Extremal Length of Union

Proposition
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
 * $$\frac{1}{\lambda(\Gamma_1\cup \Gamma_2)} \leq \frac{1}{\lambda(\Gamma_1)} + \frac{1}{\lambda(\Gamma_2)}.$$

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

Then
 * $$\frac{1}{\lambda(\Gamma_1\cup \Gamma_2)} = \frac{1}{\lambda(\Gamma_1)} + \frac{1}{\lambda(\Gamma_2)}.$$

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
 * $$ L( \Gamma_1, \rho_1)=L(\Gamma_2,\rho_2)=1. $$

We define a new metric by $$\rho := \max(\rho_1,\rho_2)$$. Then $$L(\Gamma,\rho)\geq 1$$ and $$A(\rho)\leq A(\rho_1)+A(\rho_2)$$.

Hence
 * $$\frac{1}{\lambda(\Gamma)} \leq \frac{A(\rho)}{L(\Gamma,\rho)} \leq A(\rho) = A(\rho_1) + A(\rho_2) = \frac{1}{L(\Gamma_1,\rho_1)} + \frac{1}{L(\Gamma_2,\rho_2)}.$$

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 $$L(\Gamma,\rho)=1$$.

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
 * $$ \rho_j(z)|dz| = \begin{cases} \rho(z)|dz| & z\in A_j \\ 0 |dz| & \text{otherwise}.\end{cases}$$

Then $$A(\rho)=A(\rho_1)+A(\rho_2)$$ and $$L(\Gamma_1,\rho_1),L(\Gamma_2,\rho_2)\geq 1$$.

Hence
 * $$ A(\rho)=A(\rho_1)+A(\rho_2) \geq \frac{A(\rho_1)}{L(\Gamma_1,\rho)} + \frac{A(\rho_2)}{L(\Gamma_2,\rho)} \geq \frac{1}{\lambda(\Gamma_1)} + \frac{1}{\lambda(\Gamma_2)}.$$

Taking the infimum over all metrics $$\rho$$, we see that
 * $$\frac{1}{\lambda(\Gamma_1\cup \Gamma_2)} \geq \frac{1}{\lambda(\Gamma_1)} + \frac{1}{\lambda(\Gamma_2)}.$$

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