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 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.

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 $\phi P(A) = \int\limits_E P(y,A)\phi(\mathrm dy)$.