Here is a problem I am trying to solve.
Let $G$ be a compact group acting continuously on a locally compact metrix space $X$. Suppose $\mu$ is an $G$- invariant ergodic Borel probability measure of $X$. Prove that there exists some $x\in X$ such that $\mu = (\phi^{x})_{*} m_{G}$, where $m_{G}$ is the normalized Haar measure of $G$ and $\phi^{x}: G \to X, g\mapsto g.x$.
Here is some attempt I have tried. For any $x\in X$, consider the set $A_x = \{g.x:g\in G\}$. Then it is a $G$-invariant measurable set so it is either null or full measure. If $G$ acts freely, then both case are easy to handle. However, I have no clue how to deal with the case that $G$ doesn't act freely.
As @Keen-ameteur has suggested, here is my reasoning for the case of free action. If there exists some $x\in X$ such that $A_x$ has full measure, then $G \cong A_x$ since the action is free. We can define a measure $m$ on $G$ by $m(B) = \mu (B.x)$, which is well-defined, regular(by homeomoephism) and is $G$-invariant. So $m$ must be the Haar measure on $G$.
If otherwise $\mu (A_x) = 0$ for all $x\in X$, we conclude that $\mu(X) = \mu(\cup_{x\in X} A_x) = 0$ which is a contradiction. Here I use free action to guarantee that $A_x$ is open for each $x$, so $\cup_{x\in X}A_x$ is open and thus measureable.
If the action is not free, I tried to quotient out the stablizer $L$ of a point. In the case that $\mu(A_x)=1$ for some $x$, we can still define a measure on the coset space $L \backslash G$ which is left $G$-invariant. However, I am not sure if such measure is unique (up to scalar) on $L\backslash G$, which is not necessarily a group.
On the other hand, if $\mu(A_x)=0,\forall x\in X$, I failed to arrive rigorously at a contradiction, since there can be uncountably many orbits in $X$.
Any hint is highly appreciated!!!