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.

• 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 – Thomas Jul 14 '16 at 15:31
• Your result holds for measurable function provided your measures are positive and countably additive. – ibnAbu Jul 14 '16 at 22:12
• In general it is not clear how you choose $f_n$ to converges to $g$ from below. Moreover, which convergence are you referring to? – user99914 Jul 18 '16 at 21:49
• @ArcticChar, as a first step assume $g$ is the characteristic function of a closed interval. Then we can build such $f_n$ easily, right? – PPR Jul 18 '16 at 23:56
• 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 – user99914 Jul 19 '16 at 3:59

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).