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Space Average: If μ(X) is finite and nonzero, we can consider the space or phase average of ƒ:

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But is it comparable to the time average?

Shouldn't the space average of f be compiled for a particular time, seeing that f is dependent on t (or T^k)? Or would the space average of f be equal for all values of time t?

(hmm, f is integrated and averaged across time, and then f is integrated and averaged across space...)

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up vote 2 down vote accepted

$f$ is just a function $X \rightarrow X$, there is no dependence on time. The "time" comes from composing $f$ with iterations of $T$. The strong ergodic theorem (as described in the Wikipedia article) says that if $(T,\mu)$ is an ergodic system, then $$ \frac{1}{\mu(X)} \int f \, d\mu = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} (f \circ T^k)(x),$$ for $\mu$-almost-every $x \in X$. So the LHS is the space average of $f$, and the RHS is the time-average of the maps $f \circ T^k$ applied to the point $x$. So, in other words, we are looking at the orbit $\{f(x), f(Tx), f(T^2x), \ldots\}$, and for almost every $x$, averaging these points is simply equal to averaging the values of $f$ across the whole domain $X$. Intuitively, this means that ergodicity tells you that the orbit of $x$ under $T$ visits everywhere in $X$ in some uniform fashion.

Please comment if you have further questions.

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