Some (hand-wavy) context: Suppose that we have found the invariant measures to a Markov Process, $(X_t)_{t=1}^\infty$, on a compact set $S$ by the following the process itself, i.e. looking at the weak-$\ast$ limit points of empirical measures $\frac{1}{t}\sum_{s=0}^{t-1} \delta_{X(t)}(\cdot )$.
Question: Are we guaranteed that there are no other invariant measures to the process $(X(t))_{t=1}^{\infty}$? As in, we can follow the process (say with any initial condition $X(0) = x$) and find the invariant measures to this process by following the process itself, but are we missing any others? Let's say $\Omega$ is actually just $\mathbb{R}^n$.
Attempt: I tried to do something in a more general format. I took an identity map between $\imath: (\Omega, \phi, \tilde \mu) \to (\Omega, \phi, \mu) $, where $\tilde \mu$ and $\mu$ are both invariant measures to the process $X(t)$, and $\phi$ is a measure-preserving map for both spaces. Then I used Birkhoff's Ergodic Theorem to say that:
$\mathbb{E}(\mathbb{1}_A | \tilde I) = \lim_{t\to\infty}\frac{1}{t}\sum_{s=0}^{t-1} \mathbb{1}_A(\phi^s(x)) = \frac{1}{t}\sum_{s=0}^{t-1} \mathbb{1}_A(\imath( \phi^s(x))) = \mathbb{E}(\mathbb{1}_A | I)$,
where $I$ and $\tilde I$ are the invariant sets of $\mu$ and $\tilde \mu$, respectively. Then if we took the expectation of both sides of the equation, then we'd get $\mu(A) = \tilde \mu(A)$ for all $A$ measurable. In particular, In particular if $\tilde A \in \tilde I$ was an invariant set of $\tilde \mu$, then $\mu(\tilde A) = \tilde \mu(\imath^{-1}(\tilde A)) = \tilde \mu((\phi^{-1} \imath^{-1}(\tilde A))) = \tilde \mu((\phi^{-1}(\tilde A)) = \mu(\phi^{-1}(\tilde A))$, so that $\tilde A$ is an invariant set of $I$ as well. But something seems a bit iffy about what I'm doing and I wonder if there's just an easy answer.
Take a Markov chain with states $(a,b,c)$ and connections $a\rightarrow b$, $a\rightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $\mu_1=(0,1,0)$, $\mu_2=(0,0,1)$ and $\mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $\mu_1$ and $\mu_2$. Yet, $\mu_3$ is not achievable by starting from any state in $i\in(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.