Let $M$ be a Riemannian manifold and $\phi^t$ an Anosov flow on $M$. If $\phi^t$ is measure preserving (with respect to any Borel-measure on $M$), it is ergodic. Does anybody have a proof of that statement?
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2$\begingroup$ A measure preserving Anosov flow need not be ergodic, see page 10 (free access) of Introduction to Smooth Ergodic Theory by L. Barreira and Y. Pesin. $\endgroup$– user539887Commented May 1, 2018 at 10:08
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An Anosov flow has many periodic orbits. Take the mean of two Dirac measures on two distinct periodic orbits and you get an invariant probability measure that is not ergodic. The standard proof given by Anosov of the ergodicity of the Lebesgue measure, assuming it is invariant, uses the product structure of the measure (which in that case is given by the Fubini theorem). The sum of two Dirac measures of course does not have such a structure.
