The ergodic theorem

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In this section we will prove the pointwise ergodic theorem for ergodic transformations. A crucial ingredient in its proof is the maximal ergodic theorem, which is concerned with the maximum value $${\mathsf{M}}_f(x)=sup\{{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^{n}x):N\in \mathbb{N}\}$$ of the ergodic averages of a function $f:X\to \mathbb{R}$.

Theorem (Maximal ergodic theorem)

Let $(X,\mathcal{B},\mu )$ be a probability space. Fix a measure-preserving map $T:X\to X$ and fix $f:X\to \mathbb{R}$ measurable and integrable. Then $$\int 1_{\{{\mathsf{M}}_f>\lambda \}}\cdot (f-\lambda )\mathsf{d}\mu \ge 0$$ for every $\lambda \in \mathbb{R}$.

Proof:

Fix $\lambda \in \mathbb{R}$. We will first prove this in the case that $f$ is bounded, say $|f(x)|\le K$ for all $x\in X$. Fix $R\in \mathbb{N}$ and write $${\mathsf{M}}_{f,R}(x)=max\{{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^{n}x):1\le N\le R\}$$ for all $x\in X$. Put $$E(R)=\{x\in X:{\mathsf{M}}_{f,R}(x)\ge \lambda \}$$ and let us analyse for each $x\in X$ the sum $$\sum _{n=0}^{P-1}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )$$ for $P\in \mathbb{N}$ which we imagine is much larger than $N$. Whenever $T^n(x)$ is outside $E(R)$ the summand has the value zero. If $T^j(x)$ is not in $E(R)$ and $T^{j+1}(x)$ is in $E(R)$ then we must have $$f(T^{j+1}x)+\cdots +f(T^{j+r}x)\ge r\lambda$$ for some $1\le r\le R$ by definition of ${\mathsf{M}}_{f,R}$. But this means that the portion $$\sum _{n=j+1}^{j+r}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )$$ of the above sum is non-negative. Indeed $$1_{E(R)}(x)\cdot (f(x)-\lambda )\ge f(x)-\lambda$$ holds for all $x\in X$ because points $x$ outwith $E(R)$ satisfy $f(x)\le \lambda$.

The long sum $$\sum _{n=0}^{P-1}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )$$ is therefore no smaller than $$\sum _{n=P-R}^{P-1}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )$$ because any good portion has length at most $R$ and any full portion has a non-negative summation. Now $$|\sum _{n=P-R}^{P-1}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )|\le R(K+\lambda )$$ so $$\sum _{n=0}^{P-1}{1}_{E(R)}(T^{n}x)\cdot (f(T^{n}x)-\lambda )\ge -R(K+\lambda )$$ and we can integrate both sides to get $$-R(K+\lambda )\le P\int 1_{E(R)}\cdot (f-\lambda )\mathsf{d}\mu$$ because $\mu$ is $T$ invariant. Dividing by $P$ and noting that $P\in \mathbb{N}$ was arbitrary gives $$0\le \int 1_{E(R)}\cdot (f-\lambda )\mathsf{d}\mu$$ and we may then apply the dominated convergence theorem to take the limit as $R\to \mathrm{\infty }$ obtaining $$0\le \int 1_{\{{\mathsf{M}}_f>\lambda \}}\cdot (f-\lambda )\mathsf{d}\mu$$ as desired.

It remains to deduce the result without the assumption that $f$ is bounded. Given any measurable and integrable function $f:X\to \mathbb{R}$ define for each $K\in \mathbb{N}$ the function $${\varphi }_K=f\cdot 1_{Q(K)}$$ where $Q(K)=\{x\in X:|f(x)|\le K\}$. The sequence ${\varphi }_K(x)$ converges pointwise to $f(x)$ and one can apply the dominated convergence theorem to finish. ▮

Theorem (Pointwise ergodic theorem)

Let $(X,\mathcal{B},\mu )$ be a probability space. Fix a measure-preserving map $T:X\to X$ that is ergodic. For every measurable and integrable function $f:X\to \mathbb{R}$ there is a set $\mathrm{\Omega }\in \mathcal{B}$ with $\mu (\mathrm{\Omega })=1$ and $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))=\int f\mathsf{d}\mu$$ for all $x\in \mathrm{\Omega }$.

Proof:

Fix $f:X\to \mathbb{R}$ measurable and integrable. Since we do not know whether the limit of the sequence $$N\mapsto {\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))$$ exists we will work with its limits superior and inferior. Our goal is to prove for every $k\in \mathbb{N}$ that the set $$B(k)=\{x\in X:\underset{N\to \mathrm{\infty }}{lim sup}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^{n}x)>\int f\mathsf{d}\mu +{\displaystyle \frac{1}{k}}\}$$ has zero measure. Indeed, if that is the case then $$\mu (X\setminus \bigcup _{k\in \mathbb{N}}B(k))=1$$ and we have $$\underset{N\to \mathrm{\infty }}{lim sup}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))\le \int f\mathsf{d}\mu$$ for almost every $x\in X$. Repeating the above argument with $-f$ in place of $f$ then gives $$\int f\mathsf{d}\mu \le \underset{N\to \mathrm{\infty }}{lim inf}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))$$ for almost every $x\in X$ as well and concludes the proof.

Fix $k\in \mathbb{N}$ and suppose that $B(k)$ has positive measure. The set $B(k)$ is $T$ invariant. Indeed $$|{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^{n}x)-{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(Tx))|\le {\displaystyle \frac{|f(x)|+|f(T^{N}x)|}{N}}$$ so it suffices to prove that $$\{x\in X:{\displaystyle \frac{|f(T^{N}x)|}{N}}\to 0\}$$ has zero measure. Fix $ϵ>0$. The set $$A(N)=\{x\in X:|f(T^{N}x)|\ge Nϵ\}$$ has measure equal to that of $$\{x\in X:|f(x)|\ge Nϵ\}$$ because $\mu$ is $T$ invariant. But $$\sum _{N=1}^{\mathrm{\infty }}ϵ\mu (A(N))\le \int |f|\mathsf{d}\mu$$ so the set of points $x$ that belong to infinitely many of the sets $A(N)$ has zero measure by the Borel-Cantelli lemma. In other words, almost every $x\in X$ has the property that $${\displaystyle \frac{|f(T^{N}x)|}{N}}\to 0$$ and therefore $$\mu (B(k)\mathrm{△}{T}^{-1}B(k))=0$$ holds. Ergodicity then forces $\mu (B(k))=1$.

We are looking for a contradiction to $\mu (B(k))=1$. The crucial quantity to consider is the maximum value $${\mathsf{M}}_f(x)=sup\{{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^{n}x):N\in \mathbb{N}\}$$ of the ergodic averages. Note that $$B(k)\subset \{x\in X:{\mathsf{M}}_f(x)\ge \int f\mathsf{d}\mu +{\displaystyle \frac{1}{k}}\}$$ because the supremum of the values of a sequence cannot be smaller than its limit superior. Thus the set $$C(k)=\{x\in X:{\mathsf{M}}_f(x)\ge \int f\mathsf{d}\mu +{\displaystyle \frac{1}{k}}\}$$ also has $\mu (C(k))=1$. The crucial ingredient now is the inequality $$\int 1_{C(k)}\cdot (f-\int f\mathsf{d}\mu -{\displaystyle \frac{1}{k}})\mathsf{d}\mu \ge 0$$ which is a special case of the maximal ergodic theorem. If we have this inequality then, as $C(k)$ has full measure, we conclude that $$0\le \int f-\int f\mathsf{d}\mu -{\displaystyle \frac{1}{k}}\mathsf{d}\mu =-{\displaystyle \frac{1}{k}}$$ which is the desired contradiction.▮