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?


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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.


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