# Weak Convergence of probability measures on the support

$(\mu_n)_{n \in \mathbb{N}}$ is a sequence of probability measures on $\mathbb{R}_+^2$. $\mu$ is also a probability measure on $\mathbb{R}_+^2$, but $\mu(\mathbb{R}_+ \times [0,x_0])=0$. So is it enough to show the convergence only for sets $B \in \mathcal{B}(\mathbb{R}\times (x_0,\infty))$ ?

Meaning, is it enough to show: $$\text{For all } B \in \mathcal{B}(\mathbb{R}\times (x_0,\infty)) \text{ with } \mu(\partial B)=0 : \mu_n(B) \longrightarrow \mu(B) \text{ as } n \rightarrow \infty$$

• Why I think that is true: $μ_n(\mathbb{R}_+^2)=μ_n(\mathbb{R}_+ \times [0,x_0])+μ_n(\mathbb{R}_+ \times (x_0,\infty))=1$. And because $\mu_n(\mathbb{R}_+\times (x_0,\infty))\rightarrow 1$, it follows that $\mu_n(\mathbb{R}_+\times[0,x_0]) \rightarrow 0$. – Kana Aug 16 '13 at 15:45

There is a property of Borel $\sigma$-algebra of the real line which say that the trace $\sigma$-algebra over a closed/open set is the same as the Borel $\sigma$-algebra of the trace topology.
Here, we take $B\in\mathcal B(\mathbb R\times (x_0,+\infty))$ for which $\mu(\partial B)=0$. There is $B'\in\mathcal B(\mathbb R^2)$ such that $B=B'\cap (\mathbb R\times (x_0,+\infty))$. Since $\partial B'\subset\partial B$, we have $\mu(\partial B')=0$, hence $\mu_n(B')\to \mu(B')$. Let $S:=\mathbb R\times [0,x_0]$. As $$|\mu_n(B)-\mu(B)|\leqslant |\mu_n(B')-\mu(B')|+\mu_n(S)+\mu(S).$$ Since $\lim_{n\to \infty}\mu_n(S)=0$ (using the assumption with $B:=\mathbb R\times [0,x_0]$), we are done.