Is analytic capacity continuous from below? EDIT: I also asked this question on mathoverflow, since it might be too specialized for math.stackexchange.com.
I've been wondering about the following, I don't know if anyone knows the answer :
For a compact set $K$ in the complex plane, define the analytic capacity of $K$ by
$$\gamma(K) := \sup |f'(\infty)|$$
where the supremum is taken over all functions $f$ holomorphic and bounded by $1$ in the complement of $K$ : $f \in H^{\infty}(\mathbb{C}_{\infty} \setminus K)$, $\|f\|_{\infty} \leq 1$. Here
$$f'(\infty) = \lim_{z \rightarrow \infty} z(f(z)-f(\infty)).$$
A theorem due to Ahlfors states that for each compact $K$, there always exists a unique function $F$, called the Ahlfors function of $K$, such that $F \in H^{\infty}(\mathbb{C}_{\infty} \setminus K)$, $\|F\|_{\infty} \leq 1$, and $F'(\infty)=\gamma(K)$.
It's not hard to show that $\gamma$ is continuous from above : if $(K_n)$ is a decreasing sequence of compact sets, then
$$\gamma(\cap_n K_n) = \lim_{n\rightarrow \infty} \gamma(K_n).$$
This essentially follows from Montel's theorem and the fact that $\gamma(E) \subseteq \gamma(F)$ whenever $E \subseteq F$.
My question is the following :
Is analytic capacity continuous from below? More precisely, if $(K_n)$ is a sequence of compact sets such that
$$K_1 \subseteq K_2 \subseteq K_3 \subseteq \dots$$
and such that $K:=\cup_n K_n$ is compact, then is it true that
$\gamma(K) = \lim_{n \rightarrow \infty} \gamma(K_n)?$
I could not find anything in the litterature.
Thank you,
Malik
 A: Your question is rather a good one.  Let me try to say what I know about it.
Its a simple consequence of the well-known fact that " Analytic capacity has following property " 

If $K_a\subseteq K_b$ then its well known fact that $\gamma(K_a)\le\gamma(K_b)$.

So by using that what we can conlude is that the sequence {$\gamma(K_n)$ } is decreasing ( if you assume that {$K_n$} is a decreasing sequence of compact subsets ) , and you can also notice that its bounded below by $\gamma(K)$.
So it now implies the following 
$$\gamma(K)\le\displaystyle\lim_{n\to\infty}\gamma(K_n)=\lim_{n\to\infty} \rm{Inf} \      \gamma(K_n).$$ 
Using your notation , let $F_n$ be the Ahlfors function of $K_n$ then its a known thing that they always form a normal family on every $\mathbb{C}\backslash K_n$ and also we have :

For every $K\subset\Omega$ there corresponds a number $M(K)$ ( $M(K)\lt\infty$ ) such that $|f(z)|\le M(K)$ for all $f\in X$ and $z\in K$ where $X\subset H(\Omega)$ for some region $\Omega$. 

So I think one can prove the above statement by using " Cauchy's Formula " and knowing some basic rules of Convergence and some background in Analysis. ( If  you are still looking a proof, I can add it, but its quite large, and I need to refer some old Analysis books for proof.  Anyway if you are looking for a proof of above statement let me know ) .
Coming back , by combining the above statement and Famous Cantor's Diagonal argument we can conclude that sub-sequence of {$F_n$} uniformly converge ( on compact subsets of $\mathbb{C}\backslash K$ ) to a function $F$ analytic and bounded by one on $\mathbb{C}\backslash K$ . So one can clearly imply that $$\displaystyle\lim_{n\to\infty} \rm{Inf} \ \gamma(K_n) = \lim_{n\to\infty}\rm{Inf}\ F^{\prime}_{n}(\infty)\le  F^{\prime}_{n}(\infty) = | F^{\prime}_{n}(\infty)|\le\gamma(K)$$
So now we reached our destination to say that , so it implies from above statements that  $$\displaystyle\lim_{n\to\infty} \ \gamma(K_n)=\gamma(K)$$
So to add something to this, in a more broader manner, you are asking for increasing sequence, so let me show you a clear idea how can you achieve it. 
I think that Analytic capacity is some sort of measure , measures the size of a set as a non-removable singularity . So it may be compared and taken as a normal measure which satisfies the property $$\mu(A_n)\to \mu(A)$$ as $n\to \infty$.
So I hope that is what you are looking for. Where $A=\bigcup^{\infty}_{n=1} A_n$ , and its nothing but increasing sequence $$A_1\subset A_2\subset A_3......$$
But I am not sure that one can consider the Analytic capacity to be a measure in a strict manner. But it can be extended to that notion.
And also something to add, the upper continuity you are talking about is perfectly defined for Hausdorff measure. There is a standard theorem which says 
" Let $E$ be a union of increasing sequence {$E_n$} of subsets of $\mathbb{C}$ then for any $S\in[0,2]$ , it follows that $$\displaystyle \lim_{n\to \infty} H^S(E_n)= H^S(E)$$. But there are some theorems that establish the relationship between analytic capacity and Hausdorff Measure, but I don't know how far it can be used to answer your theorem.
Thank you.
( Please feel free to write comments and valuable feed-back )
