invariant measures in topological dynamical systems

Suppose that X is a compact topological space(we first assume that X has enough good properties), and a dynamical system defined on X is just a transformation T: X\rightarrow X, then the importance of the existence of a measure invariant on X by T is reflected in the Poincare’s recurrence theorem: if there is a Borel measure \mu on X compatible with the topology of X, what is more, \mu is T-invariant, that is to say \mu(A)=\mu(T(A)) for all measurable set A of X, then almost all points x\in X is recurrent, that is to say, for any neighborhood U of x, the intersection of the set \{T^n(x)|n\in \mathbb{N}\} with U is infinite.

The importance of this result comes probably from statistics physics. In a system of many particles(the X is the phase space of this system), the Boltzmann measure on the phase space is invariant under the evolution of the system(which is a family of transformation parameterized by time t). So we have that almost all states are recurrent, given the time. So a thermodynamic system kicks off, then we are sure that for some time in the future, the state of the system is as close to the original state as we want. But, as what everyone learns from even elementary statistical physics, an isolated thermodynamic system will eventually go into an equilibrium state. This is a famous paradox. The key point to the paradox is that we don’t know how much time it will take to get a system to its original state. We call this time \tau. Then in general \tau is proportional to the number of possible macro states of the system, \tau \propto N, while the latter is proportional to e^n where n is the number of possible micro states of the system. But according to the Boltzmann’s entropy definition, the number of micro states of a system is proportional to exponent of the entropy of the system. So in fact we get \tau\propto e^{e^S} where S is the entropy of the system. So in fact this \tau is usually very very large.

In this example, the utilization of Poincare’s theorem leads to a paradox, which is interesting. But in general, this theorem is itself very useful.

But there is a problem with this theorem, that is it has to suppose that this transformation T does have an invariant measure. Which, in fact, is usually hard to find, not to mention that sometimes this invariant measure can be very different from our routine one, for example the Lebesgue measure.

But one surprising theorem of Krylov-Bogolioubov says that under some natural conditions on the topological space X and the transformation T, we can ensure the existence of an invariant measure. Here we will try to prove one of its forms as follows:

If X is a compact metric space, and T:X\rightarrow X is continuous, then the set of invariant probability Borel measures P(T) is not empty.

There are various Borel measures on X, but which ones are particular to use? Perhaps those that can be written down explicitly. Which kind of Borel measure can we define on any space? If we don’t have the Borel condition, then we can construct our own \sigma-algebra, and the easiest one is of course the trivial one, set the \sigma-algebra to be \{\emptyset, X\}, and define \mu(X)=1, \mu(\emptyset)=0. But the sad thing is that there is a Borel obstruction. Then perhaps an idea comes to our mind, why not try the Dirac measure? Indeed, for any space X, we can choose a point x_0\in X, and naturally define \mu=\delta_{x_0}, which is compatible with any topology possibly put on X. So now we have a good example in mind, then we can take a look at the action of T on this Dirac measure, by this we mean that we define (T_{*}\nu)(A)=\nu(T^{-1}(A))=\nu(\chi_{A}\circ T) where \nu is a Borel measure on X, A is a measurable set of X, and \chi_{A} is the characteristic function of A, and we write the integral \int f\text{d}\nu=\nu(f). Therefore we define \nu_n=T^n_{*}\delta_{x_0}. There is little reason that this sequence of measures should converge, even weakly converge(measures can be viewed as linear functionals on the continuous functions on X with the L^{\infty}, thus a sequence of measures converge means that convergence in the norm sense, while weak convergence means weak convergence of the linear functionals). But, as we usually do, we can take the mean of these measures. that is, we can define, \mu_n=\frac{1}{n}\sum_{i=1}^n\nu_n. This is an important step. Note that when we apply T to \mu_n, we see that T_*\mu_n=\mu_n+\frac{\nu_{n+1}-\nu_1}{n}. It is obvious that the second term on the right hand side converge to the zero measure, so T_*\mu_n is in fact very close to \mu_n. Then if, by any chance, \mu_n (weakly) converge, then we will surely have (we will prove this point) that the limit measure is invariant.

Since X is a compact metric space, we know that the space of continuous functions C(X) with the L^{\infty} norm is separable. In fact, we can choose a basis \mathfrak{B} for the topology \mathfrak{T} on X such that \mathfrak{B} is a countable set and each element in it has compact closure, this is possible since X is compact. For each element B\in \mathfrak{B}, we can construct a countable set of functions approximating \chi_B from above and from below. Then since simple functions (linear combination of finite characteristic functions) can be used to approximate any continuous function, then we get that C(X) is separable. This done, we can now show that the unit ball of C(X)^* is in fact weakly compact. In fact, this is a result from functional analysis. If \nu_n is a sequence of continuous linear functionals in the unit ball, suppose that f_n is a sequence of elements in C(X) which are dense in C(X). Then for f_1, the sequence \nu_n(f_1) is bounded, thus it exists a subsequence \nu_{g_1(n)}(f_1) which converge. Now consider \nu_{g(n)}(f_2), which is again a bounded sequence, thus we can get a sub-subsequence \nu_{g_2(g_1(n))}(f_2) so that it converge. We can continue this process, and finally we can set g(n)=g_n(g_{n-1}...(g_1(n)...)), so now \nu_{g(n)}(f_k) converge for any fixed k. So we define \nu(f_k)=\lim_n\nu_{g(n)}(f_k). And for any f\in C(X), we can, using the density of \{f_k\} define \nu(f), which can be shown to be well defined. Thus we define a limit measure $\latex \nu$ for \nu_{g(n)}, hence we showed the weak compacity of the unit ball of C(X)^*.

So now continue our reasoning, having defined \mu_n, noting that these measures are already in the unit ball of C(X)^*, thus has a subsequence, \mu_{g(n)} which weakly converge to a measure \mu. To show that T_*\mu_{g(n)} weakly converge to $\latex T_*\mu$, we have to show that for all continuous function f\in C(X), we must have that \mu_{g(n)}(f\circ T)\rightarrow \mu(f\circ T). This is obvious since T is continuous, so is f\circ T(the continuity of T) is crucial. But, as we have remarked, T_*\mu_{g(n)} is in fact close to \mu_{g(n)}, so we have that T_*\mu=\mu, which shows that \mu is an invariant measure on X. Thus we complete the proof for the above result.

There are several questions to ask. The first is whether P(T) has only one element, in other words, whether T_*: C(X)^*\rightarrow C(X)* has only one fixed point. The second question is whether all invariant Borel measures is of the form \mu=\lim_n \mu_{g(n)}(weak convergence). For the second question, there is a related concept, that is generic points. A point x\in X is a generic point for a measure \nu under the transformation T if \nu=\lim_n\frac{1}{n}(\sum_{i=1}^nT^n_*\delta_x)(the weak convergence). So we can ask if an invariant measure can have a generic point. Note that in general the answer is no. For example, consider the unit circle, X=[0,1]/(0\equiv1), and T: X\rightarrow X, x\mapsto -x. Then we can see that \nu=\frac{1}{4}(\delta_0+\delta_{1/4}+\delta_{2/4}+\delta_{3/4}) is an invariant measure. But we can not find a point x\in X such that \nu=\lim_n \frac{1}{n}\sum_{i=1}^nT^i_*\delta_x. So in the set of invariant measures, there are some measures more special than others.

This leads to another important concept in dynamical systems, the ergodicity. There is a result which roughly says that if the invariant measure \mu can’t be decomposed into a non-trivial convex linear combination of another two invariant measures, then the only measurable set invariant under T is either of measure 0 or of measure 1(we always make the measures normalized such that the measure of X is 1). We can deduce from this that if \mu is ergodic, then it has generic points(vice versa, and it is simple to see).

The uniqueness question leads to the concept of unique ergodicity, which means that \#P(T)=1. For this concept, we have another equivalent description: for any point x\in X, any continuous function f\in C(X), the limit \lim_n\frac{1}{n}\sum_{i=1}^nf(T^i(x)) is independent of x. We will not prove this result this time, perhaps later we will give a proof of it.

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