Extremal Length of Union

Theorem
Let $X$ be a Riemann surface.

Let $\Gamma_1$ and $\Gamma_2$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$.

Then the extremal length of their union satisfies:
 * $\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } \le \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$

Suppose that additionally $\Gamma_1$ and $\Gamma_2$ are disjoint in the following sense: there exist disjoint Borel subsets:
 * $A_1, A_2 \subseteq X$ such that $\ds \bigcup \Gamma_1 \subset A_1$ and $\ds \bigcup \Gamma_2 \subset A_2$

Then
 * $\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } = \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$

Proof
Set $\Gamma := \Gamma_1\cup \Gamma_2$.

Let $\rho_1$ and $\rho_2$ be conformal metrics as in the definition of extremal length, normalized such that:
 * $\map L {\Gamma_1, \rho_1} = \map L {\Gamma_2, \rho_2} = 1$

We define a new metric by:
 * $\rho := \map \max {\rho_1, \rho_2}$

Then:
 * $\map L {\Gamma, \rho} \ge 1$

and:
 * $\map A \rho \le \map A {\rho_1} + \map A {\rho_2}$

Hence:

Taking the infimum over all metrics $\rho_1$ and $\rho_2$, the claim follows.

Now suppose that the disjointness assumption holds, and let $\rho$ again be a Borel-measurable conformal metric, normalized such that $\map L {\Gamma, \rho} = 1$.

We can define $\rho_1$ to be the restriction of $\rho$ to $A_1$, and likewise $\rho_2$ to be the restriction of $\rho$ to $A_2$.

By this we mean that, in local coordinates, $\rho_j$ is given by
 * $\map {\rho_j} z \size {\d z} = \begin {cases}

\map \rho z \size {\d z} & : z \in A_j \\ 0 \size {\d z} & : \text {otherwise} \end {cases}$

Then:
 * $\map A \rho = \map A {\rho_1} + \map A {\rho_2}$

and:
 * $\map L {\Gamma_1, \rho_1}, \map L {\Gamma_2, \rho_2} \ge 1$

Hence:

Taking the infimum over all metrics $\rho$, we see that:
 * $\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } \ge \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$

Together with the first part of the Proposition, this proves the claim.