Equivalent condition for measure convergence A sequence of finite (probability) measures $(\mu_n)$ on $\mathbb R^d$ converges to a finite measure $\mu$ if for every $f \in C_b(\mathbb R^d)$, 
$$ \int_{\mathbb R^d} f(x) \mu_n(dx) \to \int_{\mathbb R^d} f(x) \mu (dx). $$
Textbook says that the following are equivalent:
(1) $\mu_n \to \mu$. 
(2) For every $B \in \mathcal B(\mathbb R^d)$ which is a $\mu$-continuity set, i.e. $\mu(\partial B) = 0$, we have $\mu_n(B) \to \mu(B)$. 
I readily proved that (1) implies (2) and for the converse, I tried to approximate arbitrary continuous bounded function to simple functions consisting of $\mu$-continuity sets and failed. Is it impossible? If then what argument do I have to use?
 A: The proof in Kallenberg works essentially like this:
Let $G$ be an open set and define $G^\varepsilon = \{x \in G \mid U_\varepsilon(x) \subset G\}$. Note that $G^\varepsilon \subset \overline{G^\varepsilon} \subset G^{\varepsilon - \delta}$ for $0 < \delta < \varepsilon$, so $G^\varepsilon$ is a $\mu$-continuity set in all the continuity points of the map $\varepsilon \mapsto \mu(G^\varepsilon)$. Since this map is decreasing in $\varepsilon$, it has only countably many discontinuities and we can find a null sequence $\varepsilon_m$ so that $G^{\varepsilon_m}$ is a $\mu$-continuity set for every $m$. This implies
$$\liminf\limits_{n \to \infty} \,\mu_n(G) \ge \liminf \limits_{n \to \infty}\,  \mu_n(G^{\varepsilon_m}) = \lim \limits_{n \to \infty} \mu_n(G^{\varepsilon_m}) = \mu(G^{\varepsilon_m}).$$
Using $G = \bigcup \limits_{m = 1}^\infty G^{\varepsilon_m}$ and the continuity of measures from below we get $\liminf \limits_{n \to \infty} \, \mu_n(G) \ge \mu(G)$ for any open set $G$.
Assume $f \ge 0$ is a bounded continuous function. Then by Fubini's theorem and Fatou's lemma
$$\int f \, d\mu = \int\limits_0^\infty \mu(f > t) \, dt \le \int\limits_0^\infty \liminf \limits_{n \to \infty}\, \mu_n(f > t) \, dt \le \liminf \limits_{n \to \infty} \int\limits_0^\infty \mu_n(f > t) \, dt = \liminf \limits_{n \to \infty} \int f \,d\mu_n.$$
Now let $f$ be a continuous function with $|f| \le c < \infty$. Applying the aforementioned inequality to $c \pm f$ yields the desired result.
It is probably also possible to approximate $f$ by simple functions consisting of $\mu$-continuity sets. Essentially you would need to prove that the set $\{t \mid \{f > t\} \text{ is } \mu-\text{continuous}\}$ is dense in $\mathbb{R}$.
