# Is Birkhoff's ergodic theorem true for $L_\infty$?

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?

• Are you asking the question: Suppose $T$ is a measure-preserving transformation of $(X,\mathcal B,\mu)$ where $\mu(X)=\infty$ and $f\in L_\infty(X)$, then does $\frac 1n(f(x)+\ldots+f(T^{n-1}x))$ converge for a.e. $x$? (Or are you asking whether the convergence of $A_nf$ is in the $L^\infty$ sense)? – Anthony Quas Nov 13 '20 at 20:34
• I know that in norm it does not converge. I wanted to ask about pointwise convergence. – A beginner mathmatician Nov 14 '20 at 5:50

## 2 Answers

Here is a counter-example to the question I think you're asking. Let $$X$$ be the space of two-sided 0-1 valued sequences. Let $$x$$ the sequence with $$x_n=0$$ for all $$n\le 0$$; $$x_n=1$$ if $$2^k for any even $$k\ge 0$$ and $$x_n=0$$ if $$2^k 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 counter-example because the measure $$\mu$$ fails to be conservative. There are conservative counter-examples also (based on renewal sequences).

• What is the measure preserving transformation here? What is $\sigma$? Can you lease elaborate your answer? – A beginner mathmatician Nov 14 '20 at 5:51
• 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 $\frac 1n\sum_{j=0}^{n-1}f(\sigma^jz)=\frac 1n\sum_{j=0}^{n-1}f(\sigma^{j+k}x)$. 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) – Anthony Quas Nov 16 '20 at 6:52

No. For the simplest counterexample take the shift on the space of one-sided 0-1 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 $$\{\epsilon_1=0\}$$.