Helly-Bray's theorem for bounded discontinuous functions If a sequence $(\mu_n)_{n\ge 1}$ of probability measures converges weakly to a p.m. $\mu$, then $$ \lim_{n\to \infty} \int f d\mu_n=\int f d\mu $$
for every $f:\mathbb{R}\rightarrow \mathbb{R}$ bounded and with finite set of discontinuity points $D$ such that $\mu(D)=0$.
Can anyone give me a hint? I know a proof in the case where $f$ is continuous using the fact that for every $\varepsilon>0$ there exists $a<b$ such that $\mu((a,b])>1-\varepsilon$ and writing
$$ \left \vert \int f d\mu_n-\int f d\mu \right \vert \le \left \vert \int_{(-\infty,a]} f d\mu_n-\int_{(-\infty,a]} f d\mu \right \vert + \left \vert \int_{(a,b]} f d\mu_n-\int_{(a,b]} f d\mu \right \vert + \left \vert \int_{(b,+\infty)} f d\mu_n-\int_{(b,+\infty)} f d\mu \right \vert $$
Is it a good path?

Let the discontinuities of $f$ be $d_i$. Write $f=\sum_{i=1}^n f_i1_{A_i}$ where $A_i$ forms a finite partition of open intervals of $\mathbb{R}$ such that the discontinuities of $f_i$ are on the edges of $A_i$ (so together with $\{d_i\}$ they partition $\mathbb{R}$). Then for all $i$, $\int_{A_i} f_id\mu_n\rightarrow \int_{A_i}f_id\mu_n$, since each $f_i$ is continuous and bounded on $A_i$. Finally write $\int fd\mu_n=\sum_i \int_{A_i} f_id\mu_n$.
