Ergodic means for an invertible system Let $(X,B,\mu,T)$ be an invertible dynamicals system (i.e. $T^{-1}$ is measurable and exists almost everywhere)
Question 1:
is $T^{-1}$ also measure-preserving( $\mu(T(A)=\mu(A)$)?
Question 2:
if $f \in L_1(X)$
Then $\lim_{n\to\infty}1/n\sum_{i=0}^n{fT^i}=\lim_{n\to\infty}1/n\sum_{i=0}^n{fT^{-i}}$.
von Neumanns Theorem says, that the ergodic means converge to some invariant function g, but how can I see that these functions are equal? I guess somehow I have the intuition, because when I start at some value x (which orbit is not in the irregular set with measure 0), then it makes no difference in which direction ($T$ or $T^{-1}$) I start the iteration process for gaining the ergodic means...
 A: For the first question, the answer is yes. Since when $T^{-1}$ exists almost everywhere and is measurable, then $T^{-1}T(A)=A$ modulo a set of measure $0$. So $$\mu(TA)=\mu(T^{-1}T(A))=m(A).$$
For the second question, you can first check in $L^2$, then approximating an $L^1$ function by an $L^2$ function, we get the result. For $f\in L^2$, we know that the ergodic average with respect to the transformation $T$ converges to theorthogonal projection of $f$ to the space of $T$-invariant function. Now that in your case, the spaces of $T$-invariant function and $T^{-1}$-invariant function are the same. 
A: Use Birkhoff's ergodic theorem to show that there is a $T$-invariant function $f^{*}(x)$ and $T^{-1}$-invariant function $g^{*}(x)$. In fact, $g^{*}(x)$ is $T$-invariant. Define the set in which $f^{*}(x) > g^{*}(x)$ then this set is $T$-invariant. Use Birkhoff's ergodic theorem again and evaluate the integral of $f^{*}$, $g^{*}$ on the set then you can see that this set must be of measure zero.
