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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.

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I think you want $f$ in the integral, not $\phi$. // Since $f$ is defined only up to sets of measure zero, the sum on the left is not even a natural thing to consider, unless something is assumed about $p$, e.g., $\phi^n(p)$ is a Lebesgue point of $f$ for every $n$. –  user53153 Jan 18 '13 at 0:01
    
Indeed, I meant $f$. –  Ricardo Buring Jan 18 '13 at 0:09
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2 Answers

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 $0$.

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I'm trying to prove that equality for almost all $p$. For a candidate set of $p$'s of measure 1 I had in mind $\mathbb{T}^2 \backslash W^s$, where $W^s$ is the projection of the stable eigenspace of $L$ (which has measure zero). If I'm not mistaken, orbits starting in $\mathbb{T}^2 \backslash W^s$ are dense in $\mathbb{T}^2$, hence the question. –  Ricardo Buring Jan 18 '13 at 0:01
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Interesting!! If you get it, post for us! –  user52188 Jan 18 '13 at 0:05
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All systems ergodic that a see the proof of ergodicity used other ways(include your example). You can prove direct from your definition for another example? –  user52188 Jan 18 '13 at 0:17
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What you are looking for is the existence of SRB measures. The traditional Ergodic theorem only guarantees that the time and space averages are equal for $p$ a.e. w.r.t the invariant measure $\mu$, which may actually be zero measure in the phase space (e.g. take the case where $\mu$ is a point measure supported on an attracting fixed point or periodic orbit). SRB measures are physical measures, and existence of such a measure imply that the "time average=space average" holds for a Lebesgue measure>0 set of initial conditions.

Look for work by L.S. Young on SRB measures to know more. As far as I know, there are very few systems where SRB measures are rigorously known to exist. Axiom-A systems would be one of those rare class of systems.

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