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In the book "Markov Chains and Stochastic stability" by Meyn and Tweedie the measurable space $(\mathsf X,\mathcal{B}(\mathsf X))$ is said to be $\varphi$-irreducible for a Markov Chain $X$ if there exists a measure $\varphi$ on $\mathcal B(\mathsf X)$ such that $\varphi(A)>0$ implies $L(x;A)>0$ for any $x\in\mathsf X,A\in\mathcal B(\mathsf X)$. Here $$L(x;A) = \mathsf P\{\text{there is }k\ge1\text{ such that }X_k\in A|X_0 = x\}.$$

I wonder if they mean only the measures $\varphi(\mathsf X)>0$ (i.e. non-trivial). The book is from 1993 and I didn't find there a convention that all measures are assumed to be non-trivial. Moreover, I saw that sometimes they assume non-triviality of some measures explicitly in the text (e.g. the definition of small sets and further in Chapter 5).

My doubts are the following. If $\varphi(\mathsf X) = 0$ then clearly it is an irreducibility measure. Moreover, as in Proposition 4.2.2 it's written that the maximal irreducibility measure $\psi$ is equivalent to $$ \psi'(A) = \int\limits_\mathsf X K_{a_{\frac12}}(y,A)\varphi(dy) $$ for any finite irreducibility measure $\varphi$ for $$ K_{a_\frac12}(y,A) = \sum\limits_{n=0}^\infty \frac{1}{2^{n+1}}\mathsf P\{X_n\in A|X_0=y\}. $$ Clearly, any trivial measure is finite (unless there is a special convention) and hence the maximal irreducibility measure $\psi$ is always trivial which of course is not true.

I also wonder if there Markov Chains which are not $\psi$-irreducible. The confusion here is that $\psi$ is not a fixed measure, but a maximal irreducible measure which depends on Markov Chain.

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How is Markov chain splitting technique useful for inferring ergodicity of a Markov Chain? I am quite confused after reading Meyn Tweedie book. – user24367 Jun 18 '13 at 13:20
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Luckily my library had this book, so I just took a quick look at it. You seemed to have answered your own questions. You are correct in noticing that according to their definition, the trivial zero measure $\mu$ is always an irreducibility measure, because it trivially satisfies the property that whenever $\mu(A)>0$, $L(x;A) > 0$ for any $x\in X$ and any $A \in B(X)$.

You are also correct that if the maximal irreducibility measure were equivalent to the measure $\psi^\prime(A) = \int_X \varphi(dy) K_{a_{\frac{1}{2}}}(y,A)$ for any finite irreducibility measure $\varphi$ as in their proposition 4.2.2, then since the trivial measure is irreducible, this integral can beget the trivial measure. And any measure equivalent to the trivial measure is trivial, where equivalence is defined here:

I believe this must be a mistake, and it would be easily fixed by only allowing non-trivial measures as irreducibility measures.

To answer your question about a Markov chain without a maximal irreducibility measure, consider this. Proposition 4.2.2 guarantees the existence of a maximal irreducibility measure as long as there exists any irreducibility measure. So such a Markov chain would have to have no non-trivial irreducibility measures. For an example of such, let $X=\{a,b\}$, and let $B(X) = \{\emptyset, \{a\},\{b\},\{a,b\}\}$. Let the Markov chain just jump from $a$ to $a$ and $b$ to $b$ with probability one. If $\varphi$ is a nontrivial measure ($\varphi(X)>0$), it must assign ${a}$ or ${b}$ positive mass to satisfy additivity. But $L(b,\{a\})=L(a,\{b\})=0$, so $\varphi$ can't satisfy the definition of an irreducibility measure.

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I understand that ergodic chains converge to the stationary distribution in total variation. But then this is true for any Harris recurrent chain. So how is the technique of chain splitting useful in proving that a chain is ergodic? – user24367 Jun 18 '13 at 13:24

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