Let $T_{1},T_{2},\ldots$ be positive distributions on $\Omega\subseteq\mathbb{R}^{n}$ satisfying $T_{j}\rightarrow T$ in distribution sense (so $T$ is also a positive distribution and $T_{j}\left(\phi\right)\rightarrow T\left(\phi\right);\forall\phi\in C_{0}^{\infty}\left(\Omega\right)$). We all know that if $F$ is a positive distribution then $F\left(\phi\right)$ is well-understood for any $\phi\in C_{0}\left(\Omega\right).$ My question is:
Is it still true that $T_{j}\left(\phi\right)\rightarrow T\left(\phi\right);\forall\phi\in C_{0}\left(\Omega\right)?$
(Here $C_{0}\left(\Omega\right):=\left\{ f:\Omega\rightarrow\mathbb{R}\;\textrm{is continuous},suppf\Subset\Omega\right\} $)
Fix $\phi\in C_{0}\left(\Omega\right)$. Since $\left\{ T_{j}\right\} $ and $T$ are positive distributions, there are positive Radon measures $\left\{ \mu_{j}\right\} $ and $\mu$ so that $$ T_{j}\left(\Psi\right)=\intop_{\Omega}\Psi d\mu_{j};T\left(\Psi\right)=\intop_{\Omega}\Psi d\mu;\forall\Psi\in C_{0}\left(\Omega\right). $$ We want to prove that $$ \intop_{\Omega}\phi d\mu_{j}\rightarrow\intop_{\Omega}\phi d\mu. $$ Set $K:=supp\left(\phi\right),$ by hypothesis we have that $\left\{ \mu_{j}\left(K\right)\right\} $ is bounded since $$ \left|\intop_{K}d\mu_{j}-\intop_{K}d\mu\right|\rightarrow0,\;as\;j\rightarrow\infty. $$
Now we can complete the problem by choosing a sequence $\left\{ \phi_{k}\right\} \subset C_{0}^{\infty},$ s.t. $\phi_{k}\rightrightarrows\phi$ uniformly on $K,$ and the triangle inequality $$ \left|\intop_{K}\phi d\mu_{j}-\intop_{K}\phi d\mu\right|\leq\left|\intop_{K}\phi d\mu_{j}-\intop_{K}\phi_{k}d\mu_{j}\right|+\left|\intop_{K}\phi_{k} d\mu_{j}-\intop_{K}\phi_{k}d\mu\right|+\left|\intop_{K}\phi_{k} d\mu-\intop_{K}\phi d\mu\right|. $$
Note that the Radon measure of a compact set is always finite.