atomic ergodic measure

Question

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.

Answer 1

Assume that $m(x) >0$. The orbit $O(x)$ of $x$ is $T$-invariant, and $m(O(x)) \geq m(x) > 0$. By ergodicity, $m(O(x)) = 1$.