Statement of 'de Rham's theorem on closed current' In this paper, page $4$, in the second line, the author quote the following: 

in fact thanks to de Rham's theorem on closed currents, if $u \in \mathcal{C_n},$ there exists a distribution $\omega$ on $\mathbb{R^n}$ such that $d\omega = u.$

The author quoted the theorem can be found in  
Georges de Rham, Variétés différentiables. Formes, courants, formes harmoniques, Hermann,
Paris, 1973, Troisième édition revue et augmentée, Publications de l’Institut de Mathématique
de l’Université de Nancago, III, Actualités Scientifiques et Industrielles, No. 1222b.
MR 0346830
or 
John Horváth, Topological vector spaces and distributions. Vol. I, Addison-Wesley Publishing
Co., Reading, Mass.-London-Don Mills, Ont., 1966. MR 0205028
But I do not know France and have no access to the second book. I googled online de Rham theorem, but I have the theorem on manifold, which I think is non-related. 
If someone can provide me the statement of the theorem in the context above, I would appreciate much. 
 A: Here's a couple definitions I'm going to use (just to be on the same page): The space of $k$-currents on $\mathbb{R}^n$ is $C_k$. The $boundary$ of a current $u \in C_k$ is defined as
$$
\partial u (\beta) = u (d \beta), \qquad \forall \beta \in \Omega^k_c (\mathbb{R}^n),
$$
and I call $u$ $closed$ if $\partial u = 0$, and $exact$ if there is $v \in C_{k + 1}$, such that $\partial v = u$ (as distributions). Since $d^2 = 0$, we also get $\partial^2 = 0$ as well, and hence every exact current is closed.
The reverse of this last statement is exactly the theorem you're looking for (I think/hope/assume/etc.):
$\textbf{Theorem (de Rahm-Poincare)}:$ Let $n > k > 0$. On $\mathbb{R}^n$ every closed $k$-current is exact, that is if $u \in C_k$ satisfies $\partial u = 0$, then there is $v \in C_{k + 1}$, such that $\partial v = u$. The statement is false for $k = n$.
Here's a proof-sketch:
Case I: Assume $0< k < n$. Now $u$ is either identically zero, then $v = 0$ will do, or it's not, then let $\alpha_0 \in \Omega^k_c (\mathbb{R}^n)$ be such that
$$
u(\alpha_0) = 1.
$$
Now we have the splitting $\Omega^k_c (\mathbb{R}^n) = \mathbb{R} \alpha_0 \oplus \ker (u)$. Note that $\alpha_0$ cannot be closed, because otherwise, by Poincare's Lemma for compactly supported forms on $\mathbb{R}^n$, we'd have a $\beta \in \Omega^{k - 1}_c (\mathbb{R}^n)$ such that $d \beta = \alpha_0$, and thus
$$
1 = u(\alpha_0) = u(d\beta) = \partial u (\beta) = 0,
$$
since $u$ was assumed to be closed, and $k < n$. Now we get the following two non-zero subspaces in $\Omega^{k + 1}_c (\mathbb{R}^n)$:
$$
\mathbb{R} d \alpha_0 \quad \& \quad d ( \ker (u) )
$$
which intersect in the zero $(k + 1)-$ form only. Thus we can find a third subspace $V$, such that
$$
\Omega^{k + 1}_c (\mathbb{R}^n) = \mathbb{R} d \alpha_0 \oplus d ( \ker (u) ) \oplus V,
$$
and define $v$ as
$$
v (x d \alpha_0 + y + z) = x
$$
where $y \in d ( \ker (u)$ and $z \in V$. Now if $\alpha \in \Omega^k_c (\mathbb{R}^n)$ any, then then I can always write it as $\alpha = x \alpha_0 + \beta$, where $\beta \in \ker (u)$, and thus setting $y = d \beta$ gives
$$
\partial v (\alpha) = v (d \alpha) = v (x d \alpha_0 + y) = x = u (\alpha),
$$
hence $\partial v = u$.
Case II: $k = n$. If an $n$-current is exact, then it's zero, because there are no non-zero $(n + 1)$-currents. Now let $u$ be the integration over $\mathbb{R}^n$. This is closed by Stoke's theorem, but definitely not identically zero, thus not exact.
$\blacksquare$
Note that this is still a sketch, you still need to worry about things like whether $v$ is continuous or not (which is essentially the same as worrying about the choice of $V$). But I think the idea is there. For the Poincare lemmas see Bott-Tu.
