It seems to me that a considerably simpler proof of Birkhoff's ergodic theorem can be obtained for bounded observables than for more general $L^1$ observables. Therefore, I feel like it would be nicer to prove Birkhoff's ergodic theorem by starting off with the bounded case and then extending to the general case, than by "directly" tackling the general case. But can this extension from $L^\infty$ to $L^1$ be done in any "straightforward" elementary manner?

Let me give one possible version of how to make this question precise:

Let $(X,\mathcal{X},\mu)$ be a probability space. A Markov operator on $L^1(\mu)$ is a linear, monotone, unity-preserving, integral-preserving function $P \colon L^1(\mu) \to L^1(\mu)$.

Suppose we have a sequence $(P_n)_{n \geq 1}$ of Markov operators on $L^1(\mu)$ such that (1) for every $f \in L^\infty(\mu)$ the following statements hold:

• $P_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \, d\mu\ $ as $n \to \infty$;
• for all $m,n \geq 1$, $\|nP_n(f) - mP_m(f)\|_{L^\infty(\mu)} \leq |m-n|\|f\|_{L^\infty(\mu)}$;

and (2) for every $f \in L^1(\mu)$ there exists a value $P_\infty[f] \in \overline{\mathbb{R}}$ such that $\,\liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{=} P_\infty[f]$.

(Obviously, if $f \in L^\infty(\mu)$ then $P_\infty[f]=\int_X f \, d\mu$. We will also see from the proof of the result in the Note below that the same holds for all $f \in L^1(\mu)$ with $f \geq 0$.)

Do we necessarily have that for all $f \in L^1(\mu)$, $P_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \, d\mu\ $ as $n \to \infty$?

(Since $P_n(-f)=-P_n(f)$, this is equivalent to saying that for all $f \in L^1(\mu)$, $P_\infty[f]=\int_X f \, d\mu$.)

Remark. I expect that if the answer is yes, then not all the conditions given above will be necessary to prove it.

Note. As below, it is not hard to show [under considerably weaker conditions than those given above] that for each $f \in L^1(\mu)$, $\ P_n(f) \overset{L^1(\mu)}{\to} \int_X f \, d\mu\ $ as $n \to \infty$.

Proof: Without loss of generality take $f \overset{\mu\textrm{-a.s.}}{\geq} 0$. For each $k>0$ we have $P_n(f \wedge k) \overset{\mu\textrm{-a.s.}}{\to} \int_X f \wedge k \; d\mu\ $ as $n \to \infty$. Hence, by monotonicity of Markov operators it is clear that $$ \liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{\geq} \int_X f \, d\mu. $$ But due to the integral-preservation of Markov operators, we also have that $\int_X P_n(f) \, d\mu = \int_X f \, d\mu$ for all $n$; and so, since $P_n(f) \overset{\mu\textrm{-a.s.}}{\geq} 0$ (by monotonicity of Markov operators), Fatou's lemma gives that $$ \int_X \liminf_{n \to \infty} P_n(f) \, d\mu \leq \int_X f \, d\mu. $$ Hence it follows that $$ \liminf_{n \to \infty} P_n(f) \overset{\mu\textrm{-a.s.}}{=} \int_X f \, d\mu. $$ The result is then an immediate consequence of the following Scheffé-like lemma.

Lemma. Given a sequence $(Y_n)_{n \in \mathbb{N}}$ of integrable random variables $Y_n$ that is uniformly bounded below, if $\ Y \!:=\! \underset{n \to \infty}{\liminf} Y_n\ $ is integrable and $\mathbb{E}[Y_n] \to \mathbb{E}[Y]$ as $n \to \infty$, then $Y_n \overset{L^1}{\to} Y$.

Proof of Lemma. Without loss of generality take $Y=0$. We have that $Y_n \wedge 0$ converges pointwise to $0$ as $n \to \infty$, and so since $(Y_n)$ is uniformly bounded below, the dominated convergence theorem can be applied to give that $\mathbb{E}[Y_n \wedge 0] \to 0$ as $n \to \infty$. Since $|Y_n|=Y_n-2(Y_n \wedge 0)$ and $\mathbb{E}[Y_n] \to 0$ as $n \to \infty$, it follows that $\mathbb{E}[|Y_n|] \to 0$ as $n \to \infty$. So we are done.