PomeProposition 2.6.3.54258F39AC23 Let $(\mu _ i)$ be a net of finite Radon measures on $\Omega $ converging vaguely to $\mu $. Suppose that the net is eventually uniformly tight: for every $\varepsilon >0$ there are a compact set $K\subset \Omega $ and an index $i_0$ such that $\mu _ i(\Omega \setminus K)<\varepsilon $ for all $i\geq i_0$. Then $\mu $ is finite and $\mu _ i\to \mu $ narrowly. In particular, $\mu _ i(\Omega )\to \mu (\Omega )$.
Proposition 2.6.354258F39AC23
Proof. Let $\varepsilon >0$, and choose $K$ and $i_0$ as above. Since $\Omega \setminus K$ is open, proposition 2.6.2 gives $\mu (\Omega \setminus K)\leq \varepsilon $, and since $\mu (K)<\infty $, the measure $\mu $ is finite. Let $f$ be a bounded continuous function on $\Omega $ and $\eta $ a cutoff function for $K$ in $\Omega $. Then $f\eta \in C_ c(\Omega )$, so $\int f\eta \, \mathrm d\mu _ i\to \int f\eta \, \mathrm d\mu $, while $1-\eta $ takes values in $[0,1]$ and vanishes on $K$, so that for $i\geq i_0$
\[ \left|\int f(1-\eta )\, \mathrm d\mu _ i\right|\leq \| f\| _\infty \varepsilon , \qquad \left|\int f(1-\eta )\, \mathrm d\mu \right|\leq \| f\| _\infty \varepsilon . \]
Hence $\varlimsup _ i|\int f\, \mathrm d\mu _ i-\int f\, \mathrm d\mu |\leq 2\| f\| _\infty \varepsilon $. Letting $\varepsilon \downarrow 0$ proves narrow convergence, and $f=1$ gives the last assertion. $\square$
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