recurrence relations - Markov-Chain with general state space - recurrent sets

I have an irreducible Markov Chain $(z_n )_{n\in \mathbb N }$ with state space $X$ and with transition-probability-kernel $K$, so $K(x,\cdot)$ is a probability measure (on the $\sigma$-Algebra $\mathcal{B}(X)$).
$\tau_B$ is the first entry time of the chain into the set $B$, ie $\tau_B :=\mathop{min}\{i\in \mathbb N | i\geq 1, z_i\in B\}$.

A set $B\in\mathcal{B}(X)$ was defined as recurrent, if

$\mathbb P_x\{\tau_B<\infty\}=1\qquad\forall x\in X$,

with $\mathbb P_x$ being the law of the chain starting in $x$.
Now lets assume there is a recurrent set $B$ and a set $C$ so that for some $\alpha >0$

$K(x,B)\leq\alpha K(x,C)\qquad \forall x\in X$.

I am quite sure, that $C$ has to be recurrent itself, because the chain would visit $C$ once for every $\frac{1}{\alpha}$-times it visits $B$. But how do I formally prove it?

Answer

$\alpha \geq 1$ means the chance to hit $C$ at any time is greater than the chance to hit $B$, so the chance to hit it in finite time will be greater or equal and we are done. In the case $\alpha <1$ let's say the stopping time of the $k$th-entry to $B$ is $\tau_B (k)$.

With the (strong) Markov property we get

$\mathbb{P}_x \{\tau_B (k)<\infty\}=1\quad\forall k\in \mathbb{N}_1$ which we shall call statement (I) and

$\mathbb{P}_x \{\tau_B (k+1)-\tau_B (k)<\infty\}=1\quad\forall k\in \mathbb{N}_1$ which we call statement (II).

Now because of $K(x,B)\leq αK(x,C)\quad \forall x\in X$,
we have

$ \mathbb{P}_x \{\tau_C > \tau_B (k) \}\leq \prod_{i=1}^{k} (1-\alpha\sum_{j=1}^{\infty} \mathbb{P}_x \{ z_{\tau_B (i)+j} \in B;\ z_{\tau_B (i)+m} \not\in B\ \forall m

as the sum is always equal to 1 (because of (II) ). One can understand it like this: each time $B$ was hit (for a total of $k$ times) there also was a chance to hit $C$, but $C$ was dodged.

Taking $k\rightarrow \infty$ together with (I) gives $\mathbb{P}_x \{\tau_C =\infty\}=0\quad\forall x\in X$.