zero extension of positive currents are always positive In Demailly's Complex Analytic and Differential Geometry page 139:
He said the trivial (zero) extension of the positive current $T$ (on $X\setminus E$), which denoted by $\tilde T$ is always positive on $X$. It seems like quite obvious, but I came across some obstacle when verifying that claim.
Notice that a current $T\in\mathcal D^{'}_{p,p}(X)$ is said to be positive if $\langle T, u\rangle\geqslant 0$ for all test forms $u\in\mathcal D_{p,p}(X)$ that are strongly positive at each point.Another way of stating the definition is:
$T$ is positive if and only if $T \wedge u \in \mathcal D_{0,0}^{'}(X)$ is a positive measure for all strongly positive forms $u \in\mathcal C_{p,p} ^{\infty}(X)$.
This is so because a distribution $S\in\mathcal D^{'}(X)$ such that $S(f)\geqslant 0$ for every non-negtive function $f\in\mathcal D (X)$ is a positive measure.
Here is my thought: 
First, select arbitrary $u \in\mathcal C_{p,p} ^{\infty}(X)$, we have $u \in\mathcal C_{p,p} ^{\infty}(X\setminus E)$, then we would like to show that
$$\langle \tilde T\wedge u,f\rangle=\langle  T\wedge u,f\rangle\geqslant 0, \qquad f\in\mathcal D(X),$$
due to the construction of $\tilde T$. However, supp $(f)\cap (X\setminus E)$ isn't compact in $X\setminus E$. Then, how can I infer LHS is non-negative?
Any help and suggestion are appreciated. Thanks a lot!
 A: You are totally right : as for distributions, it is easier to define restrictions rather than extensions for currents. This is why the local boundedness of $\lVert T \rVert$ near $E$ is crucially required in this theorem. The key is now you can locally approximate your test fonctions by smooth functions supported outside $E$.
I take the same notations as in the proof of Demailly. In particular, you have a smooth non-negative function $(\theta \circ v_k)$ such that the current $(\theta \circ v_k) T$ weakly converges to $\tilde T$ (I think this limit is the way to understand $\tilde T$).
Let $f \in \mathcal D(X)$ with non-negative values, and $u \in \mathcal C_{p,p}^{\infty}(X)$ be strongly positive. As $fu$ is a test form (smooth, compactly supported), weak convergence gives you the following :
$$\langle \tilde T, fu \rangle = \lim_{k\to \infty} \langle (\theta \circ v_k) T,
 f u \rangle $$
You conclude with duality :
$$ \langle \tilde T \wedge u , f \rangle = \lim_{k \to \infty} \langle T \wedge u, (\theta \circ v_k) f \rangle \geqslant 0 $$
Since $(\theta \circ v_k) f$ is a non-negative function.
