Extremal Length of Composition

Proposition
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 $$\alpha_j := \sqrt{A(\rho|_{E_j})}$$ (for $$j=1,2$$). We may assume that both values are positive. If we define
 * $$\rho_j := \frac{\rho|_{E_j}}{\alpha_j},$$

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.