Conclusion about measurable functions from knowledge about continuous functions 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.
 A: 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). 
