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I constructed the following proof after doing a bit of Googling of proofs of the ergodic theorem. (In particular, it combines ideas from the paper https://doi.org/10.1214/074921706000000266 of Keane and Peterson - which, to be fair, is already a pretty impressively short proof even for general $L^1$ - with proofs I've seen in online lectures notes on ergodic theory; but, of course, I also use boundedness to help get a very short proof.)

I will prove the theorem for ergodic transformations, but it is not at all difficult to extend (in the appropriate manner) to more general measure-preserving transformations.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. Define the monotone sequences \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} For $N \geq 2$, $$\ m_N^{[g]} - g = \max\big\{ \, 0 \ , \ g \circ T \ , \ \ldots\ldots \ , \ (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} = (m_{N-1}^{[g]})^+ \circ T.$$ Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now $\ \bigcup_{N=1}^\infty \! E_N^{[g]} = \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}$, which is clearly a $\mu$-full measure set. Hence by the dominated convergence theorem, taking the limit of the above as $N \to \infty$ gives $\mathbb{E}_\mu[g] \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

I constructed the following proof after doing a bit of Googling of proofs of the ergodic theorem. (In particular, it combines ideas from the paper https://doi.org/10.1214/074921706000000266 of Keane and Peterson - which, to be fair, is already a pretty impressively short proof even for general $L^1$ observables - with proofs I've seen in online lectures notes on ergodic theory; but, of course, I also use boundedness to help get a short proof.)

I will prove the theorem for ergodic transformations, but it is not at all difficult to extend (in the appropriate manner) to more general measure-preserving transformations. I have tried to write out the proof sufficiently explicitly to be smoothly readable in about 2-5 minutes.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. Define the monotone sequences \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} For $N \geq 2$, \begin{align*} (m_N^{[g]} - g)(x) &= \max\big\{ \, 0 \ , \ g(T(x)) \ , \ \ldots\ldots \ , \ g(T(x)) \!+\! \ldots \!+\! g(T^{N-1}(x)) \, \big\} \\ &= (m_{N-1}^{[g]})^+(T(x)). \end{align*} Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now \begin{align*} \bigcup_{N=1}^\infty \! E_N^{[g]} &= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(g)(x) > 0 \} \\ &= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}, \end{align*} which is a $\mu$-full measure set by definition of $\overline{f}$. Hence by the dominated convergence theorem, $\int_X g \, d\mu \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. For each $N \geq 1$ define \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} It is clear that for $N \geq 2$, $\ m_N^{[g]} - g = (m_{N-1}^{[g]})^+ \circ T$. Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now $\ \bigcup_{N=1}^\infty \! E_N^{[g]} = \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}$, which is clearly a $\mu$-full measure set. Hence by the dominated convergence theorem, taking the limit of the above as $N \to \infty$ gives $\mathbb{E}_\mu[g] \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. Define the monotone sequences \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} For $N \geq 2$, $$\ m_N^{[g]} - g = \max\big\{ \, 0 \ , \ g \circ T \ , \ \ldots\ldots \ , \ (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} = (m_{N-1}^{[g]})^+ \circ T.$$ Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now $\ \bigcup_{N=1}^\infty \! E_N^{[g]} = \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}$, which is clearly a $\mu$-full measure set. Hence by the dominated convergence theorem, taking the limit of the above as $N \to \infty$ gives $\mathbb{E}_\mu[g] \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

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?

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.

It seems to me that a considerably simpler proof [see below] 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?

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.

Short proof of BET for bounded observables.

I constructed the following proof after doing a bit of Googling of proofs of the ergodic theorem. (In particular, it combines ideas from the paper https://doi.org/10.1214/074921706000000266 of Keane and Peterson - which, to be fair, is already a pretty impressively short proof even for general $L^1$ - with proofs I've seen in online lectures notes on ergodic theory; but, of course, I also use boundedness to help get a very short proof.)

I will prove the theorem for ergodic transformations, but it is not at all difficult to extend (in the appropriate manner) to more general measure-preserving transformations.

Theorem. Let $(X,\mathcal{X},T,\mu)$ be an ergodic measure-preserving dynamical system and let $f \colon X \to \mathbb{R}$ be a bounded measurable function. Then $P_N(f):=\frac{1}{N} \sum_{i=0}^{N-1} f \circ T^i \, \overset{\mu\textrm{-a.s.}}{\to} \, \mathbb{E}_\mu[f]\,$ as $N \to \infty$.

Proof. Since $\overline{\lim}_{N \to \infty} P_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. For each $N \geq 1$ define \begin{align*} m_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\ E_N^{[g]} := & \, \{ x \in X : m_N^{[g]}(x)>0 \}. \end{align*} It is clear that for $N \geq 2$, $\ m_N^{[g]} - g = (m_{N-1}^{[g]})^+ \circ T$. Integrating over $E_N^{[g]}$ gives \begin{align*} \int_{E_N^{[g]}} g \, d\mu \ &= \ \int_{E_N^{[g]}} m_N^{[g]} \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &= \int_X (m_N^{[g]})^+ \, d\mu - \int_{E_N^{[g]}} (m_{N-1}^{[g]})^+ \circ T \, d\mu \\ &\geq \int_X (m_N^{[g]})^+ \, d\mu - \int_X (m_N^{[g]})^+ \circ T \, d\mu \ = \ 0. \end{align*} Now $\ \bigcup_{N=1}^\infty \! E_N^{[g]} = \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P_n(f)(x) > \overline{f} - \varepsilon \}$, which is clearly a $\mu$-full measure set. Hence by the dominated convergence theorem, taking the limit of the above as $N \to \infty$ gives $\mathbb{E}_\mu[g] \geq 0$, i.e. $\mathbb{E}_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. QED.

Remark. Contained within the above is a proof of a "maximal ergodic theorem" that does not rely on $g$ being bounded but only on $g$ being $\mu$-integrable; in other words, the above proof goes through without modification for any $f \in L^1(\mu)$ for which it is known that $\overline{f}$ is finite. But I see no trivial way of showing that $\overline{f}$ is finite except when $f$ is bounded.