# 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 $$a such that $$\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?

Provided that you know that the set discontinuities $$D_f$$ of $$f$$ is measurable, then one may use the Portmanteau theorem as follows:
(i) Claim ($$\mu_n\circ f^{-1}\stackrel{n}{\Longrightarrow}\mu\circ f^{-1}$$: For any closed set $$F\subset \mathbb{R}$$, we have $$f^{-1}(F)\subset\overline{f^{-1}(F)}\subset D_f\cup f^{-1}(F)$$ If $$\mu(D_f)=0$$ then $$\mu(f^{-1}(F))=\mu(\overline{f^{-1}(F)})$$. By the Portmanteau theorem \begin{align} \limsup_n\mu_n (f^{-1}(F))\leq\limsup_n\mu_n(\overline{f^{-1}(F)})\leq \mu(\overline{f^{-1}(F)})= \mu(f^{-1}(F)) \end{align} This shows that $$\mu_n\circ f^{-1}\stackrel{n}{\Longrightarrow}\mu\circ f^{-1}$$.
Now, let $$\phi(x)=((-M)\vee x)\wedge M$$ where $$M=\|f\|_u$$. As $$f=\phi\circ f$$ and $$\phi\in\mathcal{C}_b(\mathbb{R})$$, by part (i) \begin{align*} \int f\,d\mu_n&=\int \phi\circ f\,d\mu_n=\int \phi \,d\mu_n\circ f^{-1} \xrightarrow{n\rightarrow\infty}\int \phi\,d\mu\circ f^{-1}=\int \phi\circ f\,d\mu=\int f\, d\mu. \end{align*}
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 $$\int fd\mu_n=\sum_i \int_{A_i} f_id\mu_n$$.
• How can I prove the convergence $\int_{A_i} f d\mu_n \to \int_{A_i} f d\mu$? The version of Helly-Bray's theorem I know assumes that the function is continuous in $\mathbb{R}$. May 10, 2021 at 21:26
• @Curious: You can instead think of it as restricting $\mu_n,\mu$ onto the set $A_i$, on which $f$ is continuous. May 10, 2021 at 23:14