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I'm trying to prove that following :

Let $(X,T)$ by a dynamical system such that $T$ is an isometry. Then $(X,T)$ est uniquely ergodic iff $(X,T)$ is minimal.

So far, I've been unsuccessful on both sufficient and necessary condition.

On the "then" statement, I did the following: let $\nu$ be the unique invariant measure on $(X,T)$. Suppose $x \in X$ has not a dense orbit in $X$. Then, there exists some open set $V$ such that $T^n(x)$ never encounters $V$ as $n$ runs through $\mathbb{N}$. By considering $\mu_n = (1/n) \sum_{k = 0}^{n-1} \delta_{T^kx}$, then there exists a subsequence $(\mu_{n_k})_k$ converging to $\mu$ a $T-$invariant measure. By unique ergodicity, this means that $\mu = \nu$. From this, it follows easily that $\nu(V) = 0$. At first, I would have liked to use that because $T$ is isometric, any of its invariant measure must have total support, but this seems to be false anyway...

Thanks a lot!

Hermès
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  • I'm afraid I'm misunderstanding some assumptions and that this is too simplistic - but once you've shown a $T$-invariant measure $\nu$ assigns $0$ to an open set, you should be able to show it is the zero measure. – Matthew C Nov 01 '18 at 17:47
  • All the assumptions are there. – Hermès Nov 01 '18 at 18:22
  • Ok, then it should be true that the measure is zero and you have your contradiction (T maps balls $B(x,\varepsilon)$ to balls so if you take an arbitrary open set $U$, you can show that $\nu(U) =0$ by relating the measure of a ball inside $U$ to a ball inside $V$). – Matthew C Nov 01 '18 at 19:00
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    That's what I wanted to do. But nothing, a priori, assure us that for instance the ball $B$ does move under $T$. It could be fixed. – Hermès Nov 02 '18 at 11:27
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    Good point. Here's a hopefully better idea: Choose $y \in V$ and construct the same type of measure. This will be an invariant measure distinct (actually disjoint) from $\mu$; because of the isometry property, there is a fixed distance, holding for all $n$, from $T^n(y)$ to $V^C$. – Matthew C Nov 02 '18 at 14:06

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