Example of a function $h$ such that $\overline{\partial}h$ is not a measure Everything is in the title. Do you have an example of a function $h \in L^1(\mathbb{C})$ such that $\overline{\partial}h$ in the sense of distributions is not a (finite complex) measure ?
(additionnaly, is there a nice characterization of such functions ?)
I ask the question because I am reading an article where the author defines an integrable function to be "regular" if $\overline{\partial}h$ is a finite complex measure. But the following reasonning seems to prove that it is automatic : 
It is known (Bers density theorem) that meromorphic functions whith simple poles are dense in $L^1(\mathbb{C})$. For such a function, its $\overline{\partial}$ is a sum of Diracs. Let $h \in L^1$ be arbitrary, and $h_n$ a sequence of meromorphic functions with simple poles converging to $h$. Then the $\overline{\partial}h_n$ are a sequence of complex measures, and $\overline{\partial}$ being continuous from $L^1$ to the distributions, it converges to a distribution $u$. But that distribution must be of order 0, and therefore is a measure, and is $\overline{\partial}h$.
Am I missing something ? Is my reasonning correct ? 
 A: It helps to consider the situation on $\mathbb R$. Take $f(x)=\frac{\sqrt{|x|}}{x}\chi_{[-1,1]}$. This is an $L^1$ function, and smooth functions are dense in $L^1$. This means we have a sequence of smooth functions $f_n$
converging to $f$ in $L^1$. The derivatives $f_n'$ are measures (absolutely continuous ones) which converge to $f'$ 
as distributions. The problem is that $f_n'$ give more and more (signed) mass to a small neighborhood of $0$ (because 
$f_n$ changes rapidly there), and in the limit we do not have a distribution of zero order. Indeed, if
$f'$ had order zero, it would be represented by a signed Radon measure on the line, which would imply that 
$f$ has a representative of bounded variation (it does not). 
In the complex plane, try $\frac{1}{z\sqrt{|z|}}\chi_{|z|\le 1}$. 
Simply put, the distributional limit of Radon measures is not necessarily a Radon measure. 
To make this even more explicit, consider $\mu_N=\sum_{n=1}^N n^{-1}(\delta_{1/n}-\delta_{-1/n})$: this is a sequence
of measure such that $\int \varphi \, d\mu_N$ converges for every test function $\varphi$, yet the distributional 
limit of $\mu_N$ is not a Radon measure. 
I keep repeating "Radon". Finiteness on compact subsets is essential here (the context being complex analysis). 
You may be able to interpret some of distributional limits as non-Radon measures, but  even then I'd be 
worried about countable additivity.
