 is well prepared to talk and I would like to have instructions here that include just read this thing or just look at the picture on my phone. Things I wrote on the board in my office, so bear with me if I forget a step or two here and there. So let me state our setting for today and the theorem that I hope to prove throughout today will let x mu be a standard. It fell towards, scared you to just think of the unit interval with limit measure, that is actually perfectly general. Philosophy of ergodic theory, I mean it's kind of not starting with physics, but the mantra that people want to show was that time average and space what I mean by test-fix, by an integral in particular. We can take f to be Borel, NEL or the functions of time, time average. So what you're going to do is I'm a point in my standard Borel space, the transformation moving over here and I move over here and I do that a few times and I just record the average of the value of the function that I saw. So to be precise, I'm going to say for a natural number, N is just going to be the average of the sum from k. Space average, let me just state the theorem first, then it might become clear what we need by space average. This is the point-wise ergodic theorem, the Birkhoff, it was proved in 19th-century averages, function that exists on everywhere. It's a point-wise limit of a sequence, so it's going to be Borel also. We can think about what function this actually is. It's going to be equal to the space average, i.e. the expectation. What is sigma twiddle? This is the sigma algebra. If you're not familiar with what expectation is, this part is kind of just a secondary approximation. We're going to be most interested in this. Here's a sequence of functions, let's get point-wise levels. How do the analysts prove a statement like this? The proof that I've seen before, I think this is more or less Birkhoff's original proof, maybe with a few modifications, but what you'll see in an analysis textbook is you might define this to be the supremum of these averages, of the actual value of the averages. You're going to prove which is the statement, or does that statement as the state measure of a point, f star is larger than alpha, is less than or equal to the one norm of f. The next thing you do is just, through some ad-hoc method, somehow prove the theorem, prove the Birkhoff theorem for a dense subset of l1. And so then once you have the Birkhoff theorem on a dense subset, you pair that along with the maximum type of equality, and then you get the whole thing. So for maximum, it's going to imply for you the whole result. Please, I mean, one is just a definition. Three is not that hard. The dense subset of functions you write down are the invariant functions, or things that look like a difference like this. It's easy to show that for functions of this form, and these are going to span a dense subset of l1. The tricky thing is this one. This is very non-trivial and requires some work. So we're not going to do that work. What we're going to do is instead follow a very slick proof recently discovered by a new center young. I'll be writing down the name quickly so that if you're interested at all in this proof, or just basic background and descriptive set theory, there's loads of awesome notes on the website. So you should check it out. Let me just write it down. So in particular, she wrote this. This is the entire thing. It's one page. A descriptive set theorists' proof. So I kind of break this apart into three main ingredients. Two of which are kind of philosophical, more than anything else, and one of which is actually allowed that they need to do some work to prove. You can't either way would know or get to do something. So the first one is kind of philosophical. All standard Burrell spaces are the same. I guess this is underpinned by a theory of the proof that given any two uncountable standard Burrell spaces, there is actually a Burrell isomorphism between their corresponding sigma and algebra. That's not too difficult. In particular, there's wanting to be a point in time where we're going to really want our space to look like cancer space, and that's going to be convenient. So I'm just going to be free to do that because they're all the same. Oh, so there's no measure floating around again. So this is just a statement about the Burrell spaces. And so yeah, and then a corresponding statement is true for uncountable spaces with non-atomic. So the second one is another kind of philosophical to set theorists often something that we do. So we have this transformation floating around our probability space. Oftentimes, that transformation is too rigid to work with. So we kind of want to be a bit more flexible. And instead of just working with the transformation, we work with the equivalence relation. The equivalence relation to points is equivalent to y if milliliter found the same orbit. So there will be times when we kind of we deliberately mess with the transformation and tweak it a little bit, but in a way that still respects the equivalence relation. And that added flexibility, the ability to tweak the transformation that we're thinking about and even time can be useful. So those first two were more sort of just how should we think about this problem? The third one is actually a result. The third one, which requires... I mean, it's not a hard puzzle. I hope to prove it at that time. This is the sling and steel part, which I won't define just yet, but it's going to sort of be the technical tool. Let me elaborate a little bit on... I'll call these things under 3. Let me elaborate a little bit on 2. And I'll return to... So I'll be defined this orbit equivalence relation. Two points are equivalent if they're in the same orbit. So think of this a useful visual to keep in mind is we have the standard Burrell space. And this transformation kind of organizes the space into a whole bunch of z-lines, where I just have a point and then t of x and t inverse of x, and then it starts to give me a z-line. And so I'm breaking up the space into lots of z-lines. You might have some finite cycles. That's okay. Finite cycles for the purposes of... Oh, actually, I left out an assumption that this is an a-curiotic and a remarkable transformation. There are no finite cycles. Everything is a z-line. If you want to deal with the cycles, they're really easy. You can just deal with them in a finite way. Yes, that is a good... So think of that as breaking the circle apart into these z-lines. So the philosophy in 2 would need some terminology to back up this philosophy to deal with the equivalent relation is better. It has to be a subset of x squared on the equivalent relation with a sub-equivalence. Theoretically, you can just say it's a subset of e t. More graphically, we're breaking these z-lines apart into even smaller chunks. So a sub-equivalence relation is going to have more classes, which no one really likes, but it's just standard and now we're stuck with it. Is this abuse of the word finite? So we say that f is finite, but it's not set-theoretically finite. It's just we're breaking up the space into a bunch of finites. We're often going to modify... We're not just going to deal with any old equivalence relation. I might use the word Borel equivalent relation. There's going to be Borel in the subset of this product. You may have classes by sub-equivalence classes. Um, I mean, this definition doesn't depend on being a sub-equivalence of e t. But I mean, everything I'm looking at is going to be a sub-equivalence of e t. And we're going to be breaking these z-lines up into finite chunks. Can I answer your question? I'm going to answer your question. I'm going to answer your question. Oh, so one of my classes just has some things that are f related. So here's say that I have a t-class. E t classes look like these z-lines. And I might arbitrarily break this apart into smaller chunks. So here's a smaller chunk. Here's a smaller chunk. And so, I'm going to... So here's, like, x of y. And so I might say here that x is f related to y because they're in the same chunk. So f being finite just means that all of these chunks... Another definition that's a little bit weird, but really sort of underpins this philosophy that the equivalence relation is sometimes better to work with than the transformation itself is this following definition. So we had a measure-preserving transformation. And that just means that transformation moves the Borel set and the measure of the Borel set appears the same as it was over here. Let's talk about the notion of a measure-preserving equivalence relation. It's a little bit weird, because what does that mean? So this restricts to the... Actually, I don't even need to restrict. So let's say I have the measure of mu on my space x. It's called, so let's say, it's mu-preserving in the equivalence relation. Oh, let me give this a name. This equivalence relation is named e. Measure the philosophy here. What should we keep in mind? So a transformation gives rise to an equivalence relation. And an equivalence relation could in turn give rise to loads of different partial transformations, i.e. those partial transformations where a point has to be e-related to its image. But aside from that, I allow a lot of flexibility. So this just says whenever I have a partial Burrell function, or a partial Burrell injection from some subset of x into itself so that every point is e-related to its image, I demand that the measures stay the same. So one way of producing measure-preserving maps that we're going to do is we have our measure-preserving t from before and we're going to do something like break up the space in the chunks like this and then we're going to do something where we follow t until we're about to leave a chunk and then go back. So we're going to mess with t a little bit. It's going to look like t on a lot of the space, but we're going to have the flexibility to perturb t a little bit on the set of small measures. Oh, yes? x is the standard Burrell space. I'm suppressing the notation for think of it as the unit interval and it just comes with its signature of Burrell sets. Right, so I don't understand why you say so each x relates to y and you say that x and y are... They're just... So f is going to... f just breaks the space x so forget about the sigma algebra for a second. So x is just the set. f is going to be some equivalence relation and it demands that all the chunks are finite, all the equivalence classes are finite. So then in addition to finite we can also demand that the equivalence relation be Burrell as a subset of the product sigma algebra. But we can sort of... Those are independent constraints. Yeah, so f is going to be the set of pairs in the same chunk. So I'm going to define these of this function f that we're talking about. So now I'm going to define equivalence relation averages of the function. So m at point x is going to be 1 over the cardinality of brackets over every y in x's equivalence class. So this operation just takes a point, looks at every other point and its finite equivalence class and just takes the average. So the finiteness is p, and that definitely wouldn't make sense if the class was not finite. So then here's the level which is going to be kind of another technical tool which is going to be sort of the idea that if I'm going to cook up wacky transformations that look like this but they will still be finite, still going to preserve the measure. So if I start with a finite measure preserving equivalence relation and create a transformation of that, I don't change the measure. So the limit is at the integral of d mu is equal to the integral of m of d mu. First thing I'm going to do is the equivalence classes are all finite so let's partition the space into kind of many pieces based on the size of these equivalence classes. So this will be this joint union of x of n where x of n is going to be the union of all those classes with exactly n dollars. This will be the class of philosophy that all standard Burrell spaces will say. I mean I guess it is possible that the x n is countable but that's all that can always be worked with with this idea. What I'm going to do is let's say that I have only a ring of x n in order of type of theory or maybe x n. So this is x n and say that I have one of these classes of size n. So here is four elements of this class. I'm going to distinguish the least element and I'm going to let b be those set of points of x n is least. And this is Burrell just because this is a Burrell in here. So we'll notice that this b is what is called an a. This is a Burrell transversal of f. b meets every equivalence class exactly once. I'm going to use here the fact that we have a new preserving equivalence relation. So it's going to be the integral over b of the sum. What's really going on here this is saying that this is using the measure of t i inverse of b is the same as the measure of t. I haven't told you what t is. Let me use a different letter because it's not the same t as before. t is going to be the following order. I don't want to use t. Sigma Sigma is going to equal I'm just going to draw what Sigma is. Sigma is going to do this. Sigma this is a Burrell transformation. It's just t. This isn't this is not quite t. This linear ordering on xn has nothing to do with t. Right now it's just a I'm sorry I didn't tell you what assumptions I'm using. So f needs to be a finite finite new preserving. So all you need are these assumptions and then Islam goes through. And so I'm just going to cook up this in a sigma. Since Sigma the graph of Sigma is contained in f because I just map points to equivalent points the equality points and then if we read what that means this is going to be the same thing as I'm just adding up to the points the points in an f class I can also add up their averages. That's the same thing. And then I go back once again using the fact that these are all new and this will be the integral ones. Yeah this is over yeah this will be over xn. And so we do this for every xn. So this is where the background needed for this equivalence relation of philosophy. So now let me talk a bit about sling and steel. And what I think I'm going to do is I'm going to state sling and steel and then use it to prove the theorem and then given time I'll try to prove sling and steel. I might not have time to get some definitions. The sling and steel marker lemma was proven by sling and steel in the early 90s or the late 80s. Do you remember when I was proven? So I think the existence of things called marker set is. For this definition we're going back to our setup of xmu t is the same transformation of marker set which meets 3t. Equivalently it meets every et class because we think about equivalence. It's bi-infinite. And then writing down an annoying to write down definition I was going to draw an easy to draw picture. It's the same thing. It has to meet so here's an equivalence class it's just a z line with respect to the transformational t. Remember every t equivalence class looks like a z line because we're dealing with a neighbor and forever. The marker set a has to meet a point somewhere in this line. Bi-infinite just means are eternally many points in this direction and are eternally many points in this direction. So no class has a leftmost member of a or rightmost member of a and it meets every class infinitely often to the left and to the right. And here is the statements of the marker remember. Let's think for a bit about what is and is not a permissible marker set. Claude mentions the example earlier of irrational rotation and one of the easiest ways to cook up a non-measurable set is to try to cook up a marker set that meets each equivalence class exactly once. If you attempt to do that you're going to cook up a non-varyl set. So often times you're not going to be able to get a varyl selector. So necessarily our markers they meet the class but not just once lots of places in some way. But we can try to get those meetings to be more and more sparse and that's the content of this label. There exist marker sets in a subset whose intersection is there. So sometimes this whole system is called vanishing marker sets so slang and steel prove the existence of vanishing marker sets. And so this is good. This kind of means that we can get arbitrarily sparse markers and that's technically useful for including the proof of this theorem. So let me actually prove the theorem using the marker limit as a black box to be on the link soup less than or equal to the link. This is what we're trying to show this group. So towards a contradiction let's say yes so there are some a less than b so that the following set would write down less than b less than equal to f over loss. So notice this x prime is going to be invariant. We're going to see like more and more of the points neighborhood in its orbit. So x prime is going to be a t invariant set. So I'm going to say that x prime is going to be not just like throw, throw out the rest and then blow up the measure to get back to the problem. We're going to say that we have this bad behavior. So let's fix epsilon greater than zero this small is going to depend on like b minus a and we'll just have what it is like an actual number of z less than or equal to n less than or equal to this very small measure measure smaller and so on. Oh, even even worse I want the integral integral over z of 1 plus f can we do that? Why? Because I'm assuming that the limb soup, the limb soup is above b. So I can't always have, I can't always be in this situation for any x almost everywhere there's going to be a point where x is greater than or equal to n. The idea is that most of the time as we start looking at more and more averages we know that the limb soup is above b or at least as big as b so we're going to see n's where this is at least as big as b. So I can't no it's not, it's a less than than. It's less than but then as n as l gets larger fewer and fewer points are going to satisfy this and then it has to vanish eventually. And it has a very small measure. This is if x is in z or otherwise we're just going to set it to 1. And a useful notation will be the set of the interval related to x that's going to be x tx all the way up to t. So in particular if I'm not in the second case and I look at the averages here I know that we'll see what happens. I might want that to be the smallest one in either case unless if we're going to be counting the left or something. We'll see what happens. Use the marker lemma The marker lemma is going to have an important extra bonus in the conclusion of the marker lemma. These are going to be by infinite. So let's grab use the marker lemma. Not necessarily. I defined the marker set and then I defined by infinite. So I want to buy it on the marker set. So use the marker lemma to find it by infinite so that I want to have a really small measure but actually not just s, but I actually even fatten s up a little bit using t. So these are the union of i less than l t minus i of s. The integral over s prime less than it's going to be really small. Marker sets are useful for. So let's think about these markers. So these blocks appear by infinite we often in every z line. These are the members of s. So here's a member of s. Here's a member of s. Here, say that l is like 3 or something. So here is like s twiddle. So so there's going to be a region behind each block that looks like s twiddle. Twiddle isn't too big. The s twiddle blocks are like pretty far spaced but they do small enough things and now we're going to take the z line and split it up into a finite burrell subgroup. So again I could, lemme write down the definition but I'm going to draw a picture that's better. It is z so that s of x I'm not told you what s of x is Here's an x. Here's s of x. I'm going to start looking left until I hit the point of s and that's s of x. So there's a z so that s of x equals s of y equals s of z and x y equals z and the whole s of z to z is quote unquote. So yeah that's a mouthful lemme draw a picture. So here is a point in this and we're going to start marching forward with our transformation. So remember every point had this L of x which is just some number that we assign every point and we had had this interval corresponding to that number and that point. So here's the first point on this I'm going to draw a little interval so here this first point would be if this was a point of y this would be t sub L of y of y and then this point is going to have its own notion of interval et cetera a bunch of intervals are popping up. Oh wait here's the next point of s that's kind of bad we don't want to approach it and worse yet s total kind of reaches out further next point of s here's s total kind of reaching out that's bad so maybe here's a point and it's not in but then the end point of its interval it starts meeting so that's the last one we're going to consider and then with the equivalence relation the equivalence relation is just going to be it as follows here's a class here's a class here's a class here's a class here's a class but then everything after this is a single term so this is the picture and the one thing I want the picture to make clear is that any non-trivial class of the classes are either just going to be points or they're going to look like some one of these intervals and this is where my notes tell me to just follow the paper so we're going to see what happens this is I didn't write it down formally but it's the elements of s next to the left so you transform x to the left I march left, I march left, I march left wait, I get an element of s that's fantastic so we're definitely using the the bi-infinite assumption is creeping in so the demand of the non-trivial classes do not hit the same as at the set supporting non-trivial f classes z satisfies this way 1 plus f is less than something like 2 fs and what's worrying me is this claim that the demand of fs will be supported outside of s twiddle kind of looks like we might hit s twiddle just a little bit I'm going to use the fact that the way that I've defined I think domain here is being used the union of the non-singleton equivalence classes so let's call that the domain no, I'm silly, the domain contains okay I'm silly I can't read set conclusion the domain of f contains x less s twiddle it may also contain parts of s twiddle that's fine, that's how subsets work okay each of the equivalence classes love the form iz okay, I just can't read it's going to be small I guess this is where I had to tell you what epsilon was um, so what was epsilon so the way she does it g says epsilon at the beginning and sets it so that b minus a so this was the third of our contradiction we have a soup of b that I'm into below a so b minus a the measure of x, oh but I'm just going to say that x is a probability to be greater than epsilon absolute a absolute b okay, that seems reasonable so 2 epsilon is less than okay yeah, this is going to kind of remember what z was so remember z was the set of things it's not going to be z so f is in the domain of f z in particular that means that the f average, you're just a singleton because we set that little l of f equal to 1 in particular f of x has to be greater than or equal to b for those points so here so that's this observation for y in y and f of y just equals f of y it has to be greater than or equal to b so outside we want to claim that we're less than 2 epsilon because because so the domain of f was what's x plus y x plus y is either things in z things in things in z or things in s twiddle so the things in s twiddle to be small s twiddle is really really small all of us are not explaining this terribly well it is cooked up to work so what is the takeaway the takeaway is we have this so the takeaway is that the integral of f of t mu is greater than the integral of f of t mu minus 2 epsilon which is greater than or equal to b minus 2 epsilon absolute b and then we have to mimic the argument for limit from a and then we're going to get a contradiction we're going to get so let's see if I can solve some of this so yeah more than epsilon small and then here we're concerning epsilon to be kind of big the way that we're doing that is we're breaking up our space over here we have a lot of control over our function looks over on x less y in these equivalence classes x less because the domain of f is huge the domain of f is going to contain x less s and this is s twiddle I think this was just meant to be we said s twiddle to have really small measure x twiddle has really small measure I think they call it depth delta it really just needs to be epsilon so this is size not just measure epsilon but like measure epsilon okay so s twiddle on x less s twiddle we see most of the function z is very little of the function so this is okay so this is actually really easy I'm sorry this was not hard at all this is literally just the assumptions that we cooked into these sets how we solve this take away this take away we use the fact that f is yeah let me explain this last step a little bit we're going to use minus 2 epsilon fudge factor and we have a lot more control over this average because it's going to look like it's at least b measure of y minus 2 epsilon and the measure of y is not bigger than 1 and so that's where we get b measure of y or b yeah so then this is greater than or equal to b 2 epsilon so this is all just fudge factor the key thing here this is the key step the measure of observable step okay so apologies for bundling that a little bit let me try to I'm not going to have time to improve the marker level let me explain some of the ideas going to the marker level because that's also a cool descriptive set here are you trying to justify the box oh so we cooked in a lot of assumptions so we cooked in this set z was super small x plus y is contained in z union so we're assuming that z is super small we're also assuming that s twiddle is super small yeah so that's the relevance I can't see that that's sort of a subset of z union plus 2 and we actually cooked these up to control the integrals of 1 plus x and most of them act as as a trailer so this is just the first one the integral of I think it's this very last to sort of you know we're controlling the integral oh it's this set right I forgot an important subscript here's the integral over x here's the integral over y so I'm just integrating so I'm changing where I'm integrating over and then just forgetting the rest and calling the punch factor and then this is over y this is the way the measure of y pops up again some punch factor 5 minute 5 proof of the marker line worked earlier that we want our standard burrell spaces to be uncountable standard burrell spaces going to be anything we want so I'm going to take it to the counter space and we have this this is totally in the burrell context there's no measure so t is just an a periodic but I have these z lines I'm thinking of them as equivalence classes here's counter space counter space also has this topology that's kind of nice so I'm going to think about some given phi of y some y's closure so you might you might be a little skeptical like why not for this burrell it's because we are closure plays nice so you can talk you can reason about this entirely in terms of like the finite strings that define the neighbourhood so I say this neighbourhood contains an element of bracket y any lexicographically proceeding neighbourhood does not contain a member of bracket y and since there are only kindly many finite strings that's that number and now a rough idea a first approximation to these a n's is that a n is going to be close to those those y's so that y restricted to n is phi of y this is super close to working but there's a problem which is that the intersection over all the a n's it might not be empty it might contain it could contain if this let's say a member of the closure is actually in the class itself it might contain it, that's bad and there's no guarantee that these a n's are by infinite so now we have to tweak it to get by infinite and to get the by infinite it's really easy to tweak it a little bit further to get this property I don't want to keep you any longer so I think I'll stop there so you're too close any closed subset of cancer space is going to look like a subtree and so I have the leftmost branch of this subtree and so the a n's are requiring that the members in the equivalence class agree more and more so if I have some subtree I'm shrinking towards the left and so I'm demanding that I have more and more agreeance with the leftmost because I'm on the philosophy so the maximal inequality is not why I favor this argument is it's very natural these are very natural things to do if if you're comfortable with some ideas from descriptive set theory this idea of carving out sub-equivalence relations and reasoning about those shows up all the time in descriptive set theory so the idea that those touch styles of argument can prove something like this is kind of cool and that being said it is obviously easier if you are a more comfortable with these things also it might look less scary if I presented it in a clear fashion but that's subjective is this being applied to prove some of the that is an excellent question the issue here because it's a time that we're in question what groups have plant-wise ergodic theorems and marker lemmas are totally general you can generalize marker lemmas in all sorts of directions the bit that gets a little bit harder is the the geometry we really use this z-line geometry to carve out these finite sub-equivalences it's similar we're really touring around this idea that equivalence relations created by z-actions are hyperfinite and that's a big open question do all of the groups generate hyperfinite equivalence relations I believe so I don't want to say too much but yeah it seems like if you can get a reasonable finite sub-equivalence that does the things that you want that's the correct maybe maybe