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This is really a simple question involving swapping the order of integration. The step I'm confused about comes from a proof in Pommerenke's "Univalent Functions," and for those of you with a reference, it is Theorem 1.2.

Let $C$ be a (closed) piecewise smooth curve homologous to zero with respect to the domain $H\subset \mathbb{C}$. We want to show that

$$ \int_C \overline{h(w)} h'(w) dw = \dfrac{1}{\pi} \int_{H} |h'(w)|^2 \left( \int_C \dfrac{dz}{z-w} \right) d\Omega_w $$

where $d\Omega_w$ is the Lebesgue measure.

The first step in the proof is to swap the order of integration, but I'm not really sure why one could do that. I'm guessing it has something to do with a remark in the text about changing $H$ slightly to assume that $h$ is analytic on $\overline{H}$ rather than just on $H$.

Several things...

  1. I think I can assume that $H$ is a bounded domain.
  2. Assuming that $H$ is a bounded domain, with the additional assumption that $h$ is analytic on $\overline{H}$, I can assume that $h$ is bounded on $H$.

I'm guessing I have to use something like Fubini, but if I want to use Fubini, I'll have to show that

$$\int_C \int_H \dfrac{1}{|z-w|} d\Omega_w |dz|$$

is convergent.

I can see that the inner integral is convergent for all $z$ in $H$. But then I'm not sure why the iterated integral is convergent.

Is it true that

$$\int_H \dfrac{1}{|z-w|} d\Omega_w$$

is a continuous function in $z$? I think it's true, but I'm not sure how to show this...

share|cite|improve this question
As for "changing $H$ slightly" the point must be that your original $H$ is open. Let $D$ be the image of the zero homotopy of $C$; then $D$ is a closed subset of $H$. Now, $\mathbb C\setminus H$ is also closed, and since $\mathbb C$ is a $T_4$ space these two disjoint closed sets can be separated by disjoint open neighborhoods, and the closure of the open neighborhood of $D$ will be entirely contained in $H$, so $h$ is analytic there (and all points of interest are still within its interior). – Henning Makholm Jan 2 '12 at 0:49
Errr, sorry for not clarifying: I'm using the convention that "domain" means open and connected. – Braindead Jan 2 '12 at 1:08
Yes, I got that. Sorry if what I told was something you already knew; it was not entirely clear from the question whether this was something you were unsure about or just scene-setting for the real question. – Henning Makholm Jan 2 '12 at 1:12

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