Extremal Length of Composition

Theorem
Let $\Gamma_1$ and $\Gamma_2$ be families of (unions of) rectifiable curves on a Riemann surface $X$.

Also suppose that $\Gamma_1$ and $\Gamma_2$ are disjoint in the sense that there exist disjoint Borel sets $E_1, E_2 \subseteq X$ with $\bigcup\Gamma_1\subset E_1$ and $\bigcup \Gamma_2\subset E_2$.

Then the extremal length of the family:
 * $\Gamma := \{ \gamma_1\cup \gamma_2:\ \gamma_1\in\Gamma_1\text{ and }\gamma_2\in\Gamma_2\}$

satisfies
 * $\lambda(\Gamma) = \lambda(\Gamma_1)+\lambda(\Gamma_2)$

Proof
By the Series Law for Extremal Length, we have
 * $\lambda(\Gamma) \geq \lambda(\Gamma_1)+\lambda(\Gamma_2)$

Hence it remains only to prove the opposite inequality.

Let $\rho$ be a metric as in the definition of extremal length, normalized such that $A(\rho)=1$. We claim that:
 * $(L(\Gamma,\rho))^2 \leq \lambda(\Gamma_1) + \lambda(\Gamma_2)$

Define $\alpha_j := \sqrt{A(\rho \restriction_{E_j})}$ (for $j = 1, 2$).

We may assume that both values are positive.

If we define:
 * $\rho_j := \dfrac{\rho \restriction_{E_j}} {\alpha_j}$

then $A(\rho_j)=1$.

So we have:

In conclusion:
 * $ \lambda(\Gamma) = \sup_{\rho} L(\Gamma,\rho) \leq \lambda(\Gamma_1)+\lambda(\Gamma_2)$

as claimed.