Let $\Omega\subseteq\Bbb C^2$ be open bounded (and connected), $f:\Omega\to\Bbb C$ separately holomorphic (i.e. $f$ is holomorphic in each variable when the other is fixed).


Hartogs theorem is not allowed, so we can't say $f$ is holomorphic on $\Omega$.


Now I think that in this case $f$ is continous in $\bar\Omega$ iff $f$ is integrable in $\Omega$. In other words $f\in\mathcal{C}^0(\bar\Omega)\Leftrightarrow f\in L^1(\Omega)$.


Clearly one implication ($\Rightarrow$) is always true. But the other?


At the moment I can't get no proof nor counterexamples. Can someone help me? Many thanks



If $f$ is separately holomorphic on $\Omega$ then $f\in\mathcal{C}^0(\bar\Omega)\Leftrightarrow f\in L^1(\Omega)$
This will not work if $L^1$ refers to area measure: the function $f(w,z)=1/z$ on (let's say) $|w|,|z|<1$, $z\not= 0$ is a counterexample.


