I am currently going through time and state discrete Markov Chains.
note $\tau_{x}:=\inf\{n \in \mathbb N:X_{n}=x\}$
and note $G(x,x):=\sum\limits_{n\in \mathbb N_{0}}P^{x}(X_{n}=x)=\mathbb E^{x}[\sum\limits_{n\in \mathbb N_{0}}1_{\{X_{n}=x}\}]$, in other words $G(x,x)$ is the expected frequency with which we visit
and $P^{x}$ describes the given distribution with start in state $x$.
My definition of Recurrence for a state $x$ is that:
$P^{x}(\tau_{x}<\infty)=1$ and $P^{x}(X_{n}=x \operatorname{for infinitely many n})=1$ and $G(x,x)=\infty$
I have already shown that:
$P^{x}(\tau_{x}<\infty)=1\Rightarrow$ $P^{x}(X_{n}=x \operatorname{for infinitely many n})=1$ and $G(x,x)=\infty$
But in some proofs, for example of simple symmetrical random walks, we only show that $G(x,x)=\infty$ and then simply say $x$ is recurrent. I think it is trivial but I just want to make sure my reasoning is sound:
Since $x$ is visited infinitely often, it has to follow $X_{n}=x \operatorname{for infinitely many n}$ almost-surely, otherwise we would have a contradiction, wouldn't we? And then from $X_{n}=x \operatorname{for infinitely many n}$ almost-surely, $\tau_{x}<\infty$ almost-surely follows immediately. I know that Borel-Cantelli may be used when $G(x,x)=\infty$ only when $\{X_{n}=x\}_{n\in \mathbb N}$ is independent, but is this sequence independent?