A nonintegrable observable with divergent ergodic averages

The strict superlevel sets of $f$ are $[0,1)$ for negative levels, $(0,1)$ at level zero, and $(0,\min \{1,1\mathrm{/}a\})$ at positive level $a$ (with the ambient endpoint omitted). They are Borel, so [F1] makes the everywhere finite $f$ measurable.

On $I_k=[1\mathrm{/}(k+1),1\mathrm{/}k)$ one has $f\ge k$. Thus monotonicity and the simple-integral formula give, for every $N$, $\int fd\lambda \ge {\sum }_{k=1}^{N}k\lambda (I_k)={\sum }_{k=1}^N\frac{1}{k+1}\mathrm{.}$

The last sums are unbounded by [F3]. Hence $\int f=+\mathrm{\infty }$, so $f$ is not integrable in the sense of Integrable real and complex functions, and their integrals.

Put $f_m=\min \{f,m\}$ for $m\ge 1$. These are bounded integrable functions, $f_m\uparrow f$, and [F3] yields $c_m:=\int f_{m}d\lambda \uparrow +\mathrm{\infty }$.

For each $m$, [F4] makes $A_n{f}_m$ converge almost everywhere to a constant. Since the averages are bounded by $m$, [F5] identifies that constant as $c_m$.

Outside the countable union of the exceptional null sets in step 4.1, which is null by [F6], all these limits hold simultaneously. Since $f\ge f_m$, there ${\mathrm{lim\hspace{0.17em}inf}}_n{A}_{n}f\ge {\lim }_n{A}_n{f}_m=c_m$ for every $m$. Letting $m\to \mathrm{\infty }$ gives $A_{n}f\to +\mathrm{\infty }$.

Thus this finite measurable but nonintegrable $f$ is the promised counterexample. Countable choice is inherited from the Lebesgue/doubling suppliers; the truncations are explicit.