Locally exact differential in a disk is exact 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.
 A: 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.
A: Right after he defines locally exact he leaves it as an exercise that locally exact is equivalent to the integral over any sized rectangle being zero.  Assuming this, if we can consider a locally exact differential $P dx +Qdy$ in any disk contained in the region it is defined.  Since the integral around any rectangle is zero, we can choose rectangles with sides parallel to the $x$ and $y$ axes to define an anti derivative, in the disk, as he did in his initial discussion of exactness.  Once you have a well-defined anti-derivative, then it follows that the integral around closed loop in the disk is zero, i.e. it is exact. This means an equivalent definition of local exactness is that it is is exact in every disk.
A: 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.
