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Consider the following theorem about equivalent formulations of ergodicity. Let $S$ be a measure preserving map on a measure space $(\Omega,\mathfrak F,\mathbb P)$ and define

$$\nu_n(A,\omega) = \sum_{k=0}^{n-1} \mathbb I_{\{S^k\omega \in A\}}$$

The following statements are equivalent:

(1) $S$ is ergodic, i.e. $\lim_{n\mapsto \infty} \frac{\nu_n(A)}{n} = \mathbb P(A) \text{ a.s.} \forall A\in \mathfrak F$

(2) S is metric transitive, i.e. for all invariant events $A\in \mathfrak F$ holds $\mathbb P (A)\in \{0,1\}$

(3) $\forall f\in L^1(\mathbb P ) : f^* = \text{const. a.s.}$, where $f^*$ is the limit $\frac{1}{n}\sum_{k=0}^{n-1} f(S^k\omega) \rightarrow f^*(\omega) \text{ a.s.}$

Which of these conditions is easiest to work with to proof ergodicity practically, i.e. for an explicitly given map $S$ and a measure space?

For example, which one would easier to apply to give conditions for when the map $$S(x)=(x+a)\mod b$$ on a measure space $$([0,b),\mathfrak B([0,b)),\text{Uniform}([0,b))), a,b>0$$ is ergodic?

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  • $\begingroup$ I'm not sure I'm not missing anything, but I think the proof in your example follows the same ideas as the discussion of ergodicity in the case of rotations on the torus. $\endgroup$ Mar 26, 2013 at 19:53

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I would add a condition

(4) If $f$ is measurable and $f(Sx)=f(x)$ almost everywhere, then $f$ is actually almost everywhere constant.

In the particular case mentioned in the OP, we are reduced to determine ergodicity of rotations on the torus (with angle $\frac ab$). And condition (4) seems to be the easier in this case.

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