Equivalent invariant-set and invariant-function criteria for ergodicity

If the system is ergodic, a set in ${\mathcal{I}}^{\mathrm{\prime }}$ has by F2 a strict invariant representative differing by a null set, hence has measure zero or one. Conversely (ii) applies to strictly invariant sets. Thus (i) and (ii) are equivalent.

Assume (ii) and let $f:X\to \mathbb{R}$ be measurable and a.e. invariant. For $n\ge 1$, $k\in \mathbb{Z}$, let $E_{n,k}=f^{-1}([k\mathrm{/}n,(k+1)\mathrm{/}n))$. Measurability follows from F3. Its pullback differs from it only where $f\circ T\mathrm{\ne }f$, so it is in ${\mathcal{I}}^{\mathrm{\prime }}$. For each n these fibers partition X. Their measures are zero or one; countable subadditivity excludes all zero, and disjointness and total mass one exclude two fibers of measure one. There is therefore a unique k(n) with $\mu (E_{n,k(n)})=1$.

The set $Y={\bigcap }_{n\ge 1}{E}_{n,k(n)}$ is conull by countable subadditivity. It is nonempty since $\mu (Y)=1$. Fix one $x_0\in Y$. For any $x\in Y$, $\mathrm{\mid }f(x)-f(x_0)\mathrm{\mid }<1\mathrm{/}n$ for every n, so $f(x)=f(x_0)$ by the Archimedean property of the real numbers. Thus (ii) implies (iv). The uniquely determined k(n) require no countable choice.

For a complex a.e. invariant f, its real and imaginary parts are measurable and a.e. invariant by F5. Apply the preceding argument to both, and intersect the two conull sets; f is constant there. This proves both real and complex versions of (iv), and each implies the corresponding version of (iii).

If either version of (iii) holds and E is strictly invariant, then ${\mathbf{1}}_E$ is an everywhere invariant measurable function. A constant indicator on a conull nonempty set must have constant value zero or one, so E is null or conull. Thus (iii) implies (i). Also (iv) directly implies (ii) by the same argument applied to an indicator invariant a.e. All listed implications are now closed.