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In Walters' An Introduction to Ergodic Theory on page 34 the Birkhoff Ergodic Theorem is given as follows:

Suppose $T\colon (X,\mathfrak{B},m)\to (X,\mathfrak{B},m)$ is measure-preserving (where we allow $(X,\mathfrak{B},m)$ to be $\sigma$-finite) and $f\in L^1(m)$. Then $(1/n)\sum_{i=0}^{n-1}f(T^i(x))$ converges a.e. to a function $f^*\in L^1(m)$. Also $f^*\circ T=f^*$ a.e. and if $m(X)<\infty$, then $\int f^*\, dm=\int f\, dm$.

Then (before proving the theorem), Walters gives some remarks to this, namely:

If $T$ is ergodic then $f^*$ is constant a.e. and so if $m(X)<\infty$ $f^*=(1/m(X))\int f\, dm$ a.e. If $(X,\mathfrak{B},m)$ is a probability space and $T$ is ergodic we have $\forall f\in L^1(m)\lim_{n\to\infty}(1/n)\sum_{i=0}^{n-1}f(T^i(x))=\int f\, dm$ a.e.

So far so good. I do understand this. Then some applications are given. And there is one application I do not understand right now. Namely:

Let $T$ be a measure-preserving transformation of the probability space $(X,\mathfrak{B},m)$ and let $f\in L^1(m)$. We define the time mean of $f$ at $x$ to be $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}f(T^i(x)) $$ if the limit exists. The phase or space mean of $f$ is defined to be $$ \int_X f(x)\, dm. $$ The ergodic theorem implies these means are equal a.e. for all $f\in L^1(m)$ iff $T$ is ergodic. Since these two means areequated in some arguments in statistical mechanics it is important to verify ergodicity for certain transformations arising in physics. That application to time means and space means is more realistic in the case of a 1-parameter flow $\left\{T_t\right\}_{t\in\mathbb{R}}$ of measure-preserving transformations. The ergodic theorem then asserts $$ \lim_{T\to\infty}(1/T)\int_0^T f(T_tx)\, dt~~~~~~~(*) $$ exists a.e. for $f\in L^1(m)$ and equals $$ \int_X f\, dm $$ if the flow $\left\{T_t\right\}$ is ergodic and $(X,\mathfrak{B},m)$ is a probability space.

I have two questions to this cited application.

1.) Why is $\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}f(T^i(x))$ called the time mean of $f$ at $x$? And why is $\int_X f(x)\, dm$ called the phase or space mean of $f$?

2.) I do not see why the above cited ergodic theorem asserts (*). Can you please explain that to me? I do not know how Walters could mean that. I am totally helpless...

  • $T$ is a dynamical system that describes the progression of a point through a phase space $X$. Your first formula in question 1) is finding the average over time. Your second equation of 1) is averaging over the phase space. – David Simmons Aug 19 '14 at 12:15
  • In which way is T a dynamical system, please? –  Aug 19 '14 at 12:20
  • $T^i(x) = T\circ T\circ\cdots \circ T (x)$ (i times). Thus, $T^i$ gives the position of a point within a state space $X$ at (discrete) time $i$. – David Simmons Aug 19 '14 at 12:24
  • A dynamical system is just a mapping $\mathbb{T}\times X \to X$, where $\mathbb T$ is the 'time' space and $X$ is the phase space. – David Simmons Aug 19 '14 at 12:26
  • We defined a dynamical system as follows: A dynamical system consists of a set $M$, a group $(G,)$ and a map $\Phi\colon G\times M\to M$ with the properties (1) For all $x\in M$ iti s $\Phi(0,x)=x$ and (2) For all $x\in M$ and all $s,t\in G$ it is $\Phi(s,\Phi(t,x))=\Phi(s t,x)$. Why does this fit here in order to say: $T$ is a dynamical system? –  Aug 19 '14 at 12:29
  • I think it would be more appropriate to say that $\Phi_T\colon\mathbb{N}_0\times X\to X, \Phi_T(k,x):=T^k(x)$ is the map that belongs to the dynamical system. –  Aug 19 '14 at 12:39
  • I don't think the group property is necessary; however, the semigroup property is. Define $\hat T(i,x) = T^i(x)$. With $T^0$ defined as the identity mapping on $X$, (1) is satisfied. The cocycle property should also hold. Thus, $\hat T$ defines a dynamical system. – David Simmons Aug 19 '14 at 12:41
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    You are right. My original comment is not correct. Instead, $T$ canonically defines a dynamical system on the phase space $X$ ... – David Simmons Aug 19 '14 at 12:45
  • Okay, now I can understand why the first formula in question 1) is called the mean of time of $f$ at $x$. One looks at the states of x at the discrete time states and which value f does have in this time states. Then one builds the mean value. But in which sense is the integral $\int_X f(x), dm$ the mean over the phase space? –  Aug 19 '14 at 12:45
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    In the same way that if a random variable $X$ is defined on a probability space $(\Omega,\Sigma,\mathbb P)$, the expectation is given by $E[X] = \intop_{\Omega} X(\omega) d\mathbb{P}$, $\omega\in\Omega$. – David Simmons Aug 19 '14 at 12:57
  • Sorry, I do not see why for the phase mean I need the expectation value. –  Aug 19 '14 at 13:48
  • I'm not saying we need anything, I'm showing you the form of the expectation as a Lebesgue integral. It's the same as your integral. Thus, your integral represents the spatial average. Spatial because $X$ represents the phase space. – David Simmons Aug 19 '14 at 14:01
  • By the way: What is meant with $\left{T_t\right}{t\in\mathbb{R}}$? Is that the set $\left{X\to X, x\mapsto \Phi(t+t_0,x)\right}{t\in\mathbb{R}}$ where $\Phi\colon \mathbb{R}\times X\to X$ is the map that belongs to the dynamical system $(X,\mathbb{R},\Phi)$? –  Aug 19 '14 at 17:23
  • Yes, I think so. I'm not sure where you have gotten $t_0$ from though. – David Simmons Aug 19 '14 at 18:05
  • I think $t_0$ is just the starting time, and I think it shall be $t_0=0$. Btw: Do you have an indea concerning my second question? –  Aug 19 '14 at 22:43
  • You are correct that the ergodic theorem as you have it in front of you does not say anything about ergodic flows. But via abuse of terminology, we call a number of related results about integral equalities involving measure-preserving transformations "the ergodic theorem". The proof of the result with ergodic flows is similar to the proof in discrete time, once you exploit the semigroup property in an appropriate way. – Ian Aug 20 '14 at 12:13
  • Can you please explain that to me? I am new in this area and do not understand exactly what you mean and how this can explain the property (*). –  Aug 20 '14 at 12:16
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    If we assume $t \to f(T_t x)$ is Riemann integrable, then we can rewrite $\lim_{T \to \infty} 1/T \int_0^T f(T_t x) dt$ as $\lim_{T \to \infty} \lim_{m \to \infty} (1/m) \sum_{k=1}^m f(T_{T/m} x)$. If we assume $T$ is an integer, then we get (*) from the discrete time ergodic theorem. The correct proof doesn't exploit Riemann integrability, of course, but this should supply some intuition. – Ian Aug 20 '14 at 12:23
  • Do you have a link to the "discrete time ergodic theory"? I cannot find it. –  Aug 20 '14 at 12:24
  • The theorem at the top of your post could be called the "discrete time ergodic theorem", and this is what I mean. – Ian Aug 20 '14 at 12:25
  • Oh, I do not understand it then how you get (*). Why can we rewrite it in the way you did. Maybe we switch to chat if you prefer. –  Aug 20 '14 at 12:29
  • I just rewrote $\int_0^T f(T_t x) dt$ as a limit of Riemann sums (and canceled $T/T$). I also made a typo: it should have been $\lim_{T \to \infty} \lim_{m \to \infty} \sum_{k=1}^m f(T_{Tk/m} x)$. That is, we are just sampling $f(T_t x)$ at $m$ evenly spaced points of $[0,T]$, averaging the result, then sending $m \to \infty$, then sending $T \to \infty$. – Ian Aug 20 '14 at 12:32
  • Without Riemann: Writing the integral as the limit of integrals over simple functions? And assuming that f is real-valued without loss of generality because if it is complex-valued then writing it as two integrals for real valued functions? –  Aug 20 '14 at 12:34
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    Yes, I think a simple function argument will do it. However, I haven't written out the details of this argument before. My intuition for the extension from discrete to continuous time is mostly physical, not mathematical. – Ian Aug 20 '14 at 12:35
  • Please have a look on my own answer I posted. Would be great to hear whether I got it. –  Aug 20 '14 at 14:17

2 Answers2

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Let's see if I got it.

Because there is apperearing the Riemann-integral $\int_0^T f(T_tx)\, dt$ I think it is indeed okay to assume that $t\mapsto f\circ T_t$ is Riemann-integrable.

So one way to write the Riemann-integral is $$ \int_0^T f(T_tx)\, dt=\lim_{\Delta_P\to 0}\sum_{k=1}^{m}f(T_{\tau_k}x)(t_k-t_{k-1}), $$ where $0=t_0<\ldots <t_m=T$ is a partition of the intervall $[0,T]$, $\tau_k\in [t_k-t_{k-1}]$ and $\Delta_P:=\max_{k=1,m}\lvert t_k-t_{k-1}\rvert$.

(By the way is not the best idea to use $T$ for the map and the integer but now I stick to that.)

Another way to write the Riemann-integral is to choose the special partition $$ t_0=0, t_k=\frac{Tk}{m}, 1\leq k\leq m $$ to choose $\tau_k=t_k, 1\leq k\leq m$ (so $\Delta_P=\frac{T}{m}$) and then considering the limes $m\to 0$. This is the same.

So consider $$ \int_0^{T}f(T_tx)\, dt=\lim_{m\to\infty}\sum_{k=1}^{m}f(T_{Tk/m}x)\underbrace{(t_k-t_{k-1})}_{=T/m}=\lim_{m\to\infty}\frac{T}{m}\sum_{k=1}^{m}f(T_{Tk/m}x) $$

We can write this as $$ \lim_{m\to\infty}\frac{T}{m}\sum_{k=1}^{m}f(T_{Tk/m}x)=\lim_{m\to\infty}\frac{T}{m}\sum_{k=1}^{m}f(T^k_{T/m}x)=\lim_{m\to\infty}\frac{T}{m}\left(\sum_{k=0}^{m-1}f(T^k_{T/m}x)+f(T^m_{T/m}x)-f(x)\right)\\=\lim_{m\to\infty}\frac{T}{m}\sum_{k=0}^{m-1}f(T^k_{T/m}x)+\underbrace{\lim_{m\to\infty}\frac{Tf(T^m_{T/m}x)}{m}}_{=0}+\underbrace{\lim_{m\to\infty}\frac{Tf(x)}{m}}_{=0}\\=T\lim_{m\to\infty}\frac{1}{m}\sum_{k=0}^{m-1}f(T^k_{T/m}x)=Tf^*(x)\text{a.e.} $$ after the cited ergodic theorem. So it is $$ \lim_{T\to\infty}\frac{1}{T}\int_{0}^T f(T_tx)\, dt=\lim_{T\to\infty}\frac{1}{T}Tf^*(x)=\lim_{T\to\infty}f^*(x)=f^*(x)\text{ a.e.}, $$ i.e. the limits exists a.e. as Walters says.

Am I right?

With greetings and many thanks for your help.

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First, you should understand the statement of the theorem in order to understand the difference between the two means: Let $B$ be a set of the sigma algebra, the Birkhoff's mean measures time proportion of the orbit of a point $x$ between instant $0$ and $n-1$. When $n$ becomes big enough, this sum may ( in probabilistic way) be close to $P(B)$ ( $P(B)$ denotes the probability to be in $B$) in a given instant $n$ in that case $P(B)=1$ (your second statement), if it's the same case with any Borel set B, then the spatial mean ( phase, space..) and time mean coincidence, and our system is ergodic.

As an application you can consider the flow of the circle rotation Ra(x)=x+a mod1. The Dynamics associated to this application depend on rationality of the angle a, if the angle is irrational, the rotation is ergodic. If it's rational the periodicity of it's orbits impede the ergodicity of the system.

BallBoy
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