Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 25J (Probability and Measure)

A measurable set is invariant when ${\theta }^{-1}A=A$, or modulo a null set in the measure-theoretic version. A measurable function is invariant when $f\circ \theta =f$, similarly almost everywhere if appropriate. The map is a measure-preserving transformation when $\mu ({\theta }^{-1}A)=\mu (A)$ for every measurable set. A measure-preserving map is ergodic when every invariant measurable set has measure zero or has null complement. On a finite positive measure space, normalize the measure to obtain the equivalent probability-space definition.

As printed, ${\theta }_1(x)=1+x$ is not a self-map of $[0,1]$, so it is not an ergodic transformation of the specified space. No reduction modulo one is stated; if it were imposed it would instead be the identity, also nonergodic.

The map ${\theta }_2(x)=x^2$ is not Lebesgue measure-preserving: the inverse image of $[0,1/4]$ is $[0,1/2]$, of a different measure. It also has nontrivial invariant sets if ergodicity is defined using invariant sets alone. For example,is invariant under squaring because the logarithm inside the fractional part increases by one. Both it and its complement contain intervals of positive length, so their measures are positive. Thus ${\theta }_2$ is not ergodic under either convention.

Reflection ${\theta }_3(x)=1-x$ preserves Lebesgue measure, but $[0,1/4]\cup [3/4,1]$ is invariant of measure $1/2$, so it is not ergodic. Finally ${\theta }_4$ exchanges the two equally weighted atoms. The only invariant subsets are the empty and full sets, hence ${\theta }_4$ is ergodic despite having period two.

The Birkhoff ergodic theorem says that for a measure-preserving transformation on a finite measure space and $f\in L^1$, the averages $A_{n}f=n^{-1}{\sum }_{k=0}^{n-1}f\circ {\theta }^k$ converge almost everywhere to an invariant integrable function $f^{\ast }$. This limit is the conditional expectation of $f$ given the invariant sigma-algebra, and $\int f^{\ast }d\mu =\int fd\mu$. If the transformation is ergodic and $\mu (E)>0$, then $f^{\ast }$ equals the constant $\mu (E{)}^{-1}\int fd\mu$ almost everywhere.

Since the measure is finite and $f$ is bounded, $f\in L^1$. By Birkhoff ergodic theorem, $A_{n}f\to f^{\ast }$ almost everywhere. If $|f|\le M$, then $|A_{n}f|\le M$ and therefore $|f^{\ast }|\le M$ almost everywhere. For any finite $p\ge 1$,and this constant is integrable because $\mu (E)<\infty$. The dominated convergence theorem gives $\int |A_{n}f-f^{\ast }{|}^{p}d\mu \to 0$. Thus the ergodic averages converge in every $L^p$, $1\le p<\infty$, without assuming ergodicity.