# Markov chain dichotomy

Suppose $X$ is a complete separable metric space, and there is a continuous map $x \mapsto \mu_x$ associating to each point in $X$ a probability measure on $X$ (where we use the weak topology on the space of probabilities).

For each probability $\mu$ on $X$, let $T\mu = \int \mu_x \mathrm{d}\mu(x)$. A probability $\mu$ on $X$ is stationary if $T\mu = \mu$.

If I've done things correctly $T^n\mu$ is the distribution of the $n$-th step of a Markov chain with initial distribution $\mu$ and transition probabilities given by the family $\mu_x$.

Suppose one knows that for any $\mu$ and any open set $A$ one has $T^n\mu(A) > 0$ for all $n$ large enough.

Question: In the conditions above is it true that only one of the following statements hold?

1. There is a unique stationary probability.
2. For each probability $\mu$ and each compact sets $K$ one has $\lim\limits_{n \to +\infty}T^n\mu(K) = 0$.

If $X$ is countable then the statement above can be deduced from the law of large numbers and the Markov property. One obtains that almost surely, starting with any initial distribution, the asymptotic fraction of time spent in a state is the inverse of the expected return time to that state.

I know of some results (e.g. Derriennic's version of the zero-two law[1]) implying the dichotomy in the case where $X$ is a locally compact group, and $\mu_x$ is the translate of some fixed probability $x$ under multiplication by $x$.

I was wondering if there is a standard reference for general results of this type (or maybe some instructive counter-example). From what I can tell Harris chains are not general enough for the type of examples I have in mind. I am willing to assume $X$ is locally compact if necessary.

[1]Derriennic, Yves. "Lois “zéro ou deux” pour les processus de Markov. Applications aux marches aléatoires." Ann. Inst. H. Poincaré Sect. B (NS) 12.2 (1976): 111-129.

• Pointless nitpick: As written, you just ask to show that 1 and 2 cannot hold simultaneously, which is sort of obvious: by 2, the stationary measure is $0$ on every compact set and, thereby, identically $0$ (here is where separability and completeness come into play). On the other hand, even if $X$ consists of just 2 points with the trivial Markov chain staying in place, neither 1, nor 2 holds, so I'm a bit perplexed about what exactly was meant... – fedja Jul 13 '17 at 20:17
• Hi fedja. There's a mixing condition (right before the question) which excludes your two point example. The probability of getting from one point to the other is positive after a certain number of steps. – Pablo Lessa Jul 13 '17 at 22:34
• Indeed :-). I should be more attentive. Now it makes sense. – fedja Jul 13 '17 at 22:57

• How can you come up with $X$ compact, am I missing sth? – Henry.L Jul 13 '17 at 20:34