Let $(X_n)_{n \geq 1}$ be a recurrent irreducible Markov chain on a countable space. Let $a$ be a fixed point and $\tau$ be a stopping time such that almost surely $X_{\tau} = a$. For any $x,y$ let $G(x,y)$ be the expected number of visits to $y$, starting from $x$, for $n < \tau$:
$$ G(x,y) = \mathbf{E}_x \left[ \sum_{0 \leq n < \tau} \mathbf{1}_{X_n=y} \right] $$
We assume that $G(a,a) < \infty$ and want to show that then $G(a,x) < \infty$ for any $x$.
I have no idea how to do this, even if $\tau$ is the stopping time "first visit at $a$".
Fact: $$ \sum_{n \geq 0} \mathbf{P}_a (X_{n+1}=a, \tau > n) = \sum_{n \geq 0} \mathbf{P}_a (X_{n}=a, \tau > n) $$ This is a little bit counter-intuive because the summands are not equal. This identity follows from the fact that $X_\tau = a$ a.s. starting from $a$. The good thing is that the event $\tau >n $ does not depend on $X_{n+1}=a$.
Now the right-hand side is $G(a,a) < \infty$. Let us expand the left-hand side using Markov property: $$ \mathbf{P}_a (X_{n+1}=a, \tau > n) = \sum_b p_{ba} \mathbf{P}_a (X_n=b, \tau > n) $$ and summing over $n$ we get $$ G(a,a) = \sum_b p_{ba} G(a,b) $$ so for all $b$ such that $p_{ba} > 0$ we get $G(a,b) < \infty$.
More generally, for any $x$ there holds $$ G(a,x) = \sum_b p_{bx} G(a,b) $$ and we conclude, using irreducibility, that $G(a,x) <\infty$ for any $x$.
It seems to me that we didn't use the recurrence but please correct me if I am wrong.
This argument comes from https://www.stat.berkeley.edu/~aldous/RWG/book.pdf Proposition 2.3