Convergence of total masses of positive Borel measures is implied by their weak convergence.

In the context of a bigger proof, I would like the following to hold true:

Claim. Let $(\mu_j)_{j\in\mathbb{N}}$ be a sequence of positive Borel measures on $\mathbb{C}$ that converges weakly toward $\mu$, a probability measure on $\mathbb{C}$, that is for all $\varphi\in C_0^\infty(\mathbb{C})$, one has: $$\lim_{j\to+\infty}\int_\mathbb{C}\varphi\,\mathrm{d}\mu_j=\int_\mathbb{C}\varphi\,\mathrm{d}\mu.$$ Then, the following limit holds: $$\lim_{j\to+\infty}\mu_j(\mathbb{C})=\mu(\mathbb{C}).$$

I know that for each positive Borel measure $\nu$ on $\mathbb{C}$, one has: $$\nu(\mathbb{C})=\sup\left\{\int_\mathbb{C}\varphi\,\mathrm{d}\mu_j;\varphi\in C_0^\infty(\mathbb{C}),0\leqslant\varphi\leqslant 1\right\}.$$ Therefore, I kind of want to invert $\lim$ and $\sup$, but this does not really make sense (to me).

Any enlightenment will be greatly appreciated!

• It's not true without more assumptions. Let $\mu_j$ be a point mass at $z_j$, where $|z_j| \to \infty$, and let $\mu$ be the zero measure. – Nate Eldredge Oct 16 '17 at 17:04

I don't think this is the case. Take any sequence of measures $\mu_n$ that weakly approximates the unit point mass $\mu = \delta_0$ at the origin. Now I could perturb my approximating sequence by putting some additional mass to $\mu_n$, supported in the annulus $A_n$ consisting of complex numbers of modulus in between $n$ and $n+1$. Note that the perturbed sequence still converges weakly to $\delta_0$, but the total mass of each $\mu_n$ has changed.
It will hold for bounded sets $C$, but not necessarily for the whole space. You can approximate a bounded set from above by an open subset and from below by a compact subset. Then fit between the compact and the open a smooth function.