Ergodic decomposition

Remarks

For part 1, suppose $\mu$ is not ergodic, and let $A$ be an invariant set with $0 < \mu(A) < 1$. Then $\mu$ is the convex combination $\mu(A) \mu_A + (1 - \mu(A)) \mu_{X \setminus A}$ of the normalized restrictions, which are distinct invariant measures, so $\mu$ is not extreme. Conversely, if $\mu$ is ergodic and $\mu = t \nu_1 + (1-t) \nu_2$ with $0 < t < 1$, then $\nu_1$ is absolutely continuous with respect to $\mu$ with an invariant density, which ergodicity forces to be constant, so $\nu_1 = \mu$. Part 2 follows from Choquet theory, since $M(X, T)$ is a compact convex metrizable set whose barycentric decompositions are unique. The decomposition can also be built from the conditional measures of $\mu$ on the σ-algebra of invariant sets. Example: for the identity map, every Borel probability measure is invariant, and the ergodic ones are the point masses $\delta_x$. The decomposition of $\mu$ is $\mu = \int \delta_x \, d\mu(x)$, that is, $\tau$ is the image of $\mu$ under $x \mapsto \delta_x$. Example: for the rotation of the circle by a quarter turn, Lebesgue measure is the average over $x \in [0, 1/4)$ of the uniform measures on the four-point orbits of $x$.

Going deeper

For a student

Many theorems are first proved for ergodic systems, where time averages are constant, and then extended to all systems by the ergodic decomposition. For example, Birkhoff's ergodic theorem for a non-ergodic measure gives a limit that is constant on each ergodic component. Two distinct ergodic measures are mutually singular. By part 1, the unique invariant measure of a uniquely ergodic map is ergodic. That $M(X, T)$ is not empty is the Krylov–Bogolyubov theorem.