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 (in the unbounded component of the complement of $K$) 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

**EDIT ( 12-01-2012)** I edit the question to add what I know so far about this question :

 1. As pointed out by Fedja in the comments, analytic capacity is *comparable* to a quantity which is continuous from below, see the article "Painleve's problem and the semiadditivity of analytic capacity" by Xavier Tolsa.

 2. The answer is yes if the compact sets $K_n$ and $K$ are connected. Indeed,  for connected compact sets, analytic capacity is equal to *logarithmic capacity*, and logarithmic capacity is continuous from below.

 3. The answer is yes if $K$ is a compact set whose boundary consists of a finite number of analytic and pairwise disjoint Jordan curves, provided we replace the condition $K:=\cup_n K_n$ by the condition that each compact subset of the interior of $K$ is eventually contained in some $K_n$. This mainly follows from the fact that in this case, the Ahlfors function of $K$ extends analytically across the boundary of $K$. See for example the book "analytic capacity and measure" by Garnett, p. 18.