Tight sequence of measures

Question

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.

Answer 1

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.

Answer 2

This is really just a comment to add context to your question and the reactions but it will be too long so I am disguising it as an answer. Stripping away all trappings the relevant fact is that an equicontinuous set $A$ in the dual of a locally convex space $E$ is relatively compact for the corresponding weak topology and so that latter coincides on $A$ with the weak topology induced by any dense subspace $E_0$ of $E$. In your case, $E$ is the space of bounded, continuous functions on a locally compact space, provided with the so-called strict topology which was introduced by R.C. Buck in the 50‘s. This has the following three relevant properties which set up the connection to your question.

The dual space is the space of finite tight (or Radon) measures;

A family of measures is equicontinuous if and only if it is bounded and uniformly tight;

The space $E_0$ of continuous functions with compact supports is dense.

This shows that many of the assumptions given above are irrelevant. One can even replace local compactness by tcomplete regularity. The former condition is required to ensure the denseness of $E_0$. In the general case, one can use any subalgebra of $E$ which separates points and is such that for each element of the underlying space there is a function in $E_0$ which does not vanish there.