It is classical that Euclidean normal currents are dense in the space of all currents.
This can be achieved through mollification.


What I want to know if this is still true for metric currents.
In particular I am interested if the space of Normal 1-currents is always dense in the space of all 1-currents for any separable metric space.


Even more specifically I want to know if it is possible for a metric space to have metric currents while having no normal metric currents.


It would make sense that a positive density result could be achieved for nicely supported measures in Banach spaces, but I cannot be sure how to proceed for badly behaved metric spaces since mollifications would lead outside of the space.



Are normal metric currents dense in the space of all metric currents?
The answer is no, and it is pretty trivial apparently.
Let us look at $[0,1]\subseteq\mathbb R$ and define on it the simplest current
$$
T(fd\pi)=\int\_0^1f(t)\pi'(t)dt.
$$


Now we enumerate the rational numbers on it $\{q\_n\}\_{n\in\mathbb N}\subseteq[0,1]$ and given any positive $1>\varepsilon>0$ and any positive element $r\in\ell^1$ such that
$$
\|r\|\_1\leq\varepsilon/2
$$
so we can define the compact set
$$
K=[0,1]\setminus\bigcup\_n B(q\_n,r\_n).
$$
Since we have that $K$ is completely disconnected and $\mathcal L^1(K)>1-\varepsilon$ we have that
$$
T\_{|K}(fd\pi)=\int\_K f(t)\pi'(t)dt
$$
is a current but it is not normal anymore.
We can conclude that $K$ as a metric space has non trivial currents but cannot have normal currents since if it had any they would be supported on curves, and since $K$ is completely disconnected, that cannot happen.


