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Let $(X,\sigma,\mu)$ be a measure space, $\mu(X)=1$. End let $T:X\to X$ be a measurable map such that $$D\in\sigma\Longrightarrow \mu(T^{-1}D)=\mu(D).$$ Let $f:X\to\mathbb{R}_+$ be an integrable function. Construct a measurable function as follows $$\psi_n:=\Big(\limsup_{k\to\infty}\frac{1}{k}\sum_{i=0}^{k-1}fT^i\Big)\wedge n.$$ The paper Easy and nearly simultaneous proofs of the Ergodic Theorem and Maximal Ergodic Theorem claims (before the ergodic theorem!) that for all $n\in\mathbb{N}$ the functions $\psi_n$ are invariant: $$\psi_n T=\psi_n,\quad \mbox{a. e.}$$ Please could you push me in the direction of the proof. It must be easy but I am stuck.

It is a question for the moderators. I translated into Russian the article cited above and while I was doing it I simplified the proof and even generalized some assertions a bit. If it is interesting I can translate this new version of the article back in English and leave a reference here.

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I believe you are referring to the function $\lambda$ defined in the first paragraph of page $250$ which carries an extra $-1/n$ (this is irrelevant for the invariance). If I am not missing anything, you have the following.

Define $g_{\ell}\overset{\Delta}= \limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=0}^{k-1} fT^{i+\ell}.$

Then, $g_{\ell}=g_0$ for all $\ell\in\mathbb{N}$ since

$$g_{0} = \limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=0}^{k-1} fT^{i}= \underbrace{\limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=0}^{\ell-1} fT^{i}}_{=0}+\underbrace{\limsup_{k\rightarrow \infty} \frac{k-\ell}{k} \frac{1}{k-\ell}\sum_{i=\ell}^{k-1} fT^{i}}_{=g_{\ell}}.$$

Thus, $\psi_n T=g_1 \wedge n = g_0 \wedge n = \psi_n.$

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  • $\begingroup$ Yes, and it remains to show that $g_1=g_0$. I understand that just for $f\in L^\infty$ $\endgroup$
    – Oleg Zubelewicz
    Commented May 25, 2023 at 18:16
  • $\begingroup$ @OlegZubelewicz: You are missing that $g_{0}= \limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=0}^{k-1} fT^{i}= \limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=0}^{\ell-1} fT^{i}+\limsup_{k\rightarrow \infty} \frac{1}{k} \sum_{i=\ell}^{k-1} fT^{i}$. Since the first term is zero (regardless of whether $f\in L^{\infty}$ or not, since it is a finite sum of finite elements) and the second is $g_{\ell}$, you have that $g_0=g_{\ell}$, for any $\ell \in \mathbb{N}$. $\endgroup$
    – Augusto Santos
    Commented May 25, 2023 at 18:29
  • $\begingroup$ $A_kf:=\frac{1}{k}\sum_{j=0}^{k-1}fT^j;\quad A_k(fT)=A_kf-f/k+\frac{1}{k}fT^k$ $\endgroup$
    – Oleg Zubelewicz
    Commented May 25, 2023 at 18:43
  • $\begingroup$ @OlegZubelewicz: I am not sure I understand your latter comment, but I updated the answer to render this step clearer. $\endgroup$
    – Augusto Santos
    Commented May 25, 2023 at 19:00
  • $\begingroup$ Thanks, I asked a stupid question $\endgroup$
    – Oleg Zubelewicz
    Commented May 25, 2023 at 19:12

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