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Feb 10, 2020 at 16:11 comment added Julian Newman @julian This is indeed true; "irreducibility" here can be taken to mean e.g. "for each $\lambda$-positive measure open set $U$, $P(x,U)>0$ for all $x$". In my proof, I used the much stronger fact that for every $\lambda$-positive measure Borel set $A$, $P(x,A)>0$ for all $x$. This does imply that $\lambda$ is the unique stationary measure, without requiring additional assumptions such as the strong Feller property.
Feb 10, 2020 at 14:57 comment added julian To be utterly precise, irreducibility of the semigroup doesn't yet give you unique for the invariant measure. You need in addition strong Feller at some positive time. A famous counterexample would be the Ising model in dimension $d\geq 2$ above the critical temperature.
Jul 16, 2015 at 0:37 history answered Julian Newman CC BY-SA 3.0