Edit info : In the following we denote by "$1$" as the neutral of our group, hence "$1$"$=0$ for ($\mathbb{Z},+)$.
A random walk on $\Gamma$ is an infinite sequence $X_0, X_1,\ldots$ of $\Gamma$−valued random variables, all defined on the same probability space $(\Omega, F, P^x)$ of the form $$X_n = X_0\xi_1\xi_2\ldots\xi_n$$ where $\xi_1, \xi_2,\ldots$ are independent, identically distributed random variables that take values in $\Gamma$.
A random walk $X_n$ on a finitely generated group $\Gamma$ is said to be recurrent if the event $\{X_n = 1 \text{ for some }n\geq 1\}$ that the random walk eventually returns to its initial point has probability $1$.
Problem I'm reading this course on random walks on infinite groups. Page 18, they prove Kesten-Spitzer-Whitman Theorem :
KSW Theorem: For any random walk $X_n$ on any group,
$$\lim_{n\to\infty} \frac{R_n}{n} = \mathbb{P}[\text{no return to }1] = \mathbb{P}[X_n\neq 1 \ \forall n\geq 1] \quad \text{ almost surely}$$
Where $R_n$ is the range of the random walk, i.e. the number of distinct sites visited by time $n$. Formally, with $X_n=\xi_1\xi_2\ldots\xi_n$,
$$ R_n = \vert \{ X_0,X_1,\ldots,X_n\}\vert = \vert \{ 1, \xi_1, \xi_1\xi_2, \ldots, \xi_1\xi_2\ldots\xi_n\}\vert$$
Then, they state a corollary,
Corollary Any irreducible random walk $X_n=\sum_{i=1}^n\xi_i$ on the integers $\mathbb{Z}$ whose step distribution has finite first moment $\mathbb{E}\vert\xi_i\vert$ and mean $\mathbb{E}\xi_1=0$ is recurrent.
Question why does the corollary requires the irreducibility of the random walk? KSW theorem doesn't and I can't see any restriction in the proof.
Proof of corollary Using the strong law of large number $$ \frac{X_n}{n} = \frac{\sum \xi_i}{n} \xrightarrow{iid} \mathbb{E}\xi_1=0 \quad a.s.$$ Therefore with probability $1$, $$ \forall \varepsilon>0,\ \exists n_\varepsilon, \ \forall n\geq n_\varepsilon, \left\vert\frac{X_n}{n}\right\vert < \varepsilon $$ And with probability $1$, $$\forall \varepsilon>0,\ \exists n_\varepsilon, \ \forall n\geq n_\varepsilon, \ R_n \leq 2n\varepsilon+n_\varepsilon$$
Therefore with probability $1$, $$ \forall \varepsilon>0, \ \lim_{n\to\infty} \frac{R_n}{n} \leq 2\varepsilon$$
This is true for any arbitrary $\varepsilon$, therefore $R_n/n \to 0$ a.s., and by KSW theorem, $X_n$ is recurrent.
I also tried to see if an irreducible random walk would trivially always or never return to 1. But the constraint $\mathbb{E}\xi_1=0$ is quite strong and avoid any step distribution with only positive weight on positive integer for instance.