Recurrence and transience are class properties

Fix distinct $x,y$ with $x\to y$. By [F7], the set of $n$ with $p^{(n)}(x,y)>0$ is nonempty; choose its least element. It is positive because $p^{(0)}(x,y)=0$ by [F3]. Repeatedly decompose a positive $n$-step entry by [F8]. At each decomposition, some summand is positive, so this gives a finite route $x=z_0,z_1,\dots ,z_n=y$ with $\alpha :={\prod }_{i=1}^{n}p(z_{i-1},z_i)>0.$ This is a witness for this fixed pair only; it makes no simultaneous choice of routes. Minimality of $n$ implies $z_i\mathrm{\ne }x$ and $z_i\mathrm{\ne }y$ for $1\le i<n$, since either repeated endpoint would leave a shorter positive route from $x$ to $y$. The finite-dimensional law [F9], with initial state $x$ [F6], gives ${\mathbb{P}}_x(X_0=z_0,\dots ,X_n=z_n)=\alpha \mathrm{.}$ Call this cylinder event $C$. In particular, on $C$ the chain reaches $y$ before any positive return to $x$.

If $x=y$, recurrence of $y$ is the same assertion as recurrence of $x$; also $T_x=T_y=0$ under ${\mathbb{P}}_x$, so both hitting probabilities in the stronger claim equal one. This separates time-zero hitting from the strictly positive return used in [F11].

Suppose $x$ is recurrent and $x\mathrm{\ne }y$. By [F12], the return excursions $E_1,E_2,\dots$ are iid and finite almost surely. Let $B$ be the set of completed excursion words whose first $n$ transitions follow the route in step 1.1. Since $C$ contains no return to $x$ before time $n$, the event $\{E_1\in B\}$ agrees with $C$ except on the null event that the first return to $x$ is infinite. Hence ${\mathbb{P}}_x(E_1\in B)=\alpha$, and the same holds for each excursion by identical distribution. For every $m\ge 1$, the probability that none of the first $m$ excursions begins with the route is $(1-\alpha {)}^m$. If $T_y=\mathrm{\infty }$, none of them can begin with it; therefore ${\mathbb{P}}_x(T_y=\mathrm{\infty })\le (1-\alpha {)}^m(m\ge 1)\mathrm{.}$ Since $0\le 1-\alpha <1$, [F15] makes the right side tend to zero. Thus ${\mathbb{P}}_x(T_y<\mathrm{\infty })=1$.

Still suppose $x$ is recurrent and $x\mathrm{\ne }y$. Let $H_x$ be the indicator of the measurable future-path event that no coordinate equals $x$. Put $q={\mathbb{P}}_y(T_x=\mathrm{\infty })$. Apply [F13] at the deterministic stopping time $n$ from step 1.1. Since $X_n=y$ on $C$, the conditional future probability of avoiding $x$ is $q$, so ${\mathbb{P}}_x(C\cap \{X_{n+j}\mathrm{\ne }x\text{ for all }j\ge 0\})=\alpha q\mathrm{.}$ On this event there is no positive-time return to $x$: the route has no intermediate $x$, its endpoint $y$ is not $x$, and the future avoids $x$. If $q>0$ this contradicts [F11]. Hence $q=0$ and ${\mathbb{P}}_y(T_x<\mathrm{\infty })=1$.

Under ${\mathbb{P}}_y$, let $\tau =T_x$. Step 2.2 gives $\tau <\mathrm{\infty }$ almost surely, and $\tau \ge 1$ because $x\mathrm{\ne }y$. Apply [F13] at $\tau$ to the bounded future-path indicator ${\mathbf{1}}_{\{T_y<\mathrm{\infty }\}}$. By step 2.1, its probability from $x$ is one. Thus after the chain first reaches $x$ it reaches $y$ again almost surely. This is a positive-time return from the initial state $y$, so $y$ is recurrent by [F11].

Assume $x$ is recurrent, as required for the closure claim. Let $z\in [x]$ and suppose $p(z,w)>0$. By [F14], $x\leftrightarrow z$, so some $m\ge 0$ has $p^{(m)}(x,z)>0$ by [F7]. The one-step transition $p(z,w)>0$ and [F8] give $p^{(m+1)}(x,w)\ge p^{(m)}(x,z)p(z,w)>0,$ so $x\to w$. If $w\mathrm{\ne }x$, steps 2.1 and 2.2 give $w\leftrightarrow x$; if $w=x$, membership is immediate. Thus $w\in [x]$, proving the class is closed.

Suppose $x\leftrightarrow y$. If $x$ is recurrent, then either $x=y$ as in step 1.2 or $x\to y$ and step 3.1 shows $y$ is recurrent. If $y$ is recurrent, apply the implication of steps 2.1–3.1 to the ordered pair $(y,x)$ to get that $x$ is recurrent. This proves both directions of the equivalence. By [F11], a state that is not recurrent is transient, so the classification is shared.

If $E=\mathrm{\varnothing }$, there is no starting state and the assertions are vacuous. If $E$ has one state, its only transition row has probability one on itself; the state is recurrent and its class is closed. Zero transition weights cannot appear in the chosen route because every factor is positive; the Chapman–Kolmogorov sums otherwise include all states. For a deterministic transition map, recurrence of $x$ means its orbit returns to $x$ after some positive number of steps, so the orbit is a finite cycle; every state accessible from $x$ lies on that cycle and has the stated hitting and recurrence properties. The endpoint $x=y$ gives accessibility at time zero but recurrence still uses $T_x^{+}$ at positive time [F10, F11]; distinct communicating states use a route of length at least one. AC [A1] supplies the canonical fixed-start laws and is assumed by the renewal and strong-Markov results. The finite-route witness is selected only for each fixed pair, with no global route selection. The two recurrence implications were proved in step 4.1, so both iff cases are covered.