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As we know that weak limit of a sequence of Borel probability measures on metric space is unique. Do we have this property for general sequence of signed Borel measures on metric space? Thank you.

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    Yes, the measures are finite.2012-09-06

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We assume that $\{\mu_n\}$ converges weakly to $\nu_1$ and $\nu_2$. Taking $f=1$, we can see that $\nu_1(X)=\nu_2(X)$ and putting $\nu=\nu_1-\nu_2$, we have $\nu(X)=0\mbox{ and }\quad \forall f\in C_b(X), \int_Xf(x)d\nu(x)=0.$ Take $O$ an open subset of $X$. We can approach pointwise the characteristic function of $O$ by a sequence of continuous bounded functions, which gives, by dominated convergence and decomposition of $\nu$ that $\nu(O)=0$ for each open subset $O$. Since the open subsets forms a collections stable under finite intersections which generates Borel $\sigma$-algebra on $X$, we deduce that $\nu=0$, hence $\mu_1=\mu_2$ and the limit is unique.