3
$\begingroup$

Let $(\mu_n)$ be a sequence of positive finite Borel measures on $\mathbb{R}$. Suppose $\lim_{n \rightarrow \infty} \int h d\mu_n$ converges for every bounded, continuous function $h$ on $\mathbb{R}$. Must there exist a finite Borel measure $\mu$ on $\mathbb{R}$ such that $$ \lim_{n \rightarrow \infty} \int h d\mu_n = \int h d\mu $$ for every bounded, continuous function $h$ on $\mathbb{R}$?

My initial thought was that the Lévy–Prokhorov metric for finite Borel measures might be useful. But then I realized that Cauchy-ness of the sequence $\lim_{n \rightarrow \infty} \int h d\mu_n$ may not be the same thing as Cauchy-ness of the sequence $(\mu_n)$ in the Lévy–Prokhorov metric.

$\endgroup$
  • $\begingroup$ Are the $\mu_n$ positive measures? $\endgroup$ – Davide Giraudo Nov 20 '13 at 10:53
2
$\begingroup$

Let $X\sim N(0,1)$ and $\mu$ its distribution. Then $\mu$ is a probability measure in $\mathbb{R}$.

Define $\mu_n(A) = \mu(A \cap [n,n+1])$ and note that

$$ \int f d\mu_n = \int f\chi_{[n,n+1]}d\mu$$

but $f\chi_{[n,n+1]} \rightarrow 0$ a.s. and $f\chi_{[n,n+1]}$ is bounded once $f$ itself is bounded. Then by the Theorem of Dominated convergence ($\mathbb{R}$ has finite measure through $\mu$) we have that $$ \lim_n \int fd\mu_n = 0 $$ for all limited functions.

But if $\mu_n \rightarrow \eta$ weakly, then by Portmanteau, we should have $$ 0=\liminf_n \mu_n((-k,k))\ge\eta((-k,k))$$ for all $k$. And this implies $\eta \equiv 0 $

I know that $0$ is trivially a measure, but this shows that if you require a non-trivial measure the statement is false.

If you allow this kind of situation, I still don't know. Note that the hypothesis implies existence of limit for the characteristic functions. If you prove the sequence of measures is rigid so your statement is true (it is a Lévy's continuity for rigid sequences)

Sorry if I don't give you a answer, but I don't have the privilege to comment the questions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.