K^n_{J,j}\cap L^n_{J,j}\neq\emptyset$ and side length comparable to $\sigma_n\delta_n$. The analytic capacity $\gamma(\tilde K^n_{J,j})$ is comparable to $\sigma_n\delta_n$, so if $f^n_{J,j}$ is the Ahlfors function for $\tilde K^n_{J,j}$ then, choosing (uniformly bounded) constants $c_{J,j}^n$ appropriately, and setting $$k^n_{J,j}(z)=\frac{c^n_{J,j}}{n^3}f^n_{J,j}(z)$$ then $\Vert k^n_{J,j}\Vert_\infty\lesssim\frac1{n^3}$ and $$(k^n_{J,j})'(\infty)=\frac1{2\pi i}\int_{L^n_{J,j}}\tilde g_\mu\bar\partial\tilde g_\nu\,d\zeta\,d\bar\zeta.$$ For the estimates that follow, we will employ the following form of Schwarz’s lemma.

If $E$ is a compact set and $f\in H^\infty(\mathbb C^\ast\setminus E)$ has a double zero at infinity then $$|f(z)|\lesssim\frac{\Vert f\Vert_\infty{{\rm{diam}}}(E)^2}{d(z,E)^2}.$$\

Let us define $$h^n_{J,j}(z):=I^n_J(z)-k^n_{J,j}(z).$$ Applied in the current context, the lemma yields $$\label{Schwarzest}
    |h^n_{J,j}(z)|\lesssim\frac1{n^3}\left(1\wedge\frac{(\sigma_n\delta_n)^2}{d(z,\tilde K^n_{J,j}\cup L^n_{J,j})^2}\right).$$ Since $\bar\partial k^n_{J,j}=0$ in $\Omega$, formally we have $$\bar\partial\sum_{(n,J,j)}h^n_{J,j}=\tilde g_\mu\bar\partial\tilde g_\nu,$$ so it suffices to check the boundedness of the sum.

Fix $z\in\Omega$. For each $n\in\mathbb N$ there are at most boundedly many terms for which the minimum is one in the inequality (\[Schwarzest\]), and the summability of $\frac1{n^3}$ shows that these terms, summed over $n$, give a contribution which is controlled by the sum $\sum\frac1{n^3}$.

For remaining terms, we distinguish between the cases $\sigma_n\gtrsim d(z,K)$ and $\sigma_n\lesssim d(z,K)$. Let $n_0\in\mathbb N$ be such that $d(z,K)\asymp \sigma_{n_0}$ (the maximum principle prevails over any $z$ with $d(z,K)\gg1$). For $m\ge n_0$, all remaining terms are at (roughly) 