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For finite $G$, choose a root $r$ and an ordering of the other graph vertices. Begin with a tree consisting only of $r$. At each stage take the first graph vertex not already in the tree, run an independent simple random walk from it until its first hit of the existing tree, erase the loops chronologically, and add the resulting graph path. In chronological loop erasure, revisiting a graph vertex already in the current list deletes all list entries after that graph vertex; otherwise append the new graph vertex. The retained final graph path is simple, with new internal graph vertices and exactly one endpoint in the old tree. Adding it therefore preserves connectedness and creates no graph cycle. Graph vertices already included are skipped. On a finite connected graph each walk hits the existing tree almost surely; after finitely many attachments all graph vertices are present.

The Wilson algorithm theorem states that this tree is a uniform spanning tree, independently of the chosen root and order. The required hypotheses here are an undirected connected finite graph and transition probabilities proportional to conductance; unit conductances give simple random walk and the uniform law. This is the sampling result being described, not a claim that arbitrary walk transition probabilities would produce a uniform tree.

For an infinite connected locally finite recurrent graph, choose a fixed finite root $r$ and enumerate all graph vertices. A connected locally finite graph is countable, since its finite-radius balls are finite. Run the same algorithm successively. Recurrence and irreducibility imply that a walk from any graph vertex hits $r$ almost surely, hence hits the current tree. Every attached graph path is therefore finite. The countable intersection of these probability-one events has probability one, and the increasing union is an acyclic connected subgraph containing all graph vertices: each graph vertex receives a finite route to $r$.

This gives the uniform spanning tree of a recurrent infinite graph. It is not a uniform counting measure on all infinite spanning trees. Equivalently, its joint probability distributions on finite collections of edges are the common limits of uniform spanning trees along any graph exhaustion, using the approximations defining the free uniform spanning forest and the wired uniform spanning forest. To check that equivalence, fix a finite edge set $F$ and run the chosen enumeration through a finite prefix containing all its endpoints. Once those graph vertices are in the tree, later attachments cannot add an edge with both endpoints already included, so membership of $F$ is settled permanently. Only finitely many random walks have been used; each has a finite trajectory almost surely. All their visited graph vertices and all their graph neighbours lie in a sufficiently large exhaustion set. Couple finite-volume and infinite-volume walks with the same steps there. Whether the exterior is deleted or wired to one boundary graph vertex, these initial trajectories and their loop erasures then agree. This proves convergence of the probabilities of cylinder sets determined by those edges.

The limiting law is unchanged by the root and ordering because the laws on finite graphs have this invariance. The same coupling shows that changing the exhaustion also leaves the law unchanged. ThusThe result is a single spanning tree because the graph is recurrent. On a transient graph, infinite-volume spanning-tree limits may instead be uniform spanning forests; recurrence cannot be omitted from this conclusion.

Fix a root $r$. On a recurrent graph, a simple random walk started at any vertex hits $r$ almost surely. Run Wilson's algorithm with root $r$: successively attach the loop-erased random walk from each new starting vertex, stopped when it hits the existing tree. Every walk terminates almost surely, because that tree already contains $r$.

To compare the two finite approximations, fix a finite set of edges $K$ and put all their endpoints first in the ordering of starting vertices. In the infinite graph, these finitely many Wilson's algorithm walks have finite lengths almost surely. Their visited vertices, together with their neighbors, are therefore contained in $V_k$ for all sufficiently large $k$ on each realization.

Couple the walks in $G[V_k]$, in $G_k^{\mathrm{w}}$, and in $G$ using the same neighbor choices while they are in the interior of $V_k$. On the event just described, they encounter neither boundary, so the walks, their chronological loop erasures, and the resulting partial trees agree. Once every endpoint of $K$ has entered the partial tree, later walks cannot add any edge of $K$: they stop on their first hit of the existing tree. Thus the indicators of membership for all edges of $K$ agree in the free and wired uniform spanning trees, with probability tending to one.

The two limiting measures agree on every finite-edge event, and thereforeThis argument also shows that the common uniform spanning forest is a single spanning tree on a recurrent graph. The equality refers to probability measures, not to independently sampled forests being identical.

On a connected locally finite recurrent graph, root Wilson's algorithm at a fixed graph vertex $r$ and process a countable enumeration of the graph vertices. Every walk hits the existing tree almost surely, since it contains $r$. Each attached graph path is finite, and the union is a connected acyclic spanning subgraph. Its law is the infinite uniform spanning tree.

It is also the common free and wired finite-volume limit of a sequence along every graph exhaustion. For a finite edge set, run a finite initial segment of the chosen enumeration containing all its endpoints. Its membership is then permanently decided, because later attachments have new internal graph vertices. Only finitely many finite random-walk trajectories were used, and their graph vertices and graph neighbours fit inside all sufficiently large exhaustion sets. Coupling then gives convergence of those finite-dimensional edge laws for both boundary conventions. Finite Wilson's algorithm is root/order independent, so the limiting law is too. There is no uniform counting measure on all infinite spanning trees; the definition is this limiting probability law.