Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 16 / 5

We first prove invariance, before using any assertion that requires it. Write $A_n\phi (x)=n^{-1}{\sum }_{k=0}^{n-1}\phi (f^{k}x)$. Choose a point whose forward orbit has the prescribed convergence against every continuous function; the hypothesis gives a set of such points of full measure. For each continuous function $\phi$, telescoping givesbecause a continuous function on a compact metric space is bounded. Both test functions are continuous, so their averages converge to their prescribed integrals. Hencefor every continuous function $\phi$. The continuous-test criterion for an invariant measure therefore gives $f_{\ast }\mu =\mu$. The points in the hypothesis can now be called generic points for an invariant measure.

For ergodicity, let $B$ be a Borel set invariant modulo null sets, and set $h=1_B$, $p=\mu (B)$. Then $h\circ f=h$ almost everywhere. Since invariance preserves null preimages, deleting the union of all iterated preimages of the exceptional null set gives $h(f^{k}x)=h(x)$ for every $k\ge 0$ on one full-measure set. Thus $A_{n}h=h$ almost everywhere for every $n$.

Given $ϵ>0$, the continuous approximation of Borel indicators on a compact metric space, proved in the supporting steps below, supplies $\phi \in C(X)$ with $0\le \phi \le 1$ and $\Vert h-\phi {\Vert }_{L^1(\mu )}<ϵ$. By the hypothesis, $A_n\phi \to \int \phi d\mu$ almost everywhere. Since $0\le A_n\phi \le 1$, the dominated convergence theorem also gives convergence in $L^1(\mu )$. Invariance and the triangle inequality give the uniform estimateConsequently,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\Vert h-p{\Vert }_1=0$. Explicitly,so $p\in \{0,1\}$. Every invariant Borel set has measure zero or one, which provesThis proof controls measurable-set averages through a uniform $L^1$ estimate; convergence only against continuous test functions is not silently assumed to include indicator functions.

Here is an alternative route for the ergodicity step using the Birkhoff ergodic theorem. After invariance has been proved, the theorem identifies the almost-everywhere limit of $A_n\phi$ with ${\mathbb{E}}_{\mu }[\phi \mid \mathcal{I}]$. The hypothesis therefore givesfor every continuous function $\phi$. For an invariant Borel set $B$, approximate $h=1_B$ in $L^1$ by such a $\phi$. Since $h$ is measurable with respect to the invariant sigma-algebra, its conditional expectation is $h$ itself. The L1 contraction of conditional expectation givesArbitrarily accurate continuous approximation forces $h=\mu (B)$ almost everywhere. Thus this use of the theorem reaches the same zero-or-one criterion as the direct averaging proof above, without assuming in advance that $\mu$ is ergodic.

For a Borel set $B$ and $ϵ>0$, inner regularity supplies a compact set $K\subseteq B$ with $\mu (B)-\mu (K)<ϵ/2$. Independently, outer regularity supplies an open set $U\supseteq B$ with $\mu (U)-\mu (B)<ϵ/2$. Because $K\subseteq U$ and the measure is finite,This creates an arbitrarily small measurable transition region between a compact inner approximation and an open outer approximation. The cutoff in the next step then approximates $1_B$, rather than merely approximating the number $\mu (B)$.

For the compact set $K\subseteq B\subseteq U$ chosen in the preceding step, the Urysohn lemma gives a continuous function $0\le \phi \le 1$ equal to one on $K$ and zero on $X\setminus U$. In a metric space this cutoff can also be written explicitly, when $K$ and $X\setminus U$ are both nonempty:The two sets are disjoint and closed, so the denominator never vanishes, and their distance functions are continuous. If $K$ is empty, take $\phi =0$; if $U=X$ and $K$ is nonempty, take $\phi =1$. These choices satisfy the required boundary conditions in the exceptional cases.

The difference $\phi -1_B$ vanishes on $K$ and outside $U$, and its absolute value is at most one everywhere. ThereforeThis is the precise continuous approximation required by both proofs of ergodicity, and also gives $|\int \phi d\mu -\mu (B)|<ϵ$.