Birkhoff's theorem requires integrability

For $a<0$, $\{f>a\}=[0,1)$; for $a=0$, it is $(0,1)$; and for $a>0$ it is $(0,\min \{1,1\mathrm{/}a\})$, with the right endpoint omitted if it equals $1$. These are Borel. Hence [F1] makes the everywhere-finite function $f$ Lebesgue measurable.

On $I_k=[1\mathrm{/}(k+1),1\mathrm{/}k)$ one has $f\ge k$. For every $N$, the simple function $s_N={\sum }_{k=1}^{N}k{\mathbf{1}}_{I_k}$ therefore satisfies $s_N\le f$, and $\int s_{N}d\lambda ={\sum }_{k=1}^{N}k(\frac{1}{k}-\frac{1}{k+1})={\sum }_{k=1}^N\frac{1}{k+1}\mathrm{.}$

By [F2]–[F3], the right side is unbounded, so $\int fd\lambda =+\mathrm{\infty }$. Thus $f$ is not integrable under the integrability convention of Integrable real and complex functions, and their integrals.

For each integer $m\ge 1$, let $f_m=\min \{f,m\}$. It is measurable, bounded, and hence integrable on this probability space; moreover $f_m\uparrow f$. Monotone convergence and step 2.1 give $c_m:=\int f_{m}d\lambda \uparrow +\mathrm{\infty }\mathrm{.}$

By [F4], $A_n{f}_m$ converges almost everywhere to an invariant function and that function equals a constant almost everywhere. Since $0\le A_n{f}_m\le m$, [F5] identifies this constant as ${\lim }_n\int A_n{f}_{m}d\lambda =\int f_{m}d\lambda =c_m\mathrm{.}$

Remove the countable union of the null exceptional sets in step 4.1; it is null by [F6]. At every remaining $x$, for every $m$ and $n$, $A_{n}f(x)\ge A_n{f}_m(x)$, whence ${\mathrm{lim\hspace{0.17em}inf}}_n{A}_{n}f(x)\ge c_m\mathrm{.}$ Because $c_m\uparrow +\mathrm{\infty }$, this says $A_{n}f(x)\to +\mathrm{\infty }$.

Thus a finite-valued measurable observable can have divergent-to-infinity ergodic averages almost everywhere. This refutes the finite-limit claim and shows exactly why the $L^1$ hypothesis cannot be omitted. Countable choice is inherited from the Lebesgue and doubling-map suppliers; all truncations are explicit.