Let $\varphi : M \to M$ be a measure-preserving map of a measure space $M$ with measure $\mu$, and let $f \in L^1(\mu)$ be arbitrary. If $p$ is the starting point of an orbit that is dense in $M$, does $$\lim_{N\to\infty}\sum_{n=0}^{N-1}f(\varphi^n(p)) = \int_M f \ d\mu?$$ In general the answer is no. Are there nontrivial conditions on $M$, $\mu$, and $p$ that would make the answer a yes?
In the case that I'm interested in, $M$ is also a compact smooth manifold and $\varphi$ is smooth.
I'm looking at the case where $M$ is the 2-torus $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$, $\mu$ is the obvious measure such that $\mu(\mathbb{T}^2) = 1$, and $\varphi$ is a hyperbolic toral automorphism, i.e. a map induced by an invertible linear transformation $L : \mathbb{R}^2 \to \mathbb{R}^2$ that preserves the integer lattice (i.e. has determinant $\pm 1$) and has no eigenvalues of modulus 1. I would like to prove that such maps are ergodic, directly from the definition that the time mean equals the space mean almost everywhere. An answer to the above question might help me with that.
If you can prove that
$$\lim_{N\to\infty} \frac{1}{N}\sum_{n=0}^{N-1}f(\phi^n(p)) = \int_M f \ d\mu$$ for all p and all $f $ integrable
than you are proving more than ergodicity. You are proving that the system is uniquely ergodic. Your example is not uniquely ergodic. So I think be hard prove ergodicity by the way you want. Because you should be able selected this points where is not true the equality above, and check that the set of such points has measure $0$.