This is probably a very easy question for experts in probability or measure theory.


I have a sequence of finite measures $\mu\_{n}$ on a non-compact metric space $X$ such that $\mu\_{n}$ converges to $\mu$ in the following sense:
$$ \int\_{X}fd\mu\_{n} \to \int\_{X}fd\mu \ \ \ \ \ \text{ for all f continuous with compact support}
$$
I would like to say that $\mu\_{n}(X)\to \mu(X)$.


I know this is false in general, but I have the additional condition that for every $\epsilon>0$ there is $n\_{0}\in \mathbb{N}$ and $K\subset X$ compact such that $\mu\_{n}(K^{c})\leq \epsilon$ for every $n\geq n\_{0}$. This looks very similar to the definition of tight sequence (which guarantees the result I would like). Is this equivalent?


Additional assumptions: X is Polish and locally compact, precisely it is a closed surface with some finitely many points removed. All measures $\mu\_{n}$ and $\mu$ are area measures of Riemannian metrics (with singularities at the points removed) on X.



Tight sequence of measures
Since your space is Polish, $\mu$ is regular and there exists a compact set $C$ such that $\mu(X\setminus C)<\epsilon$ for each $\epsilon>0$. Since your space is locally compact, there is another compact set $C'$ such that $C$ is a subset of the interior of $C'$. The function given by $f\_n(x)=\max\{0,1-n\cdot d(C,x)\}$ is continuous and supported on $C'$ whenever $n>d(C,C')^{-1}$. Moreover, the sequence $\langle f\_n\rangle$ decreases pointwise to the indicator function$1\_C$. It follows that $\liminf\_n \mu\_n(X)\geq\mu(X)$ for each $\epsilon>0$ and $n$ large enough. A similar argument applied to the compact set in your tightness version shows that $\limsup\_n\mu\_n(X)\leq\mu(X)$.


That the sequence is tight follows directly from inner regularity of measures on Polish spaces. If there $n\_{0}\in \mathbb{N}$ and $K\subset X$ compact such that $\mu\_{n}(X\setminus K)\leq \epsilon$ for every $n\geq n\_{0}$, you can just pick a compact set $K\_m$ for each $m< n\_0$ such that $\mu\_m(X\setminus K\_m)\leq\epsilon$. Take $K'=K\cup\bigcup\_{m<n\_0}K\_m$, then $\mu\_n(X\setminus K')\leq\epsilon$ for all $n$.
As has been pointed out in the comments, tightness is not enough if you only have convergence of integrals for compactly supported continuous functions.


