Can someone provide an example of probability measures $\{\mu_n\}$ and $\{\nu_n\}$ such that although $\int_{\mathbb{R}}f d\mu_n - \int_{\mathbb{R}}f d\nu_n \rightarrow 0$ for all continuous real-valued functions with compact supports, for no finite interval $I=(a,b)$ does it hold that $\mu_n(I)-\nu_n(I) \rightarrow 0$.
Taking the probability measures as delta functions would probably help meet the first requirement, but if both such sequences of delta functions march off to infinity, then the second requirement will get violated. Any suggestions would be appreciated.
For any $n$ and $k = 0\ldots 2^{2n}$ $\mu_{n, k} = \delta_{\frac{k}{2^{2n}} - 2^n}$ (ie we go through $[-2^n; 2^n]$ with step $2^{-n}$) and $\nu_{n, k}(X) = \mu_{n, k}(X + \frac{1}{2^n})$.
Now let us build sequence $\mu_n$ enumerating $\mu_{n, k}$ first on $n$ and then on $k$: $\mu_{0, 0}, \mu_{0, 1}, \mu_{1, 0}, \mu_{1, 1}, \ldots, \mu_{1, 4}, \mu_{2, 0}, \ldots$, and similarly for $\nu$.
Now, as $|f(x) - f(x + 2^{-k})| < \varepsilon$ for sufficiently large $k$, we have $|\int f\ d\mu_n - \int f\ d\nu_n| < \varepsilon$ for sufficiently large $n$. However, as for any $I$ for all large enough $k$ there is some $m: |m| < 2^{2k}$ such that $\frac{m}{2^k} \in I$ but $\frac{m + 1}{2^k} \notin I$, $\mu_n(I) - \nu_n(I)$ can be equal to $-1$ for arbitrary large $n$ (and of course it can also be equal to $0$ for arbitrary large $n$).