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.

  • $\begingroup$ By weak limit, I guess $\int_Xfd\mu_n\to \int_Xfd\mu$ for all $f$ continuous and bounded. Are the measure supposed to be finite? $\endgroup$ Sep 6, 2012 at 9:48
  • $\begingroup$ Yes, the measures are finite. $\endgroup$
    – Deco
    Sep 6, 2012 at 9:50

1 Answer 1


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.


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