Locally exact differential in a disk is exact

Question

I'm reading through Ahlfors' Complex Analysis text for self study, and I found difficulty with a proof.

In chapter 4 he defines a locally exact differential as a differential who is exact in some neighborhood of every point of its domain.

In the proof of theorem 16 however, he claims that such differential is exact in every open disk contained in the region. I fail to see why is it so: Each point should have its own radius, and these can get very small (?).

I can use the fact that a locally exact differential of class $C^1$ is closed and apply Poincare's lemma, but no differentiability assumptions were made in the statement of the theorem.

Any help will be appriciated.

Answer 1

All you need is a continuous $1$-form $\omega$. Since a disk is convex, you know that $\int_{\partial R}\omega=0$ for any rectangle $R$ contained in the disk. But this is all you need to define a potential function. See Theorem 1 in the same chapter.

Answer 2

A differential $P\mathrm dx+Q\mathrm dy$ is said to be locally exact in the given region if it is exact in some neighbourhood of each point in that region.