4
$\begingroup$

Let $\mu$ and $\nu$ be two finite Borel measures on $\mathbb{R}$.

We know that if $$\int f d\mu = \int f d\nu $$ for all continuous functions $f$ then $\mu=\nu$ and so the equation above holds for all measurable functions.

Now let $\mu$ be as above and let $\{\nu_n\}_n$ be a sequence of measures on $\mathbb{R}$ such that $$ \int f d\mu = \lim_{n\to\infty}\int f d\nu_n$$ again, for all continuous functions $f$ (the existence of the limit is part of the input).

What is now necessary in order to conclude that the above equation holds for all measurable $f$ as well? I suppose some notion of convergence of measures and some usage of a dominated convergence theorem for the measure instead of the integrand, but I'm not sure..

Let $g$ be measurable and let $f_n$ be a sequence of continuous functions converging to $g$ from below. Then we have \begin{align} \int g d\mu &= \int \lim_{n\to\infty}f_n d\mu \\ &\stackrel{1}{=} \lim_{n\to\infty} \int f_n d\mu \\ &\stackrel{2}{=} \lim_{n\to\infty} \lim_{m\to\infty}\int f_n d\nu_{m} \\ &\stackrel{3}{=}\lim_{m\to\infty} \lim_{n\to\infty}\int f_n d\nu_{m} \\ &\stackrel{4}{=}\lim_{m\to\infty} \int \lim_{n\to\infty} f_n d\nu_{m} \\ &= \lim_{m\to\infty} \int g d\nu_{m} \end{align}

Where in $1$ and $4$ we used the dominated convergence theorem, in $2$ we used the hypothesis and $3$ remains to be justified.

$\endgroup$
  • 3
    $\begingroup$ the condition you wrote down can be viewed as weak convergence of the measures $\nu_n$ to $\mu$. Maybe the corresponding Wiki page will help you to get insight or further pointers to the topic: en.wikipedia.org/wiki/Convergence_of_measures $\endgroup$ – Thomas Jul 14 '16 at 15:31
  • $\begingroup$ Your result holds for measurable function provided your measures are positive and countably additive. $\endgroup$ – ibnAbu Jul 14 '16 at 22:12
  • 1
    $\begingroup$ In general it is not clear how you choose $f_n$ to converges to $g$ from below. Moreover, which convergence are you referring to? $\endgroup$ – user99914 Jul 18 '16 at 21:49
  • $\begingroup$ @ArcticChar, as a first step assume $g$ is the characteristic function of a closed interval. Then we can build such $f_n$ easily, right? $\endgroup$ – PPR Jul 18 '16 at 23:56
  • 1
    $\begingroup$ well, yes. But I am more concerned about the convergence in the general case. If $g$ is more general, I suppose you can at best find such a sequence so that $f_n$ converges to $g$ almost everywhere with respect to $dx$. But it can be quite different from a.e. with respect to $\mu$ or $\nu_n$. @PPR $\endgroup$ – user99914 Jul 19 '16 at 3:59
1
$\begingroup$

Partial answer: The assertion is false if no further assumption are imposed on $\nu_n$. Example: let $\phi$ be a continuous function supported in $[-1,1]$ and $\int \phi dx = 1$. Then the sequence of measure

$$ \nu_n = n \phi(nx) dx$$

converges weakly to the Dirac measure $\mu$ at $0$. That is, for all continuous $f$,

$$\int _{\mathbb R} f d\nu_n \to f(0).$$

However, if $g = \chi_{\{0\}}$, then $\mu(g) = 1$ and $\int_{\mathbb R} g d\nu_n = 0$ for all $n$. (Note that you might construct another counterexample by considering $\widetilde{\nu_n}$ as the Dirac measure at $\{1/n\}$).

In the above example $\nu_n$ are all absolutely continuous with respect to $dx$ and $n \phi(nx)$ has bounded $L^1$-norm, thus this sequence of measure is quite nice already.

On the other hand, if we assume that

$$ \nu_n = f_n dx,$$

where $f_n$ has bounded $L^p$-norm for some $p \in (1,\infty]$. Then a subsequence of $f_n$ converges weakly to some $f\in L^p$. That is,

$$\int_{\mathbb R} gf_n dx \to \int_{\mathbb R} gf dx,\ \ \ \forall g\in L^{p^*}.$$

(This is more or less what you want, you have to impose some integrability condition on $g$ anyway).

| cite | improve this answer | |
$\endgroup$

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.