I am struggling with an estimate in a proof of the pointwise ergodic theorem discussed in T. Ward and M. Einsiedler: Ergodic Theory with a view towards Number Theory, which is left as an exercise. I will first give a description of the proof in order to clarify notation, then point out the estimate which is giving me troubles and then state what I have ``found out'' up to now (which is not so much unfortunately):
Pointwise Ergodic Theorem: Let $(X,\mathscr{B},\mu)$ be a probability space, $T:(X,\mathscr{B})\to(X,\mathscr{B})$ measue-preserving, $f:X\to\mathbb{C}$ an integrable, measureable function, i.e. $\int_{X}|f|\operatorname{d}\mu<\infty$. Let $$A_{N}(f)(x):=\frac{1}{N}\sum_{n=0}^{N-1}f(T^{n}(x))\quad \forall x\in X, \forall N\geq 1$$ Then there exists an integrable function $f^{\ast}$ such that the sequence $(A_{N}(f))_{N\in\mathbb{N}}$ converges to $f^{\ast}$ pointwise almost everywhere.
Let's quickly discuss the structure of the proof, which tells us what we know already and which estimate is giving me trouble.
The proof: We start by assuming that $f$ is almost surely bounded. Once we have checked this case, we will use density of $L_{\mu}^{\infty}(X,\mathscr{B})$ in $L_{\mu}^{1}(X,\mathscr{B})$ to deduce the general case. We know from the mean ergodic theorem (which does not yet require the boundedness of $f$), that we can find an almost surely $T$-invariant, integrable function $F$ such that: $$F={\lim_{N\to\infty}}^{L_{\mu}^{1}}A_{N}(f)$$ i.e. $\lVert F-A_{N}(f)\rVert_{1}\stackrel{N\to\infty}{\longrightarrow}0$. The claim hence becomes: let $F$ be a representative of the $L_{\mu}^{1}\cap L_{\mu}^{\infty}$-limit of $(A_{N}(f))$, then $A_{N}(f)$ converges to $F$ pointwise almost surely. This claim is formalized as follows: $$\mu(\{x\in X; \lim\sup_{N\to\infty}|F(x)-A_{N}(f)(x)|>0\})=0$$ Note that the claim would follow from: $$\mu(\{x\in X; \lim\sup_{N\to\infty}|F(x)-A_{N}(f)(x)|>\epsilon\})<2\epsilon \quad\forall\epsilon>0$$ by some standard argumentation (here my estimate is a bit weaker than in the original reference but this should not affect the result). Fix $M\in\mathbb{N}$ and $\epsilon>0$, then we know from the maximal ergodic theorem applied to the functions $g_{1}:=F-A_{M}(f)$ and $g_{2}:=A_{M}(f)-F$ (this is why the 2 pops up in my discussion) that: $$\mu(\{x\in X;\sup_{n\geq 1}|A_{N}(F-A_{M}(f))|>\epsilon\})<2\epsilon$$ Now comes the critical estimate: The authors state that for fixed $M\in\mathbb{N}$ holds: $$\begin{equation}\exists N_{0}\in\mathbb{N}\exists C_{M}>0:\quad N\geq N_{0}\Rightarrow|A_{N}(f)-A_{N}(A_{M}(f))|\leq C_{M}\frac{\lVert f\rVert_{\infty}}{N}\end{equation}$$ Given the critical estimate and $A_{N}$-invariance of $F$, we find: $$\begin{align}\{x\in X;\lim\sup_{N\to\infty}|F(x)-A_{N}(f)(x)|>\epsilon\}=&\{x\in X;\lim\sup_{N\to\infty}|F(x)-A_{N}(A_{M}(f))|>\epsilon\}\\ \stackrel{\mu}{=}&\{x\in X;\lim\sup_{N\to\infty}|A_{N}(F-A_{M}(f))(x)|>\epsilon\}\\ \subseteq & \{x\in X;\sup_{N\geq1}|A_{N}(F-A_{M}(f))(x)|\}\end{align}$$ where for $A,B\in\mathscr{B}$ holds $A\stackrel{\mu}{=}B$ if $\mu(A\Delta B)=0$. This readily proves the claim.
I turn to you for the critical estimate: I do not manage to get the $N$ in the denominator. My idea was to do the following: I was about to check that $A_{N}(A_{M}(f))\to F$ in $L_{\mu}^{1}$ which sounds quite intuitive, i.e. both sequences have the same limit. This sounds quite obvious as the limit is $T$-invariant. Then I intended to use boundedness of $f$ to find a uniform estimate for the difference. But I always have the problem that the summation over $n$ and $N^{-1}$ cancel each other. Furthermore all those estimates are but almost surely. Could you please given me a hint on why the critical estimate holds?
Note that I am particularly interested in getting this proof to work, not in any proof of the Pointwise Convergence Theorem.
Detailed Estimate: Thanks to @Davide Giraudo, I think I managed to prove it if $f$ is honestly bounded. In what follows let $\xi:=(\xi_{n})_{n\in\mathbb{N}}$ be a bounded sequence and $U_{N}(\xi):=\sum_{n=0}^{N-1}\xi_{n}$. Using his idea, we get: $$U_{N}(U_{M}(\xi))=\sum_{n=0}^{N-1}\sum_{m=n}^{M+n-1}\xi_{m}$$ A little drawing and induction shows that for fixed $M\in\mathbb{N}$ and $N\in\mathbb{N}$ way larger than $M$, we get: $$\begin{align}U_{N}(U_{M}(\xi))=&\sum_{m=0}^{M-1}(m+1)\xi_{m}+M\sum_{m=M}^{N-1}\xi_{m}+\sum_{m=N}^{M+N-2}(M+N-1-m)\xi_{m}\\ =&U_{N}(\xi)+\sum_{m=0}^{M-1}m\xi_{m}+(M-1)\sum_{m=M}^{N-1}\xi_{m}+\sum_{m=N}^{M+N-2}(M+N-1-m)\xi_{m}\end{align}$$ so that: $$U_{N}(U_{M}(\xi))-MU_{N}(\xi)=\sum_{m=0}^{M-1}(m+1-M)\xi_{m}+\sum_{m=N}^{M+N-2}(M+N-1-m)\xi_{m}$$ As $|m+1-M|\leq M$ for $m\in\{0,\ldots,M-1\}$ and $|M+N-1-m|\leq M+|N-m|\leq 2M$ for $m\in\{N,\ldots,M+N-2\}$, we find that for $x\equiv (f(T^{n}(x)))_{n\in\mathbb{N}}$ holds: $$\begin{align}|A_{N}(A_{M}(x))-A_{N}(x)|=&\frac{1}{NM}|U_{N}(U_{M}(x))-MU_{N}(x)|\\ \leq &\frac{1}{NM}\left\{\sum_{m=0}^{M-1}2M\lVert f\rVert_{\infty}+\sum_{m=N}^{M+N-2}2M\lVert f\rVert_{\infty}\right\}\leq\frac{2M}{N}\lVert f\rVert_{\infty}\end{align}$$
It remains for me to ensure that if $f$ is almost surely bounded, the estimate still holds almost everywhere.
Range of validity of the estimate: This is in fact easy. Let $f$ lmost surely bounded, then there exists $B\subseteq\mathscr{B}$ with $\mu(B)=1$ and $f|_{B}$ bounded by $\lVert f\rVert_{\infty}$. Look at $B':=\cap_{n\geq 0}T^{-n}(B)$, then $\mu(B')=1$ as its complement is a countable union of nullsets and the estimate clearly holds on $B'$. But this requires us to change to: $$\{x\in X;\lim\sup_{N\to\infty}|F(x)-A_{N}(f)(x)|>\epsilon\}\stackrel{\mu}{=}\{x\in X;\lim\sup_{N\to\infty}|F(x)-A_{N}(A_{M}(f))|\}$$ instead of a strict equality. But this doesn't pose any problems.
Here is the idea: write $$A_N(A_M(f))=\frac MN\sum_{n=0}^{N-1}\sum_{m=n}^{M+n-1}f\circ T^m,$$ and write the inner sum as $\sum_{m=0}^{M+n-1}-\sum_ {m=0}^{n-1}$.
Permuting the sums, we get approximatively $$A_N(A_M(f))=\frac 1{MN}\sum_{m=0}^{m+N-2}(N+M-m)f\circ T^m-\frac 1{MN}\sum_{m=0}^{N-2}(N-m)f\circ T^m\\ =\frac 1N\sum_{m=0}^{N-2}f\circ T^m+\frac 1{MN}\sum_{m=N-1}^{M+N-2}(N+M-m)f\circ T^m.$$ The second term can be bounded by $\frac MN\lVert f\rVert_\infty$.