In the set-up of my previous post, let $\theta$ be a purely non-atomic finite regular measure defined on the Borel $\sigma$-algebra of the metric space $(\partial F, d)$. We say $\theta$ admits a $\epsilon$-partition if there exists a measurable finite partition $\{A_i\}_{i\leq N}$ of $\partial F$ such that $\theta(A_i) < \epsilon$ for each $i\leq N$. According to Michael's answer, for each $\epsilon\in(0, 1)$, $\theta$ admits a $\epsilon$-partition of $\partial F$. Since $\theta$ is regular, each $\epsilon$-partition can induce an finite open cover where each set has $\theta$-measure strictly less than $\epsilon$.
Then, given a sequence $\{g_n\}_{n\in\mathbb{N}}\subseteq F$ that converges to $\xi\in\partial F$, I am tempted to prove that $g_n\theta$ converges to the dirac measure $\delta_{\xi}$ weakly (where, for each Borel subset $O\subseteq\partial F$, $g_n\theta(O) = \theta(g_n^{-1}O)$). For the definition of weak convergence, check this wiki. Below is my attempt but, because of the list of equivalence under the Portmanteau's Theorem (in the wiki link above), it is wrong and I wonder if anyone could point out my mistakes.
For each $g = g_1 \cdots g_n \in F$, define:
$$ C_g = \big\{ x\in\partial F\,\vert\, x_1 = g_1, \cdots, x_n = g_n \big\} $$ Since $\partial F$ is totally bounded and $d$ is an ultra-metric, for each open $O\subseteq\partial F$ and $\epsilon\in(0, 1)$, we can find $\{g_i\}_{i\leq K}\subseteq F$ such that $O = \overset{\bullet}{\bigcup}_{i\leq K} C_{g_i}$ (a disjoint union) with $\operatorname{exp}\big( -\vert\,g_i\,\,\vert\big) < \epsilon$ for each $i\leq K$. Hence:
$$ O^c = \bigcap_{i\leq K}C_{g_i}^c = \bigcap_{i\leq K} \Big( \overset{\bullet}{\bigcup}_{\substack{g\in F\backslash\{g_i\} \\ \vert\,g\,\vert = \vert\,g_i\vert}} C_g\Big) $$ which implies that each closed set can be written as a finite disjoint union of sets of the form $C_h$.
Then I claim:
$$ \forall\,\epsilon\in(0, 1)\, \exists\,N_{\epsilon}\in\mathbb{N} \text{ such that } \forall\,h\in F, \hspace{0.3cm} \vert\,h\,\vert \geq N_{\epsilon} \hspace{0.3cm}\implies\hspace{0.3cm} \theta(C_h)<\epsilon $$ Fix $\epsilon\in(0, 1)$ and let $\{O_i\}_{i\leq L}$ be an open cover of $\partial F$ such that $\theta(O_i) <\epsilon$ for each $i\leq L$. Then there exists $\{h_i\}_{i\leq K}\subseteq F$ such that $\{C_{h_i}\}_{i\leq K}$ is a refinement of the cover $\{O_i\}_{i\leq L}$ so that $\theta(C_{h_i})<\epsilon$ for each $i\leq K$. Set $N_{\epsilon} = \max_{i\leq K}\vert\,h_i\,\vert$. Then for each $\gamma\in\partial F$, if $\gamma\in C_{h_i}$, for each $n\geq N_{\epsilon}$ we have $B(\gamma, e^{-n}) = C_{[\gamma]_n} \subseteq C_{h_i}$. Therefore $\theta(C_{[\gamma]_n}) < \epsilon$. Since for each $g\in F$ with $\vert\,g\,\vert\geq N_{\epsilon}$, we can find $\xi\in\partial F$ with $[\xi]_{\vert\,g\,\vert} = g$, the desired result follows (?)
If the proof above (or the statement I claim) is correct, then for each $h\in F$, we have $\limsup_n g_n\theta(C_h) = 0$ because the length of $g_n^{-1}h$ will become arbitrarily large, which shows that for each closed subset $B\subseteq\partial F$, $\limsup_n g_n\theta(B) = 0\leq \delta_{\xi}(B)$. Then we will have $g_n\theta$ converges to $\delta_{\xi}$ weakly, which implies that $\liminf_n g_n\theta(C_h)\geq \delta_{\xi}(C_h)$ for each $h\in F$ since $C_h$ is open. However, this is obviously wrong because $\liminf_n g_n\theta(C_h)=0$ for each $h\in F$.
