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I would like to investigate if an analog of the classical Portmanteau theorem holds for vague convergence of Radon measures.

Here are the definitions I'm using.

Let $X$ be a Hausdorff locally compact topological space, and let $\mathcal{B}(X)$ be its Borel $\sigma$-algebra. A positive measure $\mu$ on $(X, \mathcal{B}(X))$ is said to be a Radon measure if : (i) $\mu(K) < \infty$ for all compact subsets $K \subset X$, (ii) $\mu(O) = \sup \, \{\mu(K) \, / \, K \subset O, K \rm{\: is \: compact \:} \}$ for all open subsets $O \subset X$, and (iii) $\mu(A) = \inf \, \{ \mu(O) \, / \, A \subset O, O \rm{\: is \: open \:} \}$ for every Borel subset $A \subset X$.

I will say that a sequence $(\mu_n)_n$ of Radon measures on $X$ converges vaguely to a Radon measure $\mu$ if $\lim_{n \to \infty} \int_X f \, d \mu_n = \int_X f \, d \mu$ for all $f \in C_c(X)$, where $C_c(X)$ denotes the set of all continuous real-valued functions defined on $X$, with a compact support.

Now, consider the following propositions :

$(P_1)$ $(\mu_n)_n$ converges vaguely to $\mu$;

$(P_2)$ $\mu(O) \le \underline{ \lim } \, \mu_n(O)$ for all open subsets $O \subset X$;

$(P_3)$ $\mu(K) \ge \overline{ \lim } \, \mu_n(K)$ for all compact subsets $K \subset X$;

$(P_4)$ $\mu(A) = \lim \, \mu_n(A)$ for all Borel subsets $A \subset X$ with a compact closure, and satisfying $\mu(\partial A) = 0$.

I'm able to prove that $(P_1) \Leftrightarrow ((P_2) + (P_3)) \Rightarrow (P_4)$.

My questions are :

1) Are $(P_2)$ and $(P_3)$ equivalent in full generality ? If $X$ is compact, this is obvious by taking complementary sets.

2) Can one prove that $(P_4) \Rightarrow (P_1)$ ? I succeeded to prove that $(P_3) + (P_4) \Rightarrow (P_1)$. (in fact, instead of $(P_3)$, I only need that $\sup_n \mu_n(K) < \infty$ for all compact subsets $K \subset X$)

Thanks.

share|improve this question
    
In Araujo-Ginet's book Probablity measures in Banach spaces, (P_4)$\Rightarrow (P_1)$ when $X$ is a separable locally compact metric space is left as an exercise. –  Davide Giraudo Nov 2 '12 at 21:51
    
have you tried $P_4\Rightarrow (P_3)+(P_2)$? –  Dima McGreen Aug 1 '13 at 16:34

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