 Well, let's review something that we have seen with Stefano and Oliver and Lucia and Irene and David. We'll be working with a set, a sigma algebra b and mu, a measure. And just to keep it simple, we are going to assume that this is a probability measure, the probability measure is when the measure of the whole set is one. Okay, and so we, you have seen, I think with Stefano, you have seen the Birk of Ergodic theorem. I don't know if you remember it, I will. And so, you have this, and just assuming that mu was an invariant measure, invariant measure for measure of map f, well, we have put other hypothesis on f, we will review it. Then almost every point satisfies this. This limit exists for almost every point, okay, for some, let me, let's keep it like this. Okay, this is what we have seen with Stefano. Remember that? Yes. Yeah, it's for, we have seen this for every c continuous map, have you seen, or have you put other hypothesis on c? L1. Okay, thank you. Yes. Yes. So, yeah, so we should state this better so as not to avoid confusions. We first state the c, thank you. We first state the c, then you see what Amy's comment is, this is a, there exists a set of measure one, okay, for, for every c there exists a set of measure one on which this holds. But this measure, set of measure one is not universal, that's not, that's good for every c. So, this set, this is a set will depend on c. Okay, so this is what we have seen, and now we are going to see a version of a previous version, which is a version of von Neumann, it's a mean ergodic theorem, version of this, but, and this states something more or less similar, but it says, now we have this f will be measure preserving, measurable, and then for, I will explain this for people who doesn't know, who don't know, for every map in LP, there will be, there will exist some map in LP such that this, what does this mean? This means that this goes to zero, okay, what's the norm in LP, LP, just in case you are not familiar with this, if these are measurable maps, and the norm in LP, okay, so this is, this LP with its norm is a Banach space, so it is complete, with this norm it is complete space, okay, but we are, concentrate this, I mean this is a version in mean, the mean ergodic theorem, okay, here you got this for almost every point, and here you got, you got it in mean, okay, we will concentrate though in the case P equal to, what's so special about P equal to, is that this LP, in the case of P equal to, is a Hilbert space, this is a Hilbert space, what's a Hilbert space, well if you don't know it, just it's a Banach space, but which has, its norm comes from an inner product, what's the inner product in our case, in our case, this will be the inner product, okay, sorry, no, no, no, thank you, P will be greater equal than one, thank you so much, but we will focus on P equal to, okay, and so this is a Hilbert space, so we will have this inner product, and what's so nice about this, is that we can make projections, okay, and this is our idea, yes, the L2 norm like this, but with P equal to, yeah, the norm will be, okay, so let's just make a comment about what we are going to do, we are going to prove this in a very, well we haven't proved this, but just we are going to give it an easy proof of this, but before that let me, let me define the conditional expectation, just to make a few comments, we are not going to use the conditional expectation in the proof, so well we have B is our sigma algebra, but let's suppose we have a subset, a sub-sigma algebra, let's take, and we want to produce a map, which we are going to call like this, we want to produce a map that is in LP of this sub-sigma algebra, but it has the same integral, okay, this exists, is unique, and it is defined by this formula, of course, over the sets where this makes sense, I'm not going to prove this, but you can prove it if you like, then ask me, I will give you some hints, okay, the claim is that for every phi in LP, there exists a unique map, this is a map in LP, but of the sub-sigma algebra, so that's for any open set we have that this belongs to B, okay, but now for this let's call it, just to be simple, this is more restrictive because the pre-images of any open set will have to be in B prime, these are less sets, we have less sets to choose, so even so we managed to produce the same integrals, integrals over all B in B prime, okay, we're not going to use this in the proof, but what it's good to have an idea of what we are producing with this theorem in terms of the conditional expectation, we have a very interesting sigma algebra of the invariant sets, we have this, this is a sub, what are these, these are the sets, the sub-sigma algebra of invariant sets, okay, okay, and so in particular the conditional expectation sets that we have that, you can check it, this map, this is a set of F invariant sets, this map, these maps are the set of invariant maps, of course, with, so we have a characterization of orthodic maps, these all I leave you as an exercise, choose white, have the following remarks, remember the definition we have done of our codicity last week and that Amy reviewed today, I think it is, but it is, I make a confusion because I define this, like this there and the conditional expectation, it will eventually be the same, it will be the same, because it will, I mean for the Hilbert space this will be like the projection of C into the subset of invariant functions, okay, this is what we are going to see in a minute, but in that case, in this case was just coincident that I used the same notation, it will be eventually the same, okay, so F is ergodic if this subalgebra of the F invariant sets is modulo zero, the trivial sigma algebra, this is, well this modulo zero, modulo zero measure sets and less trivial can prove that F is ergodic if and only if L2 with respect to this sigma algebra consist of constant functions, obviously constant modulo zero, almost everywhere constant and the third is that this, in fact is this, not BF, B, okay, for all P in L1, this is this, well this I leave you, these are remarks, but you can, this is some of them are more advanced exercises, that's true for LP, yeah, it's true for LP and P greater equal than 1, yeah, okay, okay, so let's, more remarks, do I have time, yes, more remarks, this is our characterization of ergodic proof, just let me observe that if F is invertible, then the sigma algebra of invariant sets for F is the same sub sigma algebra of invariant sets for F to minus one, okay, this follows from definition, okay, and in this case we would have if F is invertible and since these sub sigma algebras are the same, we will have that E, the conditional expectation will be the same and in particular this, we will have this, this will be very useful, remember this for later when we are going to see the hope argument, this limit is the same if we iterate forward or backward, okay, I'll state it with P, but we are going to focus on P equal to, as N goes to infinity, so remember this when we see hope argument, yes, this uses that this is indeed this, this is going to be the conditional expectation, this limit map is going to be the conditional expectation on the sub sigma algebra of invariant maps, okay, I have a further comment which is going to be one of our exercises today, I don't want to erase this, well I don't want to but I will, pardon, F is invertible and its inverse is measurable, yeah, thank you, that's true, not only the best approximation but in the case of Hilbert space, it will be indeed the projection, which is the best approximation, okay, it will be in LP, LP is not Hilbert space, that's not have an inner product, but in the case of L2 which does have an inner product, it will be the projection onto the space of invariant functions, okay, so I have this final remark, how this mean ergodic theorem is related to the bulk of ergodic theorem which I erased, I probably you know this, but I will put it as an exercise, it's an exercise or just remember it if you have seen it some at a certain moment, if you have a sequence converging LP, then there exists some subsequence such that converges almost everywhere, okay, so in particular we obtain the bulk of ergodic theorem but for some subsequence, okay, for some subsequence we obtain this, in fact we will have, well this is going to be this, well I have not written this here but it is, you can prove it too that this is an invariant function, you can prove it, yes, it, yes but we have not proven it yet, so you, if you only have this, that this has a limit, you can prove only with this statement, you can prove this, okay, and once you have this, you can prove this, okay, so let's go into the proof, so the idea, the idea is to decompose the space, well we have the Hilbert space of L2 maps, the idea is to decompose the space into the subspace of invariant maps and in orthogonal complement, okay, and then project any map into the subspace of invariant maps, okay, and this, in this way, well we, we will prove that for maps in this subspace they go, every map goes into the projection, into the invariant space maps, so let me be more clear, so this is going to be, we are going to use this notation, H is going to be L2 and we are going to consider the Kupman operator is going to go from H to H, supername but it's the operator, it's very simple, it's just composing with F, okay, so this operator is, has the following properties and it's continuous and its norm is less than one, what's this, what's, what's U star, U star is the operator defined by this, so what we mean here is that U star is the inverse of U, oh yeah, okay, U of 1 equals 1, so this, you will have to prove it but it's, this is easy, spend time in this, so we have this and so this is a continuous operator and so we have the following decomposition, we can decompose the Hilbert space into the following space, subspaces, the kernel of U minus I plus the closure of the image of U minus I, I'm not going to prove this if, but if you have never seen this, you just, just can imagine in algebra and finite dimensional spaces you have this without this closure, okay, you have seen this for finite dimensional vector spaces in algebra, okay, so it's the same only that it's higher dimensional version of it, okay, so more exercises prove that this is an orthogonal decomposition, prove that this is orthogonal and prove that this decomposition is U invariant, if you apply U to this space it goes into this space and if you apply U to this space it goes into a space, so it is U invariant decomposition, okay, so let's see, so we have this space of all L2 maps and then we have this and then we, we will take any map here and we will apply this procedure of averaging this and we will see that this converges to the projection of this map into this subspace, okay, this is what we are going to do, but here we are going to use that this is, well, this is invariant and so first thing is that we are going to write this sum in terms of U, so let's take, let's begin by taking an L2 map and then let's write this in terms of U, so this is in fact this, okay, you apply U k times and you average this and the idea is just taking any map, the vectors and the idea is applying U and averaging and see that this converges to be a vector, the project, it converges to be a vector of this space but not any vector, just the projection of phi, how can we do this? Well, one possibility is what we are going to do, seeing this for any phi, we are going to take a phi here and see that this holds for phi here and also take f, sorry, fc here and seeing that this holds for maps here and taking fc here and seeing that this holds for maps here and since now any map can be decomposed as if this holds for this and this it holds for this, so let's take this one, we take c in curve of U minus i, what does it mean to belong to curve of U minus i? It means that U of c equals c, so this implies that c is invariant, the kernel of this, this is how, I mean, this is why we chose this in the first place because we wanted to project it into invariant maps, so that is why we chose this subspace and not other one, okay? So if this is in the kernel of U minus i, this implies that this is an invariant map and if this is an invariant map then this, what this is equal to because all of this will be c, you will be summing this n times and divided by n, so then we will have that, so this will trivially go to c which belongs, so this not only converges, it's constant, it's a constant sequence, so for the first case if we take a map here, this trivially follows, the second case is more delicate and we are only considering maps in the image and I will leave you as an exercise to see that this holds when you go to the closure, okay? So let's consider case two and with this we finish, so this implies that phi equals some U of some phi minus phi, okay? So it implies it is phi of f minus phi, but then when we evaluate this, this sum, we will get the following, see? I just applied the definition, but what kind of sum is this? Telescopic, so all the intermediate will be cancelled, so at the end we will have, and this is no other thing than U nf, sorry, this goes to 0 in Lp, this goes to 0 in Lp because, well, this is finite U divided by n and this will go to 0 and this, we have stated that this is, its norm is less than 1, okay? So this in norm will be less or equal than 1 over n and this goes to 0, so for the more advanced students, I leave you as an exercise to show that this holds also in the closure because I have only taken maps in the image, okay? So this goes to the closure and this is it, we're finished.