A version of Birkhoff's Ergodic Theorem is the following:
Theorem 1: Take $\xi\in L^{1}(\Omega,\mathcal{F},\mathbb{P})$. If $\theta$ preserves $\mathbb{P}$, then $$\dfrac{\xi(\omega)+\xi(\theta\omega)+\cdots+\xi(\theta^{n-1}\omega)}{n}\longrightarrow_{a.s,\ L^{1}}\mathbb{E}(\xi|\mathcal{J}),$$ where $\mathcal{J}$ is the invariance $\sigma-$algebra. In particular, if the system $(\Omega,\mathcal{F},\mathbb{P},\theta)$ is ergodic (so that $\mathcal{J}$ is trivial), then we have $$\dfrac{\xi(\omega)+\xi(\theta\omega)+\cdots+\xi(\theta^{n-1}\omega)}{n}\longrightarrow_{a.s,\ L^{1}}\mathbb{E}\xi.$$
A version of Kingman's Subadditive Ergodic Theorem is the following:
Theroem 2: Let $(X_{n})_{n\in\mathbb{N}}$ be a sequence of $L^{1}$ random variables and let $\theta$ be an MPT on $(\Omega,\mathcal{F},\mathbb{P})$ such that the following sub-additive property holds:$$X_{n+m}(\omega)\leq X_{n}+X_{m}(\theta^{n}\omega),\ \ n,m\in\mathbb{N}.$$ Then we have $$\dfrac{X_{n}}{n}\longrightarrow_{a.s,\ L^{1}}\inf_{n\in\mathbb{N}}\dfrac{\mathbb{E}(X_{n}|\mathcal{J})}{n},\ \text{as}\ n\rightarrow\infty.$$ In particular, if $\theta$ is ergodic, then $$\dfrac{X_{n}}{n}\longrightarrow_{a.s,\ L^{1}}\inf_{n\in\mathbb{N}}\dfrac{\mathbb{E}X_{n}}{n},\ \text{as}\ n\rightarrow\infty.$$
Now, I want to show Theorem 2 implies Theorem 1. I had some attempts but got stuck in the end:
Let $Y$ be a $L^{1}-$random variable and $\theta$ is an MPT. Define $$X_{n}:=Y(\omega)+\cdots+Y(\theta^{n-1}\omega).$$ Then, $$X_{n}(\omega)+X_{m}(\theta^{n}\omega)=Y(\omega)+\cdots+Y(\theta^{n-1}\omega)+Y(\theta^{n}\omega)+\cdots+Y(\theta^{n+m-1}\omega)=X_{n+m}(\omega).$$
Therefore, by the Sub-additive Ergodic Theorem, we have $$\dfrac{Y(\omega)+\cdots+Y(\theta^{n-1}\omega)}{n}=\dfrac{X_{n}}{n}\longrightarrow_{a.s,\ L^{1}}\inf_{n\in\mathbb{N}}\dfrac{(\mathbb{E}X_{n}|\mathcal{J})}{n}.$$
However, how do I show $$\inf_{n\in\mathbb{N}}\dfrac{(\mathbb{E}X_{n}|\mathcal{J})}{n}=\mathbb{E}(Y|\mathcal{J})?$$
Thank you!
It suffices to show that $$\mathbb{E}[Y \circ \theta^k | \mathcal{J}] = \mathbb{E}[Y | \mathcal{J}]$$ since then $\mathbb{E}[X_n | \mathcal{J}] = n \mathbb{E}[Y | \mathcal{J}]$ for any $n$.
For this, pick an $A \in \mathcal{J}$. We need to show that $$\mathbb{E}[1_A Y \circ \theta^k ] = \mathbb{E}[1_A \mathbb{E}[Y| \mathcal{J}]]$$
The right hand side is given by $\mathbb{E}[1_A Y]$. For the left hand side, note that since $A \in \mathcal{J}$, $\theta^{-k}(A) = A$. Then we have that $1_A \circ \theta^k = 1_{\theta^{-k}(A)} = 1_A$ and so the left hand side is nothing but $$\mathbb{E}[(1_A Y) \circ \theta^k] = \int_{\theta^k(\Omega)} 1_A Y d \theta_*^k \mathbb{P} = \int_{\Omega} 1_A Y d \theta_*^k \mathbb{P} = \mathbb{E}[1_A Y]$$ since $\theta^k$ is measure preserving and $\mathbb{P}(\theta^k(\Omega)) = 1$ as a result. Indeed, $$\mathbb{P}(\theta^k(\Omega)) = \mathbb{P}(\theta^{-k}(\theta^k(\Omega))) \geq \mathbb{P}(\Omega) = 1$$ since $\Omega \subseteq \theta^{-k}(\theta^k(\Omega))$.