Is Birkhoff's pointwise/individual ergodic theorem for $L_\infty.$ Clearly, it is true if the measure space is finite? What about the measure space not finite?

1$\begingroup$ Are you asking the question: Suppose $T$ is a measurepreserving transformation of $(X,\mathcal B,\mu)$ where $\mu(X)=\infty$ and $f\in L_\infty(X)$, then does $\frac 1n(f(x)+\ldots+f(T^{n1}x))$ converge for a.e. $x$? (Or are you asking whether the convergence of $A_nf$ is in the $L^\infty$ sense)? $\endgroup$ – Anthony Quas Nov 13 '20 at 20:34

1$\begingroup$ I know that in norm it does not converge. I wanted to ask about pointwise convergence. $\endgroup$ – A beginner mathmatician Nov 14 '20 at 5:50
Here is a counterexample to the question I think you're asking. Let $X$ be the space of twosided 01 valued sequences. Let $x$ the sequence with $x_n=0$ for all $n\le 0$; $x_n=1$ if $2^k<n\le 2^{k+1}$ for any even $k\ge 0$ and $x_n=0$ if $2^k<n\le 2^{k+1}$ for any odd $k\ge 0$. Then form an infinite measure by $\mu=\sum_{j\in\mathbb Z}\delta_{\sigma^j(x)}$. It's easy to check that if $f(z)=z_0$, averages fail to converge for $\mu$a.e. $z$. Notice this isn't a great counterexample because the measure $\mu$ fails to be conservative. There are conservative counterexamples also (based on renewal sequences).

1$\begingroup$ What is the measure preserving transformation here? What is $\sigma$? Can you lease elaborate your answer? $\endgroup$ – A beginner mathmatician Nov 14 '20 at 5:51

1$\begingroup$ The transformation is the shift $\sigma$. The measure is invariant as $\mu\circ\sigma^{1}=\sum_{j\in\mathbb Z}\delta_{\sigma^j(x)}\circ\sigma^{1}=\sum_{j\in\mathbb Z}\delta_{\sigma^{j+1}(x)}=\sum_{j\in\mathbb Z}\delta_{\sigma^j(x)}$. Since the measure is $\sum \delta_{\sigma^j(x)}$, if you want to prove something holds $\mu$a.e., you must show it holds for each $z=\sigma^k(x)$. If you compute $\frac 1n\sum_{j=0}^{n1}f(\sigma^jz)=\frac 1n\sum_{j=0}^{n1}f(\sigma^{j+k}x)$. It's not hard to see that this fails to converge (the limsup is $2/3$ at the ends of blocks of 1s; the liminf is 1/3) $\endgroup$ – Anthony Quas Nov 16 '20 at 6:52
No. For the simplest counterexample take the shift on the space of onesided 01 sequences $(\epsilon_1,\epsilon_2,\dots)$ endowed with any fully supported invariant measure (for instance, the (1/2,1/2) Bernoulli measure), and let $f$ be the indicator of the set $\{\epsilon_1=0\}$.