Let $P(x,A)$ be a stochastic kernel on a measurable space $(E,\mathcal E)$ and $G = \sum\limits\_0^\infty P^n$ be its potential kernel. A $\sigma$-finite measure $\phi$ is called the irreducibility measure if $\phi(A)>0$ implies $PG(x,A)>0$ for all $x\in E$. 


In [Numellin's book](http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511526237) it is written that if $\phi$ is an irreducibility measure then so is $\phi P$. I've tried to prove this fact but didn't manage to do it, so I hope you can help me with an idea.




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Just to clarify things: kernel $P$ means that $P(x,\cdot)$ is a probability measure on $(E,\mathcal E)$ for all $x$ and $P(\cdot, A)$ is a measurable function for all $E$. The product of kernels, say $P$ and $G$ is
$$
PG(x,A) = \int\limits\_E G(y,A)P(x,\mathrm dy)
$$
and $\phi P(A) = \int\limits\_E P(y,A)\phi(\mathrm dy)$.



One point on $\phi$-irreducibility
Well, $\phi P(A) > 0$ implies $\exists B\subset E: \phi(B)>0$ and $P(y,A)>0\ \forall y\in B$. This implies by definition of $\phi$ that $PG(x,B) > 0$ for all $x\in E$, whence $P^2G(x,A) = \int P(y,A) PG(x,dy) > 0$ for all $x\in E$. But $PG(x,A) \ge P^2G(x,A)$.


