1 at the point 1 it's everywhere bigger than 1 except at the point 1 but remember uniform expansivity does not need the derivative of the map to have derivative bigger than 1 everywhere it needs to satisfy the slightly weaker definition that I gave so let me leave that as an exercise good exam question this one lemmer 2 supremum overall x, y in partition sorry x, y in ohm so supremum overall omega in the partition supremum overall x, y in omega of f double prime of x of y squared less than equal to k so there exists k such that this is true ok I will leave also this as an exercise because all you need to do is calculate the first and second derivatives and check this so then the big measures are got so then we need just one more lemmer so sorry that was lemmer 2 let me call this lemmer 3 let me call this lemmer 4 is that if f this is a very general lemmer f i to i is a measurable map mu 1 is a probability measure that is absolutely continuous with respect to mu 2 mu 2 then if mu 2 is ergodic implies that mu 1 is ergodic ok which is what you were saying before so we don't need them to be equivalent we just need one to be absolutely continuous with respect to the other one sorry no no no because this is mu 1 is absolutely continuous with respect to mu 2 and we will use that remember what absolute continuity means it means that if something has mu 2 measure 0 then it's got mu 1 measure 0 that's the direction of the implication ok and so it's easy so we want to prove that mu 1 we will assume that mu 2 is ergodic we want to prove that mu 1 is ergodic so how do you prove ergodiste again let a be a measurable set and suppose that mu 1 of a is greater than 0 ok we show that mu 1 of a is full measure this is what we need to show to show that it's ergodic sorry so by absolute continuity we have that mu 2 of a is positive because if mu 2 of a was 0 then mu 1 of a would be 0 that's the direction of the implication absolute continuity and by ergodiste of mu 2 this implies that mu 2 of a equals 1 and this implies that mu 2 of i minus a equals 0 and this implies that mu 1 of i minus a equals 0 by absolute continuity and this implies mu 1 of a equals 1 so this is what I meant when I said I made a comment when I stated the theorem about the ergodiste with respect to Lebesgue even though Lebesgue measure was not invariant the fact that we were able to prove that Lebesgue measure is ergodic gives a fairly strong statement in the sense that any other measure that is invariant and absolutely continuous with respect to Lebesgue will automatically be ergodic because of this simple lem in the particular case is this measure here absolutely continuous with respect to Lebesgue so this Gauss measure because of this form it has a density here it is of the form where it is given by the integral of some density and so it is automatically absolutely continuous because if you take a z set that has zero measure with respect to Lebesgue you will get zero for this integral and so you will get zero for mu so that gives absolute continuity so mu is absolutely continuous with respect to Lebesgue and so mu is ergodic and this completes the proof of the ergodiste invariance of the Gauss measure for the Gauss map ok, so now let me complete finish with a proof I will not unfortunately I think be able to finish completely this proof but I think I want to talk about it a bit because it's really a kind of key and natural question at this point so we've studied piecewise affine maps maps like this and we've studied maps and we've proved that if you take so for piecewise affine so what we have is for piecewise affine we have that Lebesgue measure is invariant and ergodic and for piecewise c1 plus uniform expansion plus that condition on the second derivative plus the condition that f2 over f1 squared is bounded this implies that Lebesgue measure is ergodic but not necessarily invariant ok, as I mentioned last time what we're interested is invariant ergodic measures because even though the ergodiste of Lebesgue because the ergodiste of Lebesgue does not give us the Birkov's ergodic theorem to study the statistics of all this so we've applied this to the specific case of the Gauss measure in which to the Gauss map in which we know the measure explicitly and because that measure is absolutely continuous we apply this map and we get the ergodiste of that measure but what about in general a very simple situation in which these properties are satisfied is if for example you have just the map piecewise affine map like 2x mod 1 and then you perturb this a little bit so I change this map I take a small c1 perturbation for example and I just take something that is piecewise c1 almost piecewise affine but of course in this way it destroys all the structure that we have from here and it becomes something like this and it's easy to show that it's uniformly expanding because the derivative was constant equal to 2 here so if you make a small perturbation the derivative will be close to 2 this is trivially satisfied because f1 well because f2 is bounded away from zero everything is bounded away from zero and infinity so it's very easy to show that this is true so in this case we still get the Lebesgue measure ergodiste but do we have another measure that is for example absolutely continuous with respect to Lebesgue measure in this case this is the question so in this context is the much more general context the fact that Lebesgue is ergodiste so that we have an absolutely continuous invariant measure that will give us an absolutely continuous invariant and ergodiste measure so the question is when does there exist an invariant measure mu which is absolutely continuous with respect to Lebesgue measure because that is really understanding question that comes from this study you would expect that there should be because look you've just made a very small perturbation here is Lebesgue measure you make a very small perturbation Lebesgue measure is no longer invariant but you'd expect that there should be some invariant measure that is similar to Lebesgue so theorem piecewise let's see one full branch with bounded distortion property so remember assuming the bounded distortion property is more general than assuming uniformly expanding in that condition on the second and first derivatives so then f admits ergodiste invariant probability mu is absolutely continuous with respect to Lebesgue so this you could consider the kind of theorem that we have been leading up to right from the very beginning of the course because what this means is that for almost every point for Lebesgue almost every point so we have our interval here 0,1 and this means that for Lebesgue almost every point you have described the statistics of the orbit using Birkov's ergodic theorem so the basin of attraction of this measure has positive measure so this is a physical measure in the sense in which I define physical measures like at the beginning of the course so this gives a very general class of systems for which you have a physical measure in this way yes because that will satisfy this theorem so if I take for example if I take well any piecewise affine let me not take it so special let me just take a piecewise affine map like this and let me now take some small perturbation not even a small perturbation just small perturbation like this then it is a piecewise c1 full branch and it satisfies those two conditions which imply the boundary distortion property so it satisfies uniform expansivity and second derivative over first derivative square is bounded and so it particularly satisfies the boundary distortion property so this theorem applies to this class of example and much more generally what it says is that if you take any map even with a countable partition and you have this boundary distortion property which you can check for example the uniform expansivity and the second derivative then there is an absolutely continuous invariant measure probability measure so the Gauss map becomes a special case of this theorem also in general we cannot have an explicit form of this of this measure which we have in the Gauss map but that is a special case of this yes no I'm just giving some examples here as I said in the previous course if your map is not continuous if it's piecewise c1 then it's not completely clear what we mean by c1 perturbation because you could also perturb a little bit the point at which this change but it doesn't matter in this particular case it doesn't matter because any reasonable way as long as the perturbation continues to be full branch and has the boundary distortion property it doesn't apply so really we don't really want to think of these as c1 perturbations of piecewise affine maps this is just a very general class of systems I'm just giving you the example the reason why I use this example is because I say that even though in the piecewise affine the big measure is invariant and ergodic even if you make a smallest perturbation you lose the invariance of the big measure in this example you have no idea what you've got even if you make a very small perturbation but that case is covered by this theorem because any small perturbation in a reasonable sense will continue to satisfy these properties and therefore this shows that you continue to have an invariant measure the very interesting question is this measure close to the big measure when you take a kind of perturbation ergodic measure that is close for example in the weak start topology to the other measure does the measure change continuously with the map this is a very interesting question and this is studied in many dynamical systems and in this case it's fairly easy to show that yes indeed this is the case because which is what you expect because even though the big measure is not ergodic if you make small change then you know the measure will shift around but it will be close to the original measure in the weak start sense so it is also possible to show that this measure depends continuously on the map if you define the topology on this class of maps in a way that is reasonable but this is a really much more general problem and it's very well studied and in many cases we do not know the answer so I have now given you some foundations in this course this class of full branch maps are kind of a basic class of maps and really you can use these techniques to construct absolutely continuous invariant measures and other physical measures for many different dynamical systems and in general after you prove the existence of such a measure the next question is ok, does this measure depend continuously on the perturbation it's a very interesting question it's a kind of stability it's like structural stability in a topological sense structural stability means that as you change things there's no bifurcations the topological structure stays the same from an ergodic theory point of view you want to know if the measure depends continuously on the map so I will not have time to prove this but almost I will just give you a sketch in the argument and I will leave one proposition unproved because we don't have time but it's really not very difficult it's the same kind of techniques and calculations that we have been doing so far so how do we prove so what do we really need to prove here what have we already proved all we need to prove is that there exists a measure that is invariant and that is absolutely continuous because the ergodicity we get for free because we've already proved that Lebesgue measure is ergodic so if we get a measure that is absolutely continuous it will also be ergodic so all we need to prove is to find a measure that is absolutely continuous and invariant so we start with Lebesgue measure so let m is Lebesgue measure and then we define a sequence of measures mu n equals 1 over n sum fi so do you remember the definition of fi star fi star m of some set a is defined as exactly is defined as the measure of f minus i of a so remember we already used a similar kind of sequence right at the beginning of the course when we are showing the existence of invariant measures and this also gives you some clue right so by the same argument we know that any weak star limit point of this is invariant basically so by previous arguments weak star limit point is f invariant it's not completely the same because we used in the argument we used before we needed to use the fact that f is continuous which is not the case here ok, if we have time I will give you by maybe I should say by argument similar similar to previous arguments so if we have time I'll show you just the calculation in this case without using that f is continuous but the point is that it's not that difficult to show that any limit point is invariant so what we need to show and you know why there exist accumulation limit points of this do you remember by the compactness of the space of probability measures simply, right because each of these is a probability measure the average is also probability measure you have a sequence of probability measure so there exist weak star limit points by compactness so there exist limit points any such limit point is f invariant all we need to show that the limit point is absolutely continuous ok so why should it be absolutely continuous so notice that mu n is absolutely continuous with respect to the bag measure do you agree with that why is mu n absolutely continuous with respect to the bag measure from the definition because if you take a set that has zero the bag measure then its preimage will also have zero the bag measure so this will be zero so this will be zero so mu n will be zero so its absolutely continuous so that should be it does that imply that the limit yes well this is because this map I mean this is the boundary distortion implies that this map cannot send a set of a positively bag measure to a set of zero the bag measure this map if it was to send the preimage of any set of zero the bag measure has to have zero the bag measure otherwise you would have a set of zero the bag measure mapping to a set of zero the bag measure which would mean you have some kind of degener for example if the map was constant then you would do that you would send a map of positively bag measure to a single point which has zero the bag measure but this cannot happen because of the boundary distortion property the boundary distortion property says that all the relations are preserved ok so ok let me write this so since f is non-singular positive the bag measure sets map to positive so you are right this is something that perhaps I could spend a little bit more time on but I want because time is so short ok but it does depend on boundary distortion implies the map is non-singular and therefore this is absolutely continuous but is that enough if each mu n is absolutely continuous with respect to m does that imply that the limit point is absolutely continuous with respect to m your n-send is a to zero yes and that's why it's very complicated anybody disagree so suppose we try to do this exercise with the following map so suppose we take the map that you know very well zero one to zero one x goes to x over two is not full branch but we can still construct this sequence and we can still take the limit and we still have this absolute continuity property so what does the measure f star m look like for this map so Lebesgue measure is Lebesgue measure on zero one what does f star of m look like two multiplied by measure of a so let's see f star of m of a is equal to measure of f minus one of a so let's take a set here what is the measure of this set under f star m so we look at the preimage of a so we go what we need to do is we need to take diagonal so we need to look at a here and then we need to look at the preimage which is this this is f minus one of a excuse me right it's empty set exactly so what is the measure of that set if we take a then is zero so not all sets have positive measure which sets have positive measure where does this measure live zero to one half exactly because if you take a set between zero and one half then you calculate its preimage it will be somewhere and it will be twice as big but if you take a set between one half and one it has no preimage so this measure is completely living inside the interval zero to one half not only that but if you take a set between zero to one half its measure will be exactly twice the big measure of this set so what does the density of this measure look like the density of this measure looks like something like this this is two so the density df star m respect to dm is equal to two on zero one half so you get a measure that is living here and this density is equal to two in the sense that you need to multiply the levegg measure by two to get the f star m measure what if we do this again what if we look at f2 star m of a is equal to m f minus two of a so this measure lives by four so this measure here lives in zero one over four and its density is four because if you take a set in here and you look at this measure you need to pull it back twice so it becomes four times as big as it was so the f star two m measure of a is four times the levegg measure of the set a leaving on one fourth and its density is four it's still a probability measure because if you take the whole interval zero to one fourth its measure is one because you need to multiply its length by four and you get the measure equals one so you get a sequence of probability measure each one of these is absolutely continuous this one is absolutely continuous with respect to levegg in the sense that if you take a measure with a set with zero levegg measure you multiply by four you still get zero levegg measure so each one of these is absolutely continuous but what is the weak star limit of these measures that's right the weak star limit is the delta zero is the Dirac delta in zero which is the only limit you can expect because that is the only very measure that exists in this map because what does this map do every point converges to the point zero for this map so fn of x converges to zero for all x so the delta Dirac in zero is the only only invariant measure and since this sequence or in this case you don't even need to take the averages even this actual sequence itself converges to the Dirac delta converges to an invariant measure this sequence converges to an invariant measure then it has to converge to Dirac delta and you can see it explicitly this is actually a very nice example of where you see how this sequence of measures accumulates on a measure so this example which I would like you to review at home very carefully because it's an important and very useful example and I really see explicitly how these behave is an example of a sequence of absolutely continuous invariant measures not invariant, absolutely continuous measures converging in the weak start topology to a measure that is not absolutely continuous yes that's right, that's right the fact that the set of probability measures compact is just a general theorem of functional analysis that the unit ball of the continuous functions is compact in the weak start topology yes that's right, what we proved is that the set of invariant measures is also a compact subset of this compact so all of this example was just to emphasize the fact that we need some estimates here we need to use the properties of our full branch maps because what we've done here is really general I did not really use any properties I need either continuity or something to show that the limit point is f invariant but certainly the absolute I can do this in very general situations like this and the limit is not necessarily absolutely continuous even though each measure is so in relation to this it seems natural what is going wrong is that the density is blowing up so even though each map is absolutely continuous the region where this density is living is getting smaller and smaller and the density is blowing up so if we can control the density of these then we should be able to guarantee that it's absolutely continuous and this is just the proposition that I will state I will not prove it but it's not that difficult to prove so n equals d mu n by dm this is the density of each of these measures because each of these measures is absolutely continuous so it has a density and the proposition, the main proposition says that there exists k greater than zero such that what is what? soup here and ok we will not use this but let me write it for completeness we in fact have some additional properties on the density, no actually we will need them and for all n greater than equal to 1 for all x, y in i we have that h of y less than equal to k h and x times the distance between x and y which is less than equal to k squared times the distance between x and y so what this says is that these densities are uniformly bounded above and below and in fact that they uniformly lipshets independently of n uniformly in this sense so was this satisfied in the previous example of x goes to one half x were these conditions satisfied? they were not satisfied because the density was blowing up it was concentrated in smaller and smaller region it was getting larger, it was unbounded in n was here we are saying the density is uniformly bounded in n so just to conclude I will write the proof of why this proposition implies our theorem and then we will leave it as that so ok, so this is just a simple application of Askoli-Azela theorem because this is now proof of theorem assuming a position so hn bounded and equicontinues because it is uniformly lipshets so by a theorem called Askoli-Azela theorem it converges uniformly to a function satisfying the same properties so hn converges to h uniformly and h satisfies both of these conditions yes, some subsequences ok there exists a subsequence there exists a subsequence it says that this is a compact family a precompact family yes so let mu be given by mu equals integral mu of a equals integral a so we claim that this is the measure we need so what we've done is we've taken so on the one hand we had the weak star limit of these measures but weak star limit of these measures is not good enough ok, so what we do is we take the density of these measures and we show that the densities have a limit where we have much stronger information about this limit because we have these properties about the densities so we know that the densities converge to a very nice lipshets continuous bounded above and below function h which we use to define mu ok, so we took the limit of the densities rather than the limit of the measures and then clearly mu is absolutely continuous by definition and then let me since we have a few minutes left let me also show that the mu is invariant because what I said before is that the arguments that we used previously used the fact that f is continuous in this case we don't really have f continuous so we just have a slightly different calculation so let me give it here so mu is absolutely continuous now to show that mu is invariant to show also because we are not directly defining mu as a limit of weak star limit of mu n although it is essentially the same so notice that mu of a is equal by definition htm and this is equal to the integral of a of the limit of hn dm and now we used the dominated convergence theorem that allows us to take the limit out of the integral here, so this is the limit as n tends to infinity of a of hn dm hnj sorry, these are all sorry, this is all for the subsequence hnj and this is exactly equal to the limit as j tends to infinity of mu nj of a and by definition of mu nj this is equal to the limit as j tends to infinity of 1 over nj sum i equals 0 nj minus 1 of f star i m of a and this is just equal to the limit j tends to infinity of 1 over nj of the sum of i equals 0 nj minus 1 of the measure of f minus i so this does not give write exactly what we want but it's very useful to be able to write this in this way what we are saying is really that even though we've defined the measure in terms of limit of the densities this basically corresponds the measure of a is the limit of the measures of f star i m of a of these averages for each measurable set so from this we easily get the invariance of the measure so from this now to check the invariance of the measure take the measure of f minus 1 of a we can write using this formula here as the limit as j tends to infinity of 1 over nj from 0 to nj minus 1 of the measure of f minus i of f minus 1 so now I'm just applying exactly this way of writing it for the set f minus 1 of a that I have there and now if you rearrange things a little bit in the usual way you will get that this is equal to the limit as j tends to infinity of sum i equal 0 to nj minus 1 of the measure of f minus i of a plus 1 over nj the measure of f minus nj of a minus 1 over nj of the measure of a so I basically take this 1 into here and then I rearrange the sum to get this ok, this is very similar calculation to what we did last time so here in the limit so here you get of course that this the measure of f minus nj of a is less than equal to 1 always this is less than equal to 1 always as you divide by nj going to 0 the contribution of this is 0 so in the limit this is just equal to the limit as j tends to infinity of 1 over nj sum i equal 0 to nj minus 1 the measure of f minus i of a which is exactly by definition the measure of mu so this shows that mu is invariant mu is absolutely continuous so this is all we need because now we can apply the theorem that we had in the previous lecture and we know by the absolute continuity and the godiste of Lebesgue measure that mu is also ergodic so that exists in ergodic invariant absolutely continuous probability measure ah, good question how do you prove that this mu is unique why is it unique in the last 10 minutes is on this question so can you have two absolutely continuous so you are right so we still have to need that we still need to show that mu is unique show that if mu 1 is absolutely continuous in variant ergodic and mu 2 absolutely continuous with respect to and invariant ergodic implies mu 1 equals mu 2 this is what we need to show well, this is a more general statement but in particular it will imply that in fact we need something a little bit more here we need let me write it like this then either is singular with respect to mu 2 or mu 1 equals mu 2 so the reason is you can have this situation which is the example that we did at the beginning write if you remember where we had this example here remember this example here so in this example Lebesgue measure is invariant but it is not ergodic because Lebesgue measure is restricted to 0, 1 so this is 0, 1 so if we write mu 1 is equal to Lebesgue measure restricted to 0 1,5 and mu 2 equals Lebesgue measure restricted to 1,5,1 so mu 1 is Lebesgue measure normalized so mu 1 is equal to 2 times this and mu 2 is equal to 2 times that so here mu 1 is a probability measure which is absolutely continuous with respect to Lebesgue invariant and ergodic mu 2 is a ergodic probability measure ergodic invariant absolutely continuous but they are singular with respect to each other in the sense that they live in different places so this of course does not have uniqueness here you have two invariant ergodic probability measures so we need some slightly bit more information here in in this case what we've proved is a little bit more about the measure we've also proved that the density of the measure is bounded above and below everywhere whereas in this case the density of mu 1 with respect to Lebesgue measure on the interval is positive here but it's 0 in the part where it does not live mu 1 has a density which is 0 here and mu 2 has a density which is 0 here they are singular with respect to each other they live on different parts so the last missing ingredient to show the uniqueness here is to show that so we have proved density h equals mu by the m is bounded above below so if we have another measure so what we have here is we have also h1 bounded h2 bounded and then this implies that they cannot be singular and they have to be the same so why is that if mu 1 mu 2 are two such measures then by Bilkov's ergodic theorem and some phi composed with f i of x converges to the integral of phi d mu 1 for mu 1 almost every x and 1 over n the sum of phi composed with f i of x converges to phi d mu 2 almost every x so what does this mean so in fact because maybe I could have said it a little bit simplest because I hadn't prepared the answer to this question but maybe a simpler way to see is that because this density is bounded above and below then mu and m are equivalent as you said so m is also absolutely continuous so since h is bounded above and below then also the Lebesgue measure is absolutely continuous with respect to mu 1 and Lebesgue measure is absolutely continuous with respect to mu 2 and so this implies that the set of points that is mu 1 measure 1 has also positive Lebesgue measure in fact full Lebesgue measure and the set of points that has full Lebesgue measure sorry this this is because it is bounded below sorry is bounded below because the inf remember the inf of all h and x was greater than 0 that's right the inf this is the proposition that the inf of all x and all n is greater than 0 and the limit also satisfies that property so it's bounded above and below ok and so these are also absolutely continuous and so this implies that this is actually true for m almost every x and this is also true for m almost every x so using the fact that these two are equivalent you can show that this convergence holds not just for mu 1 almost every x but for Lebesgue almost every x and the same convergence has to hold for Lebesgue almost every x so the intersection of these two sets of full Lebesgue measure has to have full Lebesgue measure so it means that there is a for almost every x you have both of these convergence which mean for every continuous function this converges to this and this converges to this which means that these have to be the same for every continuous function which means that mu 1 should be equal to mu 2 I could have given a little bit more systematic essential reason for the uniqueness ok so mu 1 is equal to mu yes because if mu 1 has measure 0 well because basically you can write the density you can write m you can write 1 over h is the density of m is the density dm by d mu is 1 over h no we are assuming that both measures are in the same way oh I see you are saying mu 2 could be another measure ah you mean because I just say suppose there is some other absolutely continuously measuring measure where the density is not bounded above and below yes it doesn't matter I mean I don't know right now I don't know if I can recover the argument but if you look at this closely you will see that you only need it for mu 1 actually because mu 1 you only need because this is true for mu 1 ok ok you are right so this is true for mu 1 is that this convergence happens for Lebesgue almost every point if you have another absolutely continuously measuring measure even not knowing that you have equivalence the fact that this is true for mu almost every x will imply that it is true only for the positive Lebesgue measure for set with positive Lebesgue measure and that is enough to give the contradiction because positive Lebesgue measure full Lebesgue measure it still means that this has to be these limits have to be the same for positive Lebesgue measure set of points yes you are right to write it down formally and precisely I shouldn't assume that the other one also has density above and below but it doesn't need to be it doesn't well done yes ok so there is a few points in this last lecture that I had to kind of just sketch rather than give completely but I think it was important to finish with this theorem because you can think of the overall question which I think I mentioned at the beginning was precisely this was which dynamical systems have physical measures one of the main examples of physical measures is absolutely continuous in very energetic measures and so here we've defined a very important class of maps called full branch piecewise expanding or full branch with bounded distortion maps which always have a unique physical measure so I think at least it gives you some kind of idea of the kind of questions and the kind of techniques that I use in this area which we call ergodic theory because it uses the language of probability theory to understand systems that have some differentiable properties and we use this to prove the bounded distortion we use the derivative the second derivative and so on we use differentiable properties of the map to study the ergodic properties so I hope it's been interesting and very best wishes with your courses we can finish here