# Current induced by a locally integrable differential form: I don’t understand why it is not trivially 00

Today this question captured my attention, hence I want to generalize it.

Let XX be a complex manifold of dimension MM and let ω\omega be a (n−p,n−q)(n-p,n-q)-differential form such that in each chart it is represented by locally integrable functions. A standard example of current on XX (see for example De Rham – Differential manifolds, Chap. III example 2) is the following:

[ω]:α↦∫Xα∧ω[\omega]:\alpha \mapsto\int_X \alpha\wedge\omega
where α\alpha is a C∞C^\infty (p,q)(p,q)-form and α∧ω\alpha\wedge\omega is a locally integrable (n,n)(n,n)-form.

Who ensures that α∧ω\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=1n=1, p=0p=0, q=0q=0 and let’es examine the example of the question linked above:
ω=∂ˉ∂log|f|\omega=\partial\bar\partial\log|f|
where ff is a meromorphic function on the Riemann surface XX. Then the current [ω][\omega] is

[ω]:g↦∫Xgω[\omega]:g \mapsto\int_X g\omega

for any C∞C^\infty function gg. Here the problem: note that ω\omega is 00 almost everywhere, in particular ω\omega is supported in the set of zeroes and poles of ff i.e. in a finite set! Why is the integral ∫Xgω\int_X g\omega different from 00? It seems almost obvious to deduce that the integral of a differential form supported in a finite set is 00 because of the properties of the Riemann integral in Rn\mathbb R^n. I’ve been thinking to this fact all the day but without any solution.

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