[proofplan] For bounded functions, Birkhoff gives almost-everywhere convergence, while the $L^2$ mean ergodic theorem identifies the limit as conditional expectation. Bounded convergence upgrades this to $L^p$ convergence. For general $L^p$ functions, truncate in $L^p$ and use that both ergodic averaging and conditional expectation are contractions on $L^p$. [/proofplan] [step:Prove the result for bounded functions] Assume first that $f\in L^\infty(X,\mu)$. Then $f\in L^2(X,\mu)$ because $\mu(X)=1$. By the theorem identifying the mean-ergodic limit with conditional expectation, \begin{align*} A_Nf\to \mathbb{E}[f\mid\mathcal{I}] \end{align*} in $L^2$; this is the [Limit Is Conditional Expectation onto Invariant Sigma-Algebra](/theorems/3449). By the [Birkhoff Ergodic Theorem](/theorems/518), the same averages converge almost everywhere to an invariant function. The $L^2$ limit identifies that invariant function with $\mathbb{E}[f\mid\mathcal{I}]$. Since \begin{align*} |A_Nf-\mathbb{E}[f\mid\mathcal{I}]|\leq 2\|f\|_\infty \end{align*} almost everywhere, the [Dominated Convergence Theorem](/theorems/4) gives \begin{align*} \|A_Nf-\mathbb{E}[f\mid\mathcal{I}]\|_p\to0 \end{align*} for every $1\leq p<\infty$. [/step] [step:Use contraction estimates for truncation] For $g\in L^p(X,\mu)$, measure preservation gives \begin{align*} \|g\circ T^n\|_p=\|g\|_p \end{align*} for every $n\geq0$. Hence \begin{align*} \|A_Ng\|_p\leq \|g\|_p \end{align*} for every $N$. Conditional expectation is also a contraction on $L^p$, so \begin{align*} \|\mathbb{E}[g\mid\mathcal{I}]\|_p\leq \|g\|_p. \end{align*} [/step] [step:Approximate a general $L^p$ function by bounded functions] Let $f\in L^p(X,\mu)$ and choose bounded functions $f_m\in L^\infty(X,\mu)$ such that \begin{align*} \|f-f_m\|_p\to0. \end{align*} For every $N$ and $m$, \begin{align*} \|A_Nf-\mathbb{E}[f\mid\mathcal{I}]\|_p &\leq \|A_N(f-f_m)\|_p\\ &\quad+ \|A_Nf_m-\mathbb{E}[f_m\mid\mathcal{I}]\|_p\\ &\quad+ \|\mathbb{E}[f_m-f\mid\mathcal{I}]\|_p\\ &\leq 2\|f-f_m\|_p+ \|A_Nf_m-\mathbb{E}[f_m\mid\mathcal{I}]\|_p. \end{align*} First choose $m$ so large that $2\|f-f_m\|_p$ is small. Then let $N\to\infty$ and use the bounded case for $f_m$. It follows that \begin{align*} \|A_Nf-\mathbb{E}[f\mid\mathcal{I}]\|_p\to0. \end{align*} [/step]