Let $h,g$ be continuous, $g > 0$, and $$\lim\limits_{|x|\to \infty} \frac{|h(x)|}{g(x)} = 0$$ Let $X_n \xrightarrow[]{D} X$ and let $\mu$ denote the probability measures induced by $X_n, X$, respectively. Suppose that $\int g d\mu_n \leq C < \infty$. Then show that $\int h d\mu_n \to \int h d\mu$.

What I have so far:

First note that by the given condition, for every $n$, $h\in L^1(\mu_n)$ since $|h| \leq kg \in L^1(\mu_n)$, for some $k$. For any $\epsilon > 0$, we choose $M$ large enough, $$0 \leq \left\lvert \int h d\mu_n - \int h d\mu\right\rvert \leq \left\lvert \int_{|x| > M} h d\mu_n \right\rvert + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \left\lvert \int_{|x| > M} h d\mu \right\rvert \\ \leq \int_{|x| > M} \left\lvert h \right\rvert d\mu_n + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \int_{|x| > M} \left\lvert h \right\rvert d\mu \\ \leq \int_{|x| > M} \frac{\left\lvert h \right\rvert}{g} g d\mu_n + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \epsilon \\ \leq \epsilon \int_{|x| > M} g d\mu_n + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \epsilon\ \\ \leq \epsilon \sup\limits_n \int g d\mu_n + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \epsilon \\ \leq \epsilon C + \left\lvert \int_{|x| \leq M} h d\mu_n - \int_{|x| \leq M} h d\mu \right\rvert + \epsilon $$ where in the second line, I assumed integrability of $h$ w.r.t. $\mu$ (not sure if I can assume that) and in the third line by absolute continuity of integral since $\{|x| > M\} \to \emptyset$, which has measure 0. I can't seem to bound the middle term though. My motivation for this splitting was to somehow use continuity of $h$ on a compact interval to invoke boundedness.

  • $\begingroup$ For $M>0$ let $\chi_M: \mathbb{R} \to [0,1]$ be a continuous function such that $\chi_M(x)=1$ for all $|x| \leq M$ and $\chi_M(x)=0$ for all $|x|>M+1$. Write $$\int h = \int h \chi_M + \int h (1-\chi_M)$$ and use a very similar reasoning as you did in your computations to estimate each of the integrals; instead of $$\left| \int_{|x| \leq M} h \, d\mu_n- \int_{|x| \leq M} h \, d\mu \right|$$ you will get $$\left| \int \chi_M h \, d\mu_n - \int \chi_M h \, d\mu \right|$$ which converges to $0$ as $n \to \infty$ because of the weak convergence of $\mu_n$. $\endgroup$
    – saz
    Dec 5, 2017 at 7:22
  • $\begingroup$ Well, h is bounded on compacts and therefore $f \chi_M$ is bounded. $\endgroup$
    – saz
    Dec 5, 2017 at 11:55
  • $\begingroup$ Ah right and the boundedness on $[-M-1,M+1]^c$ comes from the fact that the product is 0 right? $\endgroup$
    – James Yang
    Dec 5, 2017 at 12:02
  • $\begingroup$ Yes, exactly... the function $f \chi_M$ has compact support. $\endgroup$
    – saz
    Dec 5, 2017 at 12:04

1 Answer 1


I will give s sketch of the proof. Your attempt above shows that you can complete this argument. To begin with the hypothesis does imply that h is integrable with respect to $\mu$. Instead of splitting integrals as above do the following: fix a positive integer N and choose a continuous function F which is 1 on $[-N,N]$, 0 outside $(-N-1,N+1)$ with its range contained in $[0,1]$. Write $ \int h d\mu_n$ as $ \int hF d\mu_n+\int h(1-F) d\mu_n$ and do a similar thing for $\int h d\mu$. $\int hF d\mu_n$ converges to $\int hFd\mu$ because hF is a bounded continuous function. To handle $\int h(1-F) d\mu_n$ bound it by $\int_{\mathbb R\setminus [-N,N]} h d\mu_n$ and use the argument you have used in your proof.


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