Extremal Length of the Empty Curve Family I'm working out the details of extremal lengths of quadrilaterals ($m(Q)$) in $\mathbb{C}$ . The standard result that $m(Q)m(Q^*)=1$ (where $Q^*$ denotes the dual quad) isn't troublesome, but I've been wondering about what happens in more hideous domains. 
For instance, if $Q=([0,1]\times i[0,1])\setminus \left(\left(\cup_{k=1}^{\infty} \{2^{-2k}\}\times i[0,\frac{3}{4}]\right) \cup \left(\cup_{j=1}^{\infty} \{2^{-(2j+1)}\}\times i[\frac{1}{4},1]\right)\right)$ with marked sides $I_1=\{0\}\times i[0,1]$ and $I_3=\{1\} \times i[0,1]$, then the curve family of crossings from $I_3$ to $I_1$ is empty.
However, mapping $Q$ conformally to the disk in the sense of prime ends, $I_1$ gets mapped to a single point, and suddenly, in the disk, there is a well-defined curve family, which has an extremal length. This leads me to want to define the extremal length of the empty set to be the extremal length of this set, but I don't think this is independent of $Q$.
If, on the other hand, you just try to go by the definition, it seems like the extremal length of the empty set should be $\infty,$ since any non-negative function will be admissible/(a proper metric).
So basically, I'd personally like for conformal invariance (of extremal lengths of quads, not general curve families) to hold without a caveat that you have to check whether the family of crossings is empty before you apply it.
Does anyone know of a source that actually discusses this?
 A: Okay: Since this has garnered some attention, I've actually managed to figure out that my hope was in vain.
For the quad I wrote, it's easy to check that all of $i[0,1]$ defines a single prime end and thus, corresponds to a single point in $B(0,1)$ - assume without loss of generality that it is $-1$. Meanwhile, the other arcs correspond to genuine arcs.
Now, any loop at $-1$ is now a crossing between the images of $I_2$ and $I_4$ (which are chosen to be the two arcs of the dual quad), implying that $m(Q^*)=0$, since for any metric $\rho$, we have $\int_{\mathbb{C}}\rho^2\textrm{d}z\geq \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^1 r\rho^2(-1+re^{i\theta})\textrm{d}r\textrm{d}\theta=\infty,$ by the $L^2$-Cauchy-Schwarz-Inequality and monotone convergence.
On the other hand $\{1\}\times i[0,1]$ gets mapped to some arc $J$ disjoint from $-1$. Thus, any crossing to $-1$ has to cross all annuli of the form $A(-1,\varepsilon, d(-1,J)),$ and as $\varepsilon\to 0$, the extremal length of these annuli tend to $\infty$. Accordingly, the image quad has extremal length $\infty$. So far, so good. This is what I wanted.
However, say that our marked sides were instead $I_1=i[\frac{1}{3},\frac{2}{3}],$ $I_2=i[0,\frac{1}{3}]$ and $I_4=i[\frac{2}{3},1],$ then each arc gets mapped to the same single point as prime ends. Accordingly, both the image of the quad and the image of the dual quad have extremal length $0$ by the statement above, even though each curve family is empty.
So, in conclusion: If BOTH curve families are empty (and this can happen), then conformal invariance fails.
