If a Markov chain is $\phi$-irreducible and has stationary distribution $\pi$, then $\phi\ll \pi$,
Proof:
We use the irreducibility of the chain to write the state space
$E = \bigcup_{n,m \in \mathbb{N}} \left\lbrace x \in E: K^n(x,A) \geq \frac{1}{m} \right\rbrace$
where $K$ is the Kernel and $A$ is a set in the corresponding Borel set $\mathcal{E}$. That is all clear.
But now here's the thing :
By countable additivity we can find $n,m \in \mathbb{N}$ and a subset $B \in \mathcal{E}$ with $\pi(B) > 0$, s.t. $K^n (x,A) \geq 1/m$ for all $x \in B$. Then it is also straightfoward. but I don't get the argument with the countable addivity, because where is here the measure, I only see a set.
You began with a set $A$ with $\phi(A)>0$, and deduced that $$K^n(x,A)\geq {1\over m} \, 1_B(x)\tag1$$ for some $m$, $n$, and $B$ with $\pi(B)>0$. Now integrate both sides of (1) with respect to $\pi$ to obtain $$\pi(A)=\pi K^n(A)\geq {1\over m} \, \pi(B)>0.$$
Since $\phi(A)>0$ implies $\pi(A)>0$, you've shown that $\phi\ll \pi$.