Exponential clocks and the uniform edge-order process

Write $g(u)=-\log (1-u)$ for $0\le u<1$. Since $0<1-u\le 1$, $g(u)$ is finite and nonnegative. It is strictly increasing: if $u<v$, then $1-u>1-v>0$, so $-\log (1-u)<-\log (1-v)$. For $t\ge 0$, logarithm inversion and monotonicity give $g(u)\le t$ exactly when $u\le 1-e^{-t}$, with $0\le 1-e^{-t}<1$.

Give $[0,1)$ its Borel sigma-algebra. For any real $a<0$, $\{u:g(u)\le a\}=\mathrm{\varnothing }$; for $a\ge 0$, this set is $[0,1-e^{-a}]$, a Borel subset of $[0,1)$. Hence [F3] makes $g$ Borel measurable. For every Borel set $B\subseteq \mathbb{R}$, the event $\{T_e\in B\}=\{U_e\in g^{-1}(B)\}$ is measurable because $U_e$ is a random label and $g^{-1}(B)$ is Borel. Thus each $T_e$ is a random variable. For distinct edges $e_1,\dots ,e_k$ and arbitrary Borel sets $B_1,\dots ,B_k$, independence of the labels gives $\mathbb{P}({\bigcap }_{j=1}^k\{T_{e_j}\in B_j\})=\mathbb{P}({\bigcap }_{j=1}^k\{U_{e_j}\in g^{-1}(B_j)\})={\prod }_{j=1}^k\mathbb{P}(U_{e_j}\in g^{-1}(B_j))={\prod }_{j=1}^k\mathbb{P}(T_{e_j}\in B_j)$. This proves independence for all Borel clock events.

For $t\ge 0$, the threshold equivalence gives $\mathbb{P}(T_e\le t)=\mathbb{P}(U_e\le p(t))=p(t)$ by the uniform threshold law in [F1]. Edge by edge it also gives $H_t=G_{p(t)}$ for every $t\ge 0$.

The transformation from labels to clocks is strictly increasing, so it preserves their strict order outside the label-tie event of probability zero. At the $m$th arrival, precisely the first $m$ edges have appeared. Connectedness is preserved thereafter, so $s_{\tau }$ is the first connected clock time, and substituting $t=s_m\ge s_{\tau }$ proves every indexed assertion. This argument evaluates a single pathwise assertion and uses no union bound at random times.