I would be interested to learn if the following generalization of the classical [Looman-Menchoff theorem](https://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem) is true.




---


Assume that the function $f=u+iv$, defined on a domain $D\subset\mathbb{C}$, is such that


1. $u\_x$, $u\_y$, $v\_x$, $v\_y$ exist almost everywhere in $D$.
2. $u$, $v$ satisfy the Cauchy–Riemann equations almost everywhere in $D$.
3. $f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$.
4. $f$ is locally integrable.



> 
> **Question:** Does it follow that $f$ is analytic *everywhere* in $D$?
> 
> 
> 




---


**Remark 1.** Condition 3 is essential (take $f=1/z$).


**Remark 2.** G. Sindalovskiĭ proved analyticity of $f$ under conditions 2-4 when the partial derivatives exist everywhere in $D$, *except on a countable union of closed sets of finite linear Hausdorff measure* ([link](https://iopscience.iop.org/article/10.1070/SM1987v056n02ABEH003041 "DOI: 10.1070/SM1987v056n02ABEH003041")).



The Cauchy–Riemann equations and analyticity
No.


Let $c$ be the [Cantor function](https://en.wikipedia.org/wiki/Cantor_function) on $[0,1]$, so that $c$ is continuous, $c' = 0$ a.e., but $c$ is not constant. Then take $u(x+iy) = v(x+iy)=c(x)c(y)$. We have $u\_x=u\_y=v\_x=v\_y=0$ a.e. so the Cauchy–Riemann equations are trivially satisfied, and $f(z)=u(z)+iv(z)$ is bounded and continuous on the unit square, but certainly not analytic.


Almost everywhere differentiability is almost never the right condition for solutions to a PDE. A better condition would be to have $u,v$ in some Sobolev space.


