 Okay, good afternoon. So if you remember so today we will continue our study of piecewise affine maps Our main theorem is that F piecewise affine full-branch map then the big measure is invariant Okay, so did you do the exercise to prove that the big measure is invariant? Yes, so we proved last time I left it as an exercise Okay, to show that the big measure is invariant It's quite simple exercise today. We're going to show that it's a garlic Which would not be difficult, but it's quite subtle and it's an interesting argument So first of all we're going to say something about the combinatorics of piecewise affine maps So remember that piecewise affine maps piecewise affine full-branch maps Have the structure where they have a certain number of branches and the key property is that each branch Maps to the whole interval in a fine way. It can be finite number or an accountable number of branches So the lemma let F a C0 Means continuous or respectively C1 C2 affine piecewise Sorry full-branch map B a piecewise full-branch map then there exists and exists Partitions Pn mod 0 a family of partitions. So for each end Pn is a partition of I such that P1 is equal to the original partition. So P1 is this partition here and for all and Greater than equal to one For all partition element omega n in Pn We have the map f n From the interior restricted to the interior Okay, Fn From the interior of omega n to the interior of I is a homeomorphism respectively C1 or C2 or affine Defil so what is this saying? This is just saying something that actually is very simple. It takes a little while to say it precisely but the P1 partition here is a partition into intervals of Into subintervals of the interval such that after one iterate f1 each interval is mapped on to the whole original And then What does that mean? We've looked at the structure already when we studied the topological structure here if you look at this interval here Each one of these intervals then because after one it that it maps to the whole thing Inside here. You can there is a Subdivision of this interval which maps exactly to the subdivision of the entire interval, right? So there is a small interval here that maps to the first one There's another small one that maps to the second one third one that maps to the third one and so on So there's a small copy because what this map does is it maps this small interval in an affine way to the whole Interval so inside the small interval There's exactly a small copy of the big partition here in such a way that each element of this Partition P2 is mapped after one iteration to an element of the partition P1 And therefore after two iterations to the whole interval like this Okay, and then you can continue you can repeat the process so that at each level Pn Corresponds to pulling back this partition n times So each element of the partition maps to an element of the previous partition Pn minus one and then After any that's exactly on to the whole thing in a bijection as a bijective way Well, in fact as a homeomorphism or C1 diffeomorphism or affine Okay, so let me write just very briefly the proof We can easily write the proof by induction So by induction for n equals one we have that This is true by assumption true by definition of f Okay, by definition of piecewise full branch For n equals one and now suppose it's true for n minus one. So let Omega n minus one Belong to Pn minus one. So there's a by induction. We suppose there's a partition Pn minus one we take one element and we have this property here that f n minus one from the interior of Omega n minus one to the interior of I is Let's say homeomorphism Is well, okay, whatever it is, right is Is Okay, let's say homeomorphism or in our case affine if you want it to be affine, okay And so what does this mean? All right What this says is that you have a map from omega n minus one to all of I this is f n Minus one and inside I we have our regional partition P Okay, so because these maps are finally or at least homomorphically to I that means that we can Subdivide this element in a way that is the images match up exactly with this partition That's what I mean when I say that we pull back this partition Okay, so we have a subdivision here Okay, that maps exactly onto this partition element and so what do you know about each one of these? partitions each one of these elements, so Subdivide omega n minus one Into subintervals into subintervals which we can call omega n Okay, so that for each Omega n subset of n minus one Fn minus one of omega n so fn minus one Maps the interval the interior of omega n to Some omega in P the interior of some omega in P All again I always like the interior just because I've not specified if these intervals are closed or open or half open And so on into the interval of some omega for some in P Yes, this is this we've already looked at that in the previous course. We're just taking small ones Well, we're defining the partition using the dynamics Okay, I'm just doing it again partly for completeness and also to emphasize the member that we Here as opposed to what we did in the previous course we we can allow countable Partitions here. This is some omega for the original partition some element of the original partition I because this is the original partition P here We have omega n minus one maps to all of I so we split omega n minus one such that each element maps to an element of the original partition This original partition therefore, what do we know About fn Where does it map omega n? To to the whole original interval right because after n minus one iterates It's maps to one of the original partition elements and by assumption the original partition elements get mapped in one iterate to the whole thing Okay, so this maps to the interior of I and of course the regularity is always the same right because we assume that this is a homomorphism Okay, and therefore when you restricted the restriction of fn minus one to a sub interval is also homomorphism on to this and Then from this you apply f once more but f from this interior of this to the whole thing is a homomorphism So the composition is a homomorphism So this is a homomorphism if we're looking at homomorphism or it's affine a composition of affine maps is affine composition of c1 maps is c1 and so on Okay, so we get the right regularity So this very nice very clean combinatorial structure is the key to the full branch to the full branch maps Right, this is a very nice. So all the iterates are themselves full branch maps. This is what we're saying If f is a piecewise affine full branch map Then all the iterates are also piecewise affine full branch maps on some finer and finer partition because Each one has an iterated maps each small interval to the whole to everything. Okay, so that is why One of the main reasons why full branch maps have such an easy structure to study Okay, so we're going to use this now to prove a goodicity. This is going to be a key property So first me let me prove one more small lemma. So f i to i piecewise affine piecewise Affine Full branch. Okay then supremum then Supremum overall omega n in Pn Of the size of the length of omega n converges to zero as n tends to infinity And why is this? Yes, but remember that the intervals decreasing is not enough to show that they decrease to zero right The intervals you can have a sequence of nested intervals decreasing but not to zero Remember we use some additional property. What do we how did we do this calculation before? The derivative is strictly bigger than one and so by the mean value theorem This is less than equal to lambda to the minus n Okay, and how do we know the derivative is bigger than one? I did not assume anything about the derivative here Excuse me f is full branch up fine. That's right Exactly, okay as long as this partition is not trivial So there's at least two elements in this partition Then both elements are smaller than the whole interval and they get mapped in an affine way to the full interval So you must have the derivative bigger than one. Okay, the more partition elements the bigger the derivative so proof Since f is is Peace wise Affine there exists lambda greater than one Such that derivative f prime of x is that an equal to lambda for all x in omega for all omega in p right and so The size of omega n is less than equal to 1 over lambda n for all n greater than 1 and for all omega in P and Okay, we have uniform bounds on how they shrink the Lebesgue measure. You don't like this Yes, thank you absolute value of the derivative Okay, so now we're ready to prove a good this city So how do we prove a good this team? Remember? What do we need to prove? So let A be some subset of I with f minus one of a Equals a so how do we prove that it's? Algorithm what do we need to prove? Zero one. So how do we prove that it's zero or one? Well, there's various ways in various situations that one standard ways the following we need to prove is a one So let's assume that it's got positive measure and Prove that it has that the measure one And that's one way to prove that I did zero one is to show that if you assume the measure is non-zero Then it has to be one Okay, so we'll assume that we have that the Lebesgue measure of A is greater than zero and now we will prove that the Lebesgue measure is equal to one Okay, so Here we have our interval Zero one Here we have our partition P and all the refinements of our partition So We assume that a has positive measure. So have you heard of Lebesgue density theorem? So Lebesgue density theorem Theorem is very interesting and quite subtle and we will not prove it here because it's a Theorem from measure theory. We cannot prove it here, but it says the following if you have a a set of measure of positive measure right then For so if measure of a is positive then For m almost every x in a We have the following property that if you take Interval x minus epsilon x plus epsilon and you look at the measure of a Inside this interval So you don't know where this set is. Okay, but you take a set a point x that belongs to a Right, and then you take a small interval here Okay, and you look at the amount of a that's in see inside this interval and you divide By the size of this interval, which is 2 epsilon this converges to 1 as epsilon goes to 0 So one way to say this is that almost every point is a Lebesgue density point of the set a Now what does it mean and why is this a little bit surprising? What it says is that if you Look at a very small region around x Then 99% of that region belongs to a Okay, if the set a is an interval, this is obvious Okay, if the set a is not an interval This is not at all obvious if it's for example a Kantor set of positive measure or somehow a more complicated set because Suppose I take a set a very strange set and I say the measure of the set is one half For example Then why should it not be that if I look at a very small scale? I always get that the measure of of that Interval of a compared to the complement is also half and half at every scale What this says is that this is not true The measure of a might be a half at a certain scale But if you focus on to a specific point then either the measure of a the more you focus the measure of a either Goes to zero or goes to one for almost every point It goes to one and for the complement for the other points It goes to zero the density of the set a Okay, so this is one of the basic theorem of measure set theory, which I acknowledge is I appreciate is not Composers really subtle actually it's not at all intuitive or obvious but we will use this in an essential way here because That's what we have here. Okay here. We have a set of positive measure and We have the Refinements of the partition So if we take almost every point with respect to Lebesgue is always in the interior of these refinements So by taking a deep enough refinement, it's like taking a sufficiently small neighborhood of the point x right so using the Lebesgue density theorem so For m almost every x in a There exists and and for all epsilon greater than zero there exists n sufficiently large such and Omega n in pn because n sufficiently large means that omega n is sufficiently small Okay, so that the Lebesgue measure of omega n intersection a over the Lebesgue measure of Omega n is greater than or equal to F So what we're doing is we're using these partitions to zoom in to very small scale And when you zoom into very small scale most of the measure Sorry, what I meant here actually is one minus epsilon what I want to write here is One minus epsilon because I wanted to be totally close to one So I'm saying that at a very small scale I can I Can focus in and I can get an interval where most of the measure of this interval belongs to a or the qu and equivalently in particular this means that the measure of Omega n Intersection the complement of a Over the measure of omega n is less than equal to epsilon Okay, this is just a complement Now, what do I know about this little omega n? So remember that I know that omega n maps to I under fn and This map is a fine Right, so here. I know that I have the complement of a inside here Okay, and so what I have is because fn Fn from omega n to I is an affine Byjection Then the measure is preserved and I have that omega n Intersection a complement over sorry Sometimes I use that and measure of omega n is precisely equal to the image of both So the measure of the image of fn of Omega n Intersection a over Measure of fn This section Compliment Okay, this of course is just the same as the measure of fn Omega n intersection a complement Over the measure of I which is equal to one and this Is by assumption less than equal to epsilon? Okay, so this is equal to one so this is less than equal to epsilon and what is this so So they call that f minus One of a equals a by assumption What's the question here? exactly here Exactly, that's what we're going to do So the reason we can do that is because f minus a equals to a so f minus one of a c equals a c Because when a set is backward invariance, so it's this complement and so f minus n of a c equals a c Okay, and so We have that Fn of Omega n intersection a c Is equal to fn of Omega n Intersection f minus n of a c and now I apply fn to both sides That fn of Omega n is the whole interval Fn of f minus n of a c is just a c Okay, so this is just equal to a c if you think about it. It's quite obvious I so the fact that f minus a c is invariant under the map means that you have a c here Okay, and some sub in this is some sub interval of this But because these maps are finely to this the part of a c that is here must map exactly to all of a c here Because this interval maps exactly to all of i so the part of this interval that belongs to a c must map to a c because a c maps to a c a maps to a Okay, so it's just that the fact that this is invariant Okay, so if you take the image of the part of a c that belongs to Omega n You apply you iterate it n times This is this maps to all of i and a c simply maps to a c and so from this Okay, and so We get that measure of a c is less than epsilon f of what? Yes You're right. It's f of minus n of a c intersected Omega n well so f Minus n of fn of a Is not this is not equal to a but if you take the pre image of a c and then you apply n You get the set itself in one direction. It works in the other direction. It doesn't Yes Yes Yes What's it? What is the difficulty here here? This equality here This equality here this one Because fn is affine fn is affine so and then affine map all the Length is preserved on an affine map the ratios of lengths are preserved Right because if this affine is stretching by a factor two hundred all the measures are stretched by a factor two hundred So the length of this is stretched is increased by a factor two hundred But the length of this is also increased by like factor two hundred So it is the ratio that is preserved It's not the numerator and the denominator. Of course, they're not preserved It's only the ratio that is preserved because f is affine And in fact, this is a key point that we will come back to because we want to generalize this theorem with removing the affine Assumption at the moment. We're using the affine assumption in a very important way So the affine means that the ratios of measures are preserved So here we have this set here This is a c intersection omega n and we have omega n Okay, now we map all of this in an affine way to this So the image of this set in here the ratio between the size of the image and the of AC of This set and this set are preserved by the affine map and here This is nothing like this is just the fact that fn of omega n is equal to the whole thing which is equal to one In this case, okay, so back to this step here We have We look at the image of this set Okay, which we which is this because AC is equal to f minus n of AC and then What is disturbing you here, you know, maybe maybe it's easier if you if we look this Let's write. Let's write this the set theoretic definition of this. Okay, this is the set of all points z such that z is in Wn and fn of Z belongs to AC This is the definition of this right so Z is in Wn and fn of Z Is an AC so when you apply fn you get AC So you're saying you need to it's not clear that this gives all of AC maybe But then this depends on this Omega n is fixed, but this is true for any omega n of course So f minus n of a of c of course has many components inside each omega n There is a small copy of AC. So this is the way to think about it. Okay is that here You have your interval I like you say here You have all these all your omega n's right and then you have your globally you have your AC Okay, which is in this set and if you take the pre-image of AC this pre-image has some compose some piece inside each AC and So it's clear that this AC will map when you map this Bijectively onto this this will map to AC because this is exactly the pre-image of AC on the left hand Okay, so Maybe it's possible that you know formally They might be necessary. Maybe that's not exactly the right way to write it when you think of the set the retic definition But the meaning is clear and the fact that it's true is clear so Because these are exactly the pre-images of AC inside each element and you map them back to AC So you're using a crucial way this property here and the conclusion is that this less than epsilon and the conclusion is that we chose epsilon arbitrary, of course Dependant we have that the n and the omega n will depend on epsilon Okay, but the bound on AC does not depend on n or omega n so for arbitrarily small epsilon We get this so this implies that the measure of AC Equals zero and this implies that the measure Equals one so we use two ingredients in the proof and it's worth spending a couple of minutes thinking about them because we will generalize this proof in the next lecture So what are the two ingredients in this very last part we used this fact to To allow us just to make this last Statement right to show that you we're estimating AC and AC has a small component inside Omega n and each of these maps to all of AC But besides that that is just part of the way we've chosen The set a and the set ac. What are the two ingredients that we used? so first of all We use this statement here. This is one ingredient Yes Sorry Exactly This is one ingredient and this is the second ingredient everything else is just you know everything else is basically just Understanding the sets and mapping and these are the really the two ingredients and what I mean by the two ingredients Is the two places in which we use the assumptions on the map? What assumption on the map? Are we using here? What property of the map are we using here? All we're using is a very weak property We're just using the fact that these will be arbitrarily small We apply the Lebesgue density theorem and to apply the Lebesgue density theorem. We need to say that for every epsilon To get this we need to say that we're in a very small neighborhood of the point x we're in a very small interval Okay, so the ingredient we use here is the fact that the size of this is going to zero okay, and In the previous lemma we showed that if F is piecewise a fine then these are going to zero But as you know these go to zero in the much more general conditions like what are the conditions? That imply that omega n goes to zero just if the derivative is bigger than one But we don't need piecewise a fine so piecewise a fine is a very strong Tool just to prove that these go to zero and so we get this Okay, so I'm we're not going to do it now But I'm just mentioning it for the future for the next lecture that it is very easy to get this part of the argument Without requiring F to be piecewise a fine all we need is to be expanding Right there to be the one and then we immediately get this part So this is very easy we can generalize we can say it's full branch C1 piecewise expanding this is more delicate This equality we certainly needs piecewise a fine if it's not piecewise a fine the ratio will not be Constant right because if it's not piecewise a fine it means even if it's expanding It might be expanding a little bit more here a little bit less here So this ratio between the image of AC and the image of the whole thing does not necessarily stay constant Okay So how do we solve this? How could we resolve this is that does the whole proof break down? Or is there something we can do if it's not piecewise a fine? It's not LeBeg invariant Exactly, that's what I'm saying. There's no way we can get this equality if it's piecewise a fine if it's not piecewise a fine We will not get the equality So the question is do we need the equality? Exactly all we need is an inequality there. We don't need an equality Piecewise a fine gives us a quality, but piecewise a fine is very strong. What do we need? We really need here just something like this and Everything else will work right because what we really want to show is that this is less than equal to epsilon Which is what gives us this? So we don't need an equality here Okay, in fact, we don't even need this we need some constant we can add some constant here Okay, as long as this constant is kind of uniform and doesn't depend on n and this we will This will be one of our main theorems in the course will be to Precisely to prove this theorem for much more general C1 Maps that are uniformly expanding and we will show this is automatic and we will study Exactly this point here will be a key point. This is called bounded distortion And it says that the map is not our fine, but is close to our fine sufficiently The distortion distortion means how different the derivative is in one place and the other is not so bad And so we still get this inequality. Okay, so this is We will come back to this I just wanted to mention this So, okay, this proves that good this city so before we go on to this generalization Again, I want to give a little application to number theory But maybe this is a very good place just to take a couple of minutes break. So let's do that Okay, so there's a very nice application of this ergodicity to number theory So do you know what the base? K expansion of a real number is remember, yes, so Let X St. Zero one K greater than equal to two integer Okay, then the base K expansion of X X is Given by X Equals X1 Over K That's X2 Okay, squared Cube okay, well X I belongs to zero K minus one. So certainly, you know when K equals 10 This is just the decimal expansion of the number Decimal expansion. So sometimes we can just write X equals Say zero point X1 X2 X3 X4 Okay base K This is just the sequence That gives the expansion of the number. Okay, so it's the usual decimal expansion, for example in the case of Base 10 and another expansion as you know, this expansion is essentially unique except for some Very special points, right? So in the decimal expansion, you know these points where you finish with 0 0 9 9 And you finish or in the other case, it's very similar the same special points So where have we encountered in the previous course? We already encountered something similar remember In the context of piecewise are fine full branch maps exactly when we did the symbolic coding right, so there's a dynamical way in which we can Study these base K expansions and I will come back to that in a second Fine So but first of all without any dynamics just from this point of view we say that so definition X is Called normal in base K if X equals Zero point X1 X2 X3 base K and and One over N of the number I from 1 to n such that X I equals J limit and tends to infinity equals 1 over K for all J So what are the normal numbers in base 10? Let's think about base 10 a second decimal Expansion, what does this definition mean? What is a? What is the definition say in the case of base 10? What are you measuring here the frequency of the digits exactly a Sympathetically in the limit the frequency of each digit should be 1 over 10 in base 10 Then the number is normal. Do they exist normal numbers? That's right. You can easily construct a normal number like that. Do they exist not normal numbers? Like what? Exactly You just have one digit if we for example you have any decimal expansion that never contains the number five or that only Contains a number zero. There's lots of non normal numbers. Okay, so what do you think is most? likely For the number to be normal or to be not normal to be abnormal you think so Anyone else thinks so? You know mathematicians will call normal. They use the word normal to you don't know if it means Normal should be normal or abnormal What the opposite should be true the opposite what there's more abnormal numbers You think it's very it's a very strong condition things a very strong condition only very few numbers Will satisfy this you think the normal numbers are very rare Why is that? Because you mean you're you're saying well if you just take a sequence of digits This frequency could be anything so what are the chances that it would be exactly one of a K? Very few chances is this your argument Anybody else have the same argument? Yes. Yes, what one yes? Yes, yes So right, so if we we use the dynamics the map Kx mod 1 and You're saying that this should allow us to show that somehow something from this converges For many numbers or for few numbers or for all numbers for many numbers Okay, we're going to see this in a second. Okay, you're along very good lines Okay, let's see in a second. Let me first There will be a dynamic we will discuss this from a dynamical system point of view. This is the interesting thing So I'm formulating the question just from the point of view of numbers and number theory But indeed the answer will come from dynamics Theory dynamics first of all, let me just ask one more question just to to to ask you the questions if a number is Normal in base 10. Do you think that it would be normal in base 9 or in some other base? You think it would be normal in base 2 What if it's normal in base 10 then it's normal in base 2 So what we're going to prove if it's the following theorem there exists a set N N in 0 1 with the Lebesgue measure of N equals 1 such that X in N implies X is normal in every base K Although I as you shall see even though the it's very surprising that N does not depend on K That's actually the easy part of the proof We'll see okay in a second. So let me remark. However, that as far as I know Nobody knows any such number So I don't think that anybody knows one number That I can tell you this number is normal in every base So this is an abstract theorem and Probably any number you choose will not belong to this set any explicit number you choose choose the number one half or Any rational number or choose the number pi or whatever number probably it will not belong to this And I don't think anyone has tried recently I think I saw some Paper in which someone was discussing this time to find some construction to find some number, but it's not It's not known. Okay, however, I think it's good that we had a little discussion about it because indeed This is quite a strong property this convergence here Okay, so let's see how we will prove it using the dynamical tools that we have so far. So first of all as you remarked It's remarkable that this set is independent of K, but Remember that if so it is enough if we can show So it is enough It is sufficient Show that for all K greater than equal to 2 there exists and K with measure of Nk equals 1 Okay, such that X in Nk Implies X Normal in base K. So why is it enough to show this? It's enough to show that for every K almost every point is Normally in base K We don't need Nk to be close because if you take the intersection of all these Nk's All we need to show that this intersection still has measure 1 and how do we know that the intersection has measure 1? No, no, no, no, you don't need no This is you don't need any of that with measure is much more simple the complement has measure 0 and The complement of the intersection will be the union of all the complements and the union of a complements of sets of measure 0 has measure 0 Okay, so we write that down indeed the measure Lebesgue measure of the interval minus N the of the interval Minus the union sorry interval minus N so let Let N N equals the union Sorry the intersection Of all the Nk K equals 2 to infinity so this is the set that we're looking for we want to show that N has measure 1 So I minus N is just equal to the Lebesgue measure of the union So the measure of the complement is just the measure of the union From K equals 2 to infinity of the complement of Nk I'd take the unions of all the complements and then you have the complements of the intersection And this is just less than equal to the sum K equals 2 to infinity of the measure of I minus Nk and this is just equal to 0 Okay, so this is the easy part so we just need to choose 1 Nk Okay, so let K greater than equal to 2 let F Zero one to zero one the F of X equals K X Not one Okay, like you said and then If you remember What is the structure here? So this is the case K equals 3 For example, if you remember we have three intervals I 0 I 1 I 2 So let I j equals the interval J over K J plus 1 Over K are these intervals here? Then you take some point here X and let X equals 0 X 1 X 2 X 3 and so on is the base K expansion of X Then where is this point? Where does this point? Exactly exactly so X belongs to I j if and only if X 1 equals J and F I of X Belongs to I j if and only if So f of X Belongs to I j or a different I j we can call it. Okay, it's still I j for any J So what is the property f of X belongs to I j if and only if X? I plus 1 equals J, which means That the frequency of digits of J in this expansion is Exactly the frequency of how many times the orbit of X belongs to I j Every time the orbit of X belongs to I j so you choose for example to I To every time the orbit of X belongs here It corresponds to the fact that there's a 2 in the base 3 expansion So the limit that we're looking for the asymptotic frequency of J inside here is Exactly in the limit the asymptotic frequency of the visits to I j and what is this limit? What is the frequency of the visits to this interval I j one of a K? Why is it one of a K? because Lebesgue measures invariant and Ergodic and by Birkhoff's ergodic theorem the frequency of the visit by the corollary remember to be closer got it there The frequency of visit converges exactly to the measure of that set for almost every point exactly So there is this is true so the frequency of J's will converge to one of a K for almost every point Well, it beg almost every point which is exactly our statement, which is what we want to prove. Okay, so Frequency of the digit J equals Frequency of visits Frequency of visits of the orbit of X to I j and this converges to one of a K for almost every X and all J and this completes the proof So there's a lot of Connections this is a result in number theory, but the proof is ergodic theory And in fact, there's quite a lot of even recently. There's been a lot of research in Ergodic theory methods in number theory to prove certain things in number and this is just one example Okay, so next time we are going to Remove the piecewise affine assumption and we're going to study full branch maps Without the piecewise affine assumptions and we're going to study the ergodicity of Lebesgue measure in those cases Okay. Thank you