TITLE: Vague convergence of measures
QUESTION [5 upvotes]: Define a subprobability measure to be a measure on Borel sigma algebra of the real line $\mathbb{R}$  with the measure for the whole real line less or equal than 1.
I was wondering about the definition of vague convergence of a sequence of subprobability measures $\{ \mu_n, n\geq 1 \}$ to another subprobability measure $\mu$. The convergence can be defined in slightly two different ways as in Chung's probability theory book p85 and p90:
(1) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b]) \rightarrow \mu((a,b])$.
(2) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b)) \rightarrow \mu((a,b))$ (the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?).
How to show these two definitions are equivalent?
A side question: 
is there a definition of vague convergence for general measures on more general sigma algebra with more general underlying space?
Thank you so much!

REPLY [2 votes]: In my previous answer, I have shown that definition (1) implies definition (2). Now I prove the converse implication.
Suppose that the condition in definition (2) holds. That is, $\mu_n ((a,b))$ converges for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $\mathbb{R}$. We are now going to specify the limit. For any $a,b \in D$, $a < b$, define
$$
\mu ((a,b)) = \mathop {\lim }\limits_{n \to \infty } \mu _n ((a,b)).
$$
Since $\mu_n$ is a sequence of subprobability measures (s.p.m.), $\mu ((a,b)) \in [0,1]$.
Next, fix any $a,b \in {\mathbb R}$, $a < b$. If $a_m$ is a decreasing sequence converging to $a$, and $b_m$ is an increasing sequence converging to $b$, $a_m,b_m \in D$, then, by the last definition, $\mu ((a_m,b_m))$ is a monotone increasing sequence, bounded from above by $1$. Hence, this sequence converges, and we can define
$$
\mu ((a,b)) = \mathop {\lim }\limits_{m \to \infty } \mu ((a_m ,b_m )) \in [0,1].
$$
This actually determines a s.p.m. $\mu$ on $\mathbb{R}$. Now, define $\tilde D = \lbrace x \in {\mathbb R}:\mu (\lbrace x \rbrace ) = 0 \rbrace$ and proceed similarly to my previous answer, based on 
$$
\mu _n ((a',b')) \le \mu _n ((a,b]) \le \mu _n ((a'',b'')),
$$
to conclude that for any $a < b$ in (the dense set) $\tilde D$, $\mu _n ((a,b]) \to \mu ((a,b])$.