Stokes theorem for a current I am struggling to understand the Stokes theorem for currents(differential form with distributional coefficients). The statement is as follows:
Lest $S$ be a current of degree $N - 1$ with compact support in $\Omega$. Then $\int_{\Omega} dS = 0$. The proof is: Let $\chi$ be a smooth cut-off function in $\Omega$ such that $\chi \equiv  1$ in a neighborhood of $K$, a compact subset of $\Omega$ containing the support of $S$. Then $\int_{\Omega} dS = \int_{\Omega} \chi dS = \langle S, d\chi \rangle = 0$.
I am not really following the logic here. First of all, how would one define the integral of a current? I know that a current of degree $p$ is a continuous linear functional on the space of smooth (n-p) forms with compact support, and they are essentially forms with distributional coefficients. My guess is that the coefficients of a top degree form will always be locally integrable, then you just integrate the coefficients over the support of the current but I am not sure whether this is the correct interpretation or not.
 A: In my opinion the statement is a tautology but at the same time abuses the notation. $S$ is of degree $N-1$ (in view of the statement the dimension must be $N$) so it 'acts' upon 1 forms (with smooth and compact coefficients). The exterior derivative is defined by duality so $dS$ acts upon $0$-forms, i.e. functions. So the stated claim may be written:
$$ "\langle dS,1 \rangle" = \langle dS, \chi \rangle 
= (-1)^N \langle S, d\chi \rangle = 0 $$
since $d\chi$ has support disjoint from that of $S$.
You may look e.g. in chapter 3 of
Currents for this type of calculations. I have put the first expression in quotation marks as it is also a slight abuse of notation  since the constant function 1 does not have compact support, but it still makes sense since $S$ is supposed to have compact support.
The above coincides with standard integrals when these are defined but the notation may be misleading.
Some more details: In $n$ dimensional Euclidean space and for (non-negative)
$p+q=n$ there is a
natural bi-linear map on smooth forms $(\alpha,\beta) \in \Omega^p({\Bbb R}^n) \times \Omega^q({\Bbb R}^n)) \rightarrow \int \alpha\wedge \beta\in {\Bbb R}$ defined when at least one of the forms (say $\beta$) has compact support.
The map, $\beta\in \Omega^q_c({\Bbb R}^n) \mapsto \langle \Lambda_\alpha, \beta\rangle =\int \alpha \wedge \beta$ then defines a continuous linear map on the space of $q$-forms with compact support (equipped with its natural Frechet space topology). The space of such linear maps is called the space of $p$-currents, which thus contains $\Omega^p({\Bbb R}^n)$ as a naturally embedded image under the map $\alpha \mapsto \Lambda_\alpha$ above.
For smooth forms $\alpha$ and $\beta$ of degree $p$ and $q$, respectively, one has $d(\alpha\wedge \beta)= d\alpha \wedge \beta + (-1)^p \alpha \wedge d\beta$. However, when $p+q=n-1$ then by Stokes,
$\int_D d(\alpha \wedge \beta) = \int_{\partial D} \alpha\wedge \beta = 0$ when the open set $D$ contains the support of $\beta$. Thus,
$$ \langle d\Lambda_\alpha, \beta \rangle = \int d\alpha \wedge \beta = -(-1)^{p} \int \alpha \wedge d\beta = (-1)^{p+1} \langle \Lambda , d\beta\rangle.$$
One then declares equality of extremities to define the exterior derivative of any $p$-current $\Lambda$ (when $p=n-q-1$). This now leads to the tautology as explained above.
