Let $X$ be a non-compact metric space with a sub shift $\sigma: X \to X$. Do I have that the the space of $\sigma$-invariant probability measures on $X$ such that $\mu (B) = \mu (\sigma^{-1}(B))$ with the weak-* is a compact space?
I have seen this post: Space of T-invariant probability measures is compact, but the theorem used asked for the compactness of the metric space.
To be more specific I am working with the the space of infinite words with a countable alphabet: $$X = \{ (x_n)_{n\in \mathbb{N}}:x_n \in I \}, \text{where $I$ is a countable infinite subset of $\mathbb{N}$} $$ and \begin{align*} \sigma: X& \to X \\ (x_n)_{n\in \mathbb{N}}& \mapsto (x_{n+1})_{n\in \mathbb{N}}. \end{align*}
No.
The usual proof that the space of invariant measures is compact uses (separability of) the dual of the space of continuous functions on X which is only separable if X is compact.
For a simple example that the space of measures is not always compact use the full shift on a countably infinite alphabet. There are measures sitting on each of the counably infinite fixed points and this sequence does not have a convergent subsequence in the weak-* topology, hence the space of measures is not compact in this particular case.