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:
 * $\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} \le \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$

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 $\displaystyle \bigcup \Gamma_1 \subset A_1$ and $\displaystyle \bigcup \Gamma_2 \subset A_2$

Then
 * $\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} = \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$

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:
 * $ L \left({\Gamma_1, \rho_1}\right) = L \left({\Gamma_2, \rho_2}\right) = 1$

We define a new metric by:
 * $\rho := \max \left({\rho_1, \rho_2}\right)$.

Then:
 * $L \left({\Gamma, \rho}\right) \ge 1$

and:
 * $A \left({\rho}\right) \le A \left({\rho_1}\right) + A \left({\rho_2}\right)$

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 $L \left({\Gamma, \rho}\right)= 1$.

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
 * $ \rho_j \left({z}\right) \ \left|{\mathrm d z}\right| = \begin{cases}

\rho \left({z}\right) \ \left|{\mathrm d z}\right| & : z \in A_j \\ 0 \ \left|{\mathrm d z}\right| & : \text{otherwise} \end{cases}$

Then:
 * $A \left({\rho}\right) = A \left({\rho_1}\right) + A \left({\rho_2}\right)$

and:
 * $L \left({\Gamma_1, \rho_1}\right), L \left({\Gamma_2, \rho_2}\right) \ge 1$

Hence:

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

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