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The effective resistance to infinity is the limit of the effective resistance from a fixed vertex to the wired complement of increasing finite sets. It is finite exactly when the finite-energy flow criterion for transience holds.

Assume first that deleting an edge leaves two transient graph components. One component supports a finite-energy unit flow from its endpoint to infinity by the finite-energy flow criterion for transience. Extend that flow by zero across the deleted edge and throughout the other component. It remains a finite-energy unit flow on the whole tree, so the tree is a transient graph.

Conversely, root a transient tree at $o$ and take its finite-energy unit electrical flow to infinity. On a tree this flow can be chosen nonnegative in every direction away from the root: it is obtained as the limit of the currents to wired boundaries, each of which sends nonnegative current into descendant subtrees. At each nonroot vertex its incoming current equals the sum of its outgoing currents, by flow conservation.

If every vertex reached by positive current had exactly one positive-current child, the unit current would run along a single infinite ray. Every edge of that ray would carry current one, so its energy of a flow would be ${\sum }_{j\ge 1}1=\infty$, a contradiction. Thus there is a vertex $v$ with two positive-current child edges, say $(v,x)$ and $(v,y)$.

The restricted flow below $x$, rescaled by its incoming current, is a finite-energy unit flow from $x$ to infinity; hence the component below $x$ is a transient graph. After removing $(v,x)$, the other component contains $v$, the edge $(v,y)$, and the descendant subtree below $y$. Send unit current along $(v,y)$ and then use the rescaled descendant flow below $y$, with zero current elsewhere. Its energy of a flow is finite, so this component is a transient graph too. ThereforeUnit edge resistances are essential here: an infinite ray with sufficiently rapidly decreasing resistances can be transient without any such splitting edge.

Let the energy of a flow be $\mathcal{E}(\theta )={\sum }_e\theta (e{)}^2$, with each unoriented edge counted once. Expanding as in part (i), dropping orientation signs, and summing over common edges givesuniformly in $r$.

Part (i), $\mathbb{E}{X}_r=1$, and the Paley-Zygmund inequality give a constant $a>0$ such that $\mathbb{P}(X_r>1/2)\ge a$ for every $r$. The preceding uniform expectation bound and Markov inequality allow a constant $L$ such thatOn this event, ${\theta }_{\omega }/X_r$ is a unit open flow from $o$ to $z_r$ with energy at most $4L$.

The events that there is such a bounded-energy unit flow from $o$ out of $B(o,r)$ are decreasing in $r$. Their intersection still has probability at least $a/2$. A diagonal compactness argument produces on this intersection a unit flow from $o$ to infinity, supported on its open cluster, with finite energy. The finite-energy flow criterion for transience makes that open cluster transient.

Thus a transient open cluster exists with positive probability. This existence event is a tail event: changing finitely many edges cannot destroy transience in every infinite component, because transience is invariant under finite graph modifications. The Kolmogorov zero-one law upgrades its probability to one.