Extremal Length of Composition/Argument for Case of Zero values

In the proof, we stated that we may assume that the values $\alpha_1$ and $\alpha_2$ can be positive. For completeness, we provide the details of the simple argument, continuing to use the notation from the proof.

The claim that is to be proved is trivial if $\lambda(\Gamma_1)$ or $\lambda(\Gamma_2)$ are infinite. So we may assume both extremal lengths are finite.

Now if $\alpha_j=0$, then it follows that

$L(\Gamma_j,\rho)=0.$

Indeed, we can choose a metric $\tilde{\rho}$ that agrees with $\rho$ on $A_j$ and has total area $\varepsilon>0$. Then

$(L(\Gamma_j,\rho))^2 = (L(\Gamma_j,\tilde{\rho}))^2\leq \varepsilon\cdot \lambda(\Gamma_j).$

Because $\varepsilon$ was arbitary, the claim follows.

So if, say, $\alpha_2=0$, then

$(L(\Gamma,\rho))^2=(L(\Gamma_1,\rho) + L(\Gamma_2,\rho))^2= (L(\Gamma_1,\rho))^2\leq \lambda(\Gamma_1)\leq\lambda(\Gamma_1)+\lambda(\Gamma_2).$

$\blacksquare$