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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?

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2 Answers 2

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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*}

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  • $\begingroup$ No, thanks, Oliver ;) $\endgroup$
    – Curious
    May 17, 2021 at 12:28
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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 $\int fd\mu_n=\sum_i \int_{A_i} f_id\mu_n$.

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  • $\begingroup$ 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 $\mathbb{R}$. $\endgroup$
    – Curious
    May 10, 2021 at 21:26
  • $\begingroup$ @Curious: You can instead think of it as restricting $\mu_n,\mu$ onto the set $A_i$, on which $f$ is continuous. $\endgroup$
    – Alex R.
    May 10, 2021 at 23:14

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