Current induced by a locally integrable differential form: I don't understand why it is not trivially $0$ Today this question captured my attention, hence I want to generalize it. 
Let $X$ be a complex manifold of dimension $M$ and let $\omega $ be a $(n-p,n-q)$-differential form such that in each chart it is represented by locally integrable functions. A standard example of current on $X$ (see for example De Rham - Differential manifolds, Chap. III example 2) is the following:
$$[\omega]:\alpha \mapsto\int_X \alpha\wedge\omega$$
where $\alpha$ is a $C^\infty$ $(p,q)$-form and $\alpha\wedge\omega$ is a locally integrable $(n,n)$-form.


*

*Who ensures that $\alpha\wedge\omega$ is integrable? Usually the integral is defined for smooth differential forms on oriented manifolds (see for example Lee's book). About this point I'm quite sure the answer will be: "integration can be extended to locally integrable forms", I just wanted  to check.

*Consider $n=1$, $p=0$, $q=0$ and let'es examine the example of the question linked above:
$$\omega=\partial\bar\partial\log|f|$$
where $f$ is a meromorphic function on the Riemann surface $X$. Then the current $[\omega]$ is


$$[\omega]:g \mapsto\int_X g\omega$$
for any $C^\infty$ function $g$. Here the problem: note that $\omega$ is $0$ almost everywhere, in particular $\omega$ is supported in the set of zeroes and poles of $f$ i.e. in a finite set! Why is the integral $\int_X g\omega$ different from $0$? It seems almost obvious to deduce that the integral of a differential form supported in a finite set is $0$ because of the properties of the Riemann integral in $\mathbb R^n$. I've been thinking to this fact all the day but without any solution.
 A: Concerning the first bullet point: Yes, integration can be extended to non-smooth forms. Locally integrable isn't quite sufficient, unless the manifold is compact. In your setting, one would typically require that $\alpha$ has compact support. Then $\alpha \wedge \omega$ is a locally integrable form with compact support, and hence integrable. One can also require fast enough decay for $\alpha$ and impose growth conditions for $\omega$ - think e.g. of of the Schwartz space of rapidly decreasing functions and its dual, the space of tempered distributions; not every locally integrable function defines a tempered distribution, it mustn't grow too rapidly at $\infty$.
Concerning the second bullet point, we interpret $\log \lvert f\rvert$ as a current, and hence don't consider the classical derivative where it exists, but the distributional derivative (I'm not sure whether there's a different term in the context of currents, but the idea is the same as with distributions and test functions). Thus when you integrate, you shift the differentiations over to $g$ via integration by parts. Evaluating such an integral then shows that $[\omega]$ is some linear combination of evaluation at the zeros and poles of $f$ (the coefficients depending on the order of the respective zeros and poles).
