# Vague convergence of absolutely continuous measure to absolutely continuous measure

Suppose we have a sequence of absolutely continuous measure $$\mu_n$$ converges vaguely to $$\mu$$, which is also absolutely continuous. Generally, we have $$\int fd\mu_n \to \int fd\mu,\ \forall f\in C_B(\mathbb{R})$$ where $$C_B$$ stands for the space of continuous bounded functions. How to find a counterexample if we take $$f$$ only bounded and Borel-measurable?

First, choose any $$\varepsilon \in (0, 1)$$ and define, for each $$n \geq 1$$, $$\varepsilon_n = \varepsilon/2^{n+1}(2^n - 1)$$, $$I_{n,k} = \left( \frac{k}{2^n} - \varepsilon_n, \frac{k}{2^n} + \varepsilon_n \right)$$ and $$I_n = \bigcup_{k=1}^{2^n-1}{I_{n,k}}.$$ Let $$\mu_n$$ be the uniform distribution on $$I_n$$ - so that its density is just the indicator function of $$I_n$$ rescaled by a factor of $$2^n/\varepsilon$$ (the measure of this set). We first notice that the $$\mu_n$$ converge weakly to the uniform distribution on $$[0, 1]$$ - call this measure $$\mu$$. Indeed, take $$f \in C_B(\mathbb{R})$$. Since $$[0, 1]$$ is compact, the restriction $$f|_{[0, 1]}$$ is uniformly continuous. Choose then any $$\delta > 0$$ and there must be $$n_0 \geq 1$$ such that, for all $$n \geq n_0$$, $$|x - y| < \varepsilon_n \implies |f(x) - f(y)| < \delta$$ whenever $$x, y \in [0, 1]$$. Therefore we have, for $$n \geq n_0$$, $$\int{f\text{d}\mu_n} = \sum_{k=1}^{2^n-1}{ \int_{I_{n,k}}{\frac{2^n}{\varepsilon}f(x)dx} } \leq \frac{2^n}{\varepsilon}\sum_{k=1}^{2^n-1}{ 2\varepsilon_n\left( f\left( \frac{k}{2^n} \right) + \delta \right) } = \frac{1}{2^n-1}\sum_{k=1}^{2^n-1}{f\left( \frac{k}{2^n} \right)} + \delta.$$ Since a continuous function is Riemann integrable, the limit of the right side is just $$\delta + \int{f\text{d}\mu}$$. Since this is true for any $$\delta > 0$$, we get that $$\limsup_n{ \int{f\text{d}\mu_n} } \leq \int{f\text{d}\mu}.$$ Similarly, we can also find that $$\int{f\text{d}\mu_n} \geq \frac{1}{2^n-1}\sum_{k=1}^{2^n-1}{f\left( \frac{k}{2^n} \right)} - \delta$$ and so $$\liminf_n{ \int{f\text{d}\mu_n} } \geq \int{f\text{d}\mu}.$$
Now, consider $$f$$ to be the indicator function of $$I := \bigcup_{n\geq 1}{ I_n }$$. Then $$\int{f\text{d}\mu} = \mu(I) \leq \sum_{n\geq 1}{ \mu(I_n) } = \sum_{n\geq 1}(2^n - 1)\frac{2\varepsilon}{2^{n+1}(2^n-1)} = \varepsilon < 1.$$ But all measures $$\mu_n$$ are supported on $$I_n$$ and therefore $$\mu_n(I) = 1$$ for all $$n$$, showing that $$\int{f\text{d}\mu_n}$$ does not converge to $$\int{f\text{d}\mu}$$.
An interesting observation is that we couldn't find much simpler sets than $$I$$. By Portmanteau Theorem, if $$A$$ is a continuity set - i.e. $$\mu(\partial A) = 0$$ - then we must have $$\mu_n(A) \xrightarrow{} \mu(A)$$. Since we demand $$\mu$$ to be absolutely continuous, this includes all intervals. In this case, we can see that $$\partial I = [0, 1]\setminus I$$, which is a closed set of empty interior and positive measure - since $$I$$ itself does not have full measure on $$[0, 1]$$ - very similar to the fat Cantor set.