Cauchy formula in Polydisks

Question

I don't understand a remark after the proof. Here's the theorem:

The proof is done by induction on $n$; starting from $n=1$ on the unitary disk in $\Bbb C$, which is the well known Cauchy formula. The inductive step is now easy:

and 'till here it's all right. My problem is that, after the theorem, appears the following remark:

The first sentence is clear: separate analiticity is needed in order to apply the $1$-dimensional case (which uses Stokes) in the inductive step. My problem is that I can't understand where the joint $\mathcal{C}^0$ regularity is used, in Fubini theorem. Doesn't to be $L^1$ suffice for our purpose?

Many thanks to all.

Answer 1

My problem is that I can't understand where the joint $\mathcal{C}^0$ regularity is used, in Fubini theorem.

It is an easy sufficient condition for the applicability of Fubini's theorem. You need some condition that ensures we can apply Fubini's theorem, and the author chose continuity, since that is an easy condition.

Doesn't to be $L^1$ suffice for our purpose?

For the application of Fubini's theorem in the proof, you need the integrability on the distinguished boundary $\partial_0 P$ of the polydisk. That follows immediately if you require $f$ to be continuous. If you only assume $f\in L^1(\Omega)$ for some neighbourhood $\Omega$ of $\overline{P}$, then that doesn't imply the integrability over $\partial_0 P$ (directly), and you need some gymnastics to obtain that. If you require "$f\in L^1(\partial_0 P)$", you're fine, but that looks like an ad hoc condition.

So I wrote this post in order to understand if they are also necessarely (why ask more if we can get the same asking less?).

If we make stronger assumptions, we can often reach the conclusion with simpler arguments.

In fact, Hartogs' theorem says that the separate analyticity is sufficient. Any further condition is redundant, the separate analyticity of $f$ alone implies that $f$ is holomorphic.

But proving that is harder.