One-dimensional simple symmetric walk is recurrent

Define $e:\mathbb{N}\to \mathbb{Z}$ by $e(0)=0$, $e(2k-1)=k$, and $e(2k)=-k$ for $k\ge 1$. Every integer occurs exactly once, so this is a bijection and $\mathbb{Z}$ is countably infinite, hence at most countable as required by the chain-law and recurrence suppliers.

For each fixed $z$, the two Dirac measures in the displayed definition of $K(z,\cdot )$ are probability measures [F2]; their weighted sum is a measure by [F3], and its total mass is $1\mathrm{/}2+1\mathrm{/}2=1$. For each fixed $A\subseteq \mathbb{Z}$, the map $z\mapsto K(z,A)$ is measurable because every subset of the discrete source $\mathbb{Z}$ is measurable [F5]. Thus [F4] makes $K$ a probability kernel. Its singleton entries agree with [F1], so this is exactly the d=1 simple symmetric walk kernel.

Under AC [A1], [F6] supplies the canonical law with initial measure ${\delta }_z$ for each fixed $z$; [F7] names it ${\mathbb{P}}_z$, and [F8] gives the probabilities of its finite path cylinders. The transition matrix of this chain is the matrix in [F9].

Fix $z\in \mathbb{Z}$ and $n\ge 0$. A sign word $({\epsilon }_1,\dots ,{\epsilon }_{2n})\in \{-1,1{\}}^{2n}$ determines the path $X_j=z+{\sum }_{i=1}^j{\epsilon }_i$ for $0\le j\le 2n$. Each such cylinder has probability $2^{-2n}$ by [F1], [F7], and [F8]. Distinct sign words give disjoint cylinders, and their endpoint equals $z$ exactly when the word has equally many $+1$ and $-1$ entries. By [F10] there are $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$ such words. Consequently $p^{(2n)}(z,z)={\mathbb{P}}_z(X_{2n}=z)=4^{-n}\left(\genfrac{}{}{0px}{}{2n}{n}\right)\mathrm{.}$ For $n=0$, this says $p^{(0)}(z,z)=1$, as required by [F9].

A sum of an odd number of $\pm 1$ increments is odd and cannot be zero. Thus $p^{(2n+1)}(z,z)=0$ for every $n\ge 0$.

Put $a_n:=p^{(2n)}(z,z)$. By step 1.4 and [F14], $\sqrt{\pi n}{a}_n\to 1$ as $n\to \mathrm{\infty }$. Hence there is $N\ge 1$ such that for $n\ge N$, $a_n\ge \frac{1}{2\sqrt{\pi }\sqrt{n}}\mathrm{.}$

For each integer $m\ge 1$, there are $2m+1$ integers in $[m^2,(m+1{)}^2)$, and for each one $n^{-1\mathrm{/}2}\ge (m+1{)}^{-1}$. Therefore ${\sum }_{n=m^2}^{(m+1{)}^2-1}{n}^{-1\mathrm{/}2}\ge \frac{2m+1}{m+1}>1.$ Infinitely many such disjoint blocks show ${\sum }_{n\ge N}{n}^{-1\mathrm{/}2}=\mathrm{\infty }$; [step 2.1] then gives ${\sum }_{n\ge 0}{p}^{(2n)}(z,z)=\mathrm{\infty }$. The odd-time terms are zero by step 1.5, so the full Green series ${\sum }_{n\ge 0}{p}^{(n)}(z,z)$ diverges. The initial $n=0$ term is $1$ and is included.

By [F11], divergence of the Green series implies that this fixed state $z$ is recurrent; by [F12] and [F13], this means exactly ${\mathbb{P}}_z(T_z^{+}<\mathrm{\infty })=1$. Since $z$ was arbitrary, the conclusion holds for every state of $\mathbb{Z}$. This uses only the forward implication of [F11], and no converse to the corollary is asserted.

The state space is fixed as $\mathbb{Z}$, which contains $0$ and infinitely many integers, so the empty-space and one-state cases are inapplicable. Every row has two distinct positive transitions, so the walk is neither absorbing nor deterministic. Zero return weights occur at every odd time by step 1.5; the time-zero Green term is $1$ but is not a positive-time return [F12, F13]. AC [A1] is used exactly for the canonical path law and the stated finite-dimensional and recurrence suppliers. The enumeration of $\mathbb{Z}$, the finite path count, and the square-block divergence require no choice. The only iff input is [F11]; the proof uses its divergence-to-recurrence direction, so the reverse direction is not part of the claim.