Let $m$ be a atomic Borel probability measure on $X$. Let $\phi$ be a homeomorphism on $X$. I want to prove that $m$ is $\phi$ Ergodic if and only if $m$ is concentrated on a single $\phi$ orbit. I tried to proceed like this. Let $x \in X$ and $O(x)$ denotes the orbit of $x$. If $O(x)\neq X$, there exists $y \in X$ such that $y$ does not belong to $O(x)$. Since the measure is atomic $y$ belongs to some atom $V_y$ with $m(V_y)>0$. So for each $y \in X-O(x)$, we get $V_y$ with $m(V_y)>0$. I could not get a contradiction.
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Assume that $m(x) >0$. The orbit $O(x)$ of $x$ is $T$-invariant, and $m(O(x)) \geq m(x) > 0$. By ergodicity, $m(O(x)) = 1$.
D. Thomine
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will atomic measure imply that $m( {x})$>0? – budi May 14 '20 at 17:34
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if the measure is atomic, then there exists some $x$ such that $m(x) >0$. – D. Thomine May 14 '20 at 18:17
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Singleton ${x}$ with $m({x})>0$ are atoms. But how can prove that there is atleast one such exists? – budi May 15 '20 at 09:48
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Do you know what an atomic measure is ? – D. Thomine May 15 '20 at 10:40
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yes. First an atom in a measure space $(X,\mathcal{M},m)$ is a measurable subset $B\in ,\mathcal{M}$ such that $m(B)>0$ and for any $A\subseteq B$ with $A \in \mathcal{M}$, either $m(A)=0$ or $m(A)=m(B)$. A measure space is atomic if every element belongs to some atom. – budi May 15 '20 at 12:44
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Oh, that's the slightly-more-general-and-infinitely-more-annoying definition of atomic measure space. Since your space is Borel, this should answer your question : https://math.stackexchange.com/questions/1473874/must-atoms-of-a-borel-measure-space-be-singletons – D. Thomine May 15 '20 at 17:05