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I'm dealing with a continuous flow on a compact metric space $X$, and $\mu$, $\nu$ are two invariant Borel probability measures on $X$. If I know that $\mu(A)=\nu(A)$ for all the invariant Borel subsets $A$, can I conclude that $\mu=\nu$?

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Ian's answer uses two two ingredients, which are not really necessary: the Birkhoff ergodic theorem and continuity of the state space. One can do it in a more direct way just in the Borel category. Let me first notice that without loss of generality one may assume that the measures $\mu$ and $\nu$ are equivalent (if not, one can pass from the couple $(\mu,\nu)$ to the couple $(2\mu+\nu,\mu+2\nu)$). Therefore, the Radon--Nikodym derivative $f=d\nu/d\mu$ is well-defined, and, since both $\mu$ and $\nu$ are invariant, it is also invariant. Now, if $\mu\neq\nu$, there is $t<1$ such that $A=\{f\le t\}$ has positive measure. Then obviously $\mu(A)\neq\nu(A)$.

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Let $\phi_t$ denote the flow map at time $t$. Suppose that $\mu$ and $\nu$ agree on every invariant Borel set,and let $f \colon X \to \mathbb{R}$ be continuous. Define $\overline{f}:=\liminf_{t \to \infty} \frac{1}{t}\int_0^tf(\phi_tx)$ for every $x \in X$, which is a Borel function and satisfies $\overline{f}(\phi_\tau x)=\overline{f}(x)$ for every $x \in X$ and $\tau>0$. The level sets $\overline{f }^{-1}[a,b]$ are invariant Borel sets, so are given equal measure by $\mu$ and $\nu$, and it follows that $\int \overline{f}d\mu=\int \overline{f}d\nu$. By the Birkhoff ergodic theorem for general invariant measures we have $\frac{1}{t}\int_0^tf(\phi_tx ) \to \mathbb{E}_\nu(f |\mathcal{I})$ almost everywhere and in $L^1$ with respect to $\nu$, where $\mathbb{E}_\nu$ denotes conditional expectation and $\mathcal{I}$ denotes the $\sigma$-algebra of flow-invariant Borel sets. In particular $\overline{f}=\mathbb{E}_\nu(f|\mathcal{I})$ almost everywhere with respect to $\nu$, and hence $\int \overline{f}d\nu = \int \mathbb{E}_\nu(f|\mathcal{I})d\nu=\int fd\nu$. By the same reasoning $\int \overline{f}d\mu = \int fd\mu$, and hence $\int fd\mu=\int fd\nu$. It follows that $\nu$ and $\mu$ agree on every continuous function and hence are equal.

(I am not particularly used to working with flows as opposed to maps, so it is possible that I may have missed a subtle point relating to measurabilty. In any event I think that the above should work anyway if we retain the same definition of $\overline{f}$ but apply the ergodic theorem in discrete time.)

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    $\begingroup$ All looks solid to me. $\endgroup$ – Anthony Quas Mar 6 '14 at 17:18
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    $\begingroup$ Like @RW says, you don't really need the assumption that the flow is topological. Suppose $\mu$ and $\nu$ are distinct $\phi$-invariant measures. So, there is a measurable set $B$ such that $\mu(B)\neq\nu(B)$. Let $f:=1_B$ be the indicator of $B$ and $\bar{f}$ as you defined. By the ergodic theorem, $\pi(\bar{f})=\pi(B)\neq\nu(B)=\nu(\bar{f})$. Since $\bar{f}$ is measurable with respect to the $\phi$-invariant $\sigma$-algebra, it follows that $\pi$ and $\nu$ are distinct when restricted to the $\phi$-invariant $\sigma$-algebra (because their associated integrations are distinct). $\endgroup$ – Algernon Mar 8 '14 at 12:59

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