How to show that the limit in Birkhoff theorem can be taken for $ n\to-\infty $ when the system is reversible?

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Assume that $ (X,\mathcal{B},m,T) $ is a measure-preserving dynamical system, where $ (X,\mathcal{B},m) $ is a probability space, $ \mathcal{B} $ denotes all the measurable sets in $ X $, $ m $ is the measure on it and $ T $ is a measure-preserving map, i.e. for any $ B\in\mathcal{B} $, we have $ m(T^{-1}B)=m(B) $. If $ T $ is reversible, i.e. $ T^{-1} $ is also a measure-preserving map, show that $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}f(T^ix)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}f(T^{-i}x) $$ for $ m $-a.e. $ x\in X $.

In view of the famous Birkhoff theorem, the esxistence of limits for the RHS and LHS is ensured. However, I cannot show that the limits are the same. Can you give me some hints or references?

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The answers using conditional expectations are right. But if you wish, you can see this fact using the Mean (Von Neumann) ergodic theorem. A good reference to this is the book of Einsiedler and Ward (your question correspond to the exercise 2.6.3). Let me explain some ideas to proof this. Let $(X, \mathcal{B},\mu)$ be a probability space and $T\colon X \to X $ a measurable automorphism. Consider $U_Tf := f\circ T $ the associated Koopman operator defined on $L^1(X,\mathcal{B},\mu)$. Note that $ U_T^j = U_{T^j}$. Let us denote the avarage of $f \in L^1$ under $T$ and $T^{-1}$ by $$A_nf := \frac{1}{n}\sum_{j=0}^{n-1}U^j_{T}f \quad \text{and} \quad A^{-1}_nf := \frac{1}{n}\sum_{j=0}^{n-1}U^j_{T^{-1}}f. $$ Note that we can restrict $U_T$ to the Hilbert space $L^2 \subset L^1$. The Mean ergodic theorem (see Theorem 2.2.1 in 1) tell us that for each $ f \in L^2(X,\mathcal{B},\mu)$, we have $$ A_nf \longrightarrow P_Tf \quad \text{and} \quad A^{-1}_nf \longrightarrow P_{T^{-1}}f \quad \text{as}\; n \to \infty \;\;\; \text{in}\; L^{2},$$
where $P_T\colon L^2 \to \mathcal{I}(T)$ denote the orthogonal projection on the closed subspace of invariant functions: $\mathcal{I}(T) := \ker(U_T - id) = \{ g \in L^2 : U_Tg = g\}$. Then, it is easy to see that $\mathcal{I}(T) = \mathcal{I}(T^{-1})$, thus $P_T = P_{T^{-1}}$. Notice that the conditional expectation as mentioned in the other answers is exactly $P_T$ in the case of $L^2$. We conclude that $A_nf$ and $A^{-1}_nf$ converge to the same function in $L^2$. Let us see that this is also true in $L^1$:

Corollary 2.2.2. (1): If $f \in L^1$, then there exists $f^* \in L^1$ such that $$\lim_{n\to\infty} A_nf = \lim_{n\to\infty}A^{-1}_nf = f^* \quad \text{in}\; L^1.$$

Steps to proof: Step 1: For each $f \in L^{\infty}$, we already know that there exists $f^* = P_Tf \in L^2$ such that $$ \lim A_nf = \lim A^{-1}_nf = f^* \quad \text{in}\; L^2.$$ Step 2: Prove that $f^* \in L^\infty$. Step 3: Using that $\|\cdot\|_1 \leq \|\cdot\|_2$, we can obtain that $$ \lim_{n\to\infty} A_nf = \lim_{n \to \infty} A^{-1}_nf = f^* \quad \text{in} \; L^1$$ Step 4: The corollary holds true for $L^\infty$ functions which is dense in $L^1$, proceed now by an approximation method to conclude to $L^1$.

Finally, note that Birkhoff theorem tell us that for every $f \in L^1$, there exists $f^* \in L^1$ such that $A_nf \to f^*$ almost everywhere and in $L^1$. So, $A^{-1}_nf$ must converge to the same function $f^*$ almost everywhere.

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Hint: As mentioned in the comments above, the Birkhoff Ergodic Theorem gives that the time averages of the observable $f$ w/r/t $T$ converge on a full $\mu$-measure set to the function $\mathbb{E}_\mu(f\,|\, \mathcal{I}(T))$ that is $f$ conditioned on the $\sigma$-algebra of $T$-invariant measurable subsets. But then $\mathcal{I}(T) = \mathcal{I}(T^{-1})$.