Absolutely Continuous Weakly Convergent Sequence Need Not Converge Strongly The following appears as an exercise in Sinai and Koralov's Theory of Probability and Random Processes.

Give an example of a family of probability measures $P_{n}$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ such that $P_{n}\Rightarrow P$ (weakly), $P_{n}$, $P$ are absolutely continuous with respect to the Lebesgue measure, yet there exists a Borel set $A$ such that $P_{n}(A)$ does not converge to $P(A)$.

My Efforts


*

*Denote the corresponding cumulative distribution functions by
$F_{n}$, $F$. Because $P_{n}\Rightarrow P$, $F_{n}(x)\to F(x)$ at
all points $x\in\mathbb{R}$ where $F$ is continuous, which is
everywhere since $F\ll\lambda$ ($\lambda$ denoting Lebesgue
measure). (Furthermore, since each $F_{n}$ is also continuous, the
pointwise convergence $F_{n}\to F$ is uniform in any compact set.)
This seems to limit our options. For example, it implies that
$P_{n}(A)\to P(A)$ whenever $A$ can be written as a finite union of
intervals. Perhaps a countable union of intervals
could work. Since we are dealing with probability measures, though,
the component intervals 'running off to $\infty$' seems unlikely to
help.

*Because $F_{n}$, $F$ are absolutely continuous with respect to
$\lambda$, they have densities $f_{n}$, $f$. If $f_{n}\to f$ too
nicely (e.g. in such a way that allowed us to apply the Monotone or
Dominated Convergence Theorem), we'd find that
$P_{n}(A)=\int_{A}f_{n}\,\mathrm{d}\lambda$ converged to
$\int_{A}f\,\mathrm{d}\lambda=P(A)$ for any Borel set $A$. Thinking
along these lines, I looked around unsuccessfully for a kind of 
counterpart to Scheffé's Theorem, which says

If $f_{n}\to f$ $\lambda$-a.e., then $P_{n}(A)\to P(A)$ for all $A\in\mathcal{B}(\mathbb{R})$ (and the rate of convergence is uniform).


*The Portmanteau Theorem tells us that if $P_{n}\Rightarrow P$, then
$P_{n}(A)\to P(A)$ for all Borel sets $A$ satisfying
$\lambda(\partial A)=0$. Fat Cantor sets are Borel sets whose
boundaries (themselves) have positive Lebesgue measure, so maybe
$P$ could be defined in terms of a fat Cantor set and each $P_{n}$
in terms of some step in its construction or something.
 A: Let $P$ be the uniform measure on $[0,1]$. Consider the family
$n^{-1}\sum_{i=1}^{n}\delta_{i/n}$. It is easy to see that
$n^{-1}\sum_{i=1}^{n}\delta_{i/n}\Rightarrow P$ (use the fact that any
$f\in C_{b}(\mathbb{R})$ is uniformly continuous on $[0,1]$). Of course,
anything involving a Dirac measure isn't going to be absolutely
continuous with respect to $\lambda$. However, we can approximate:
define $P_{n}=\sum_{i=1}^{n}2^{n+1}\mathbf{1}_{I^{n}_{i}}$ where
$I^{n}_{i}=(i/n-1/n2^{n+1},i/n)$. By a similar argument,
we can see that $P_{n}\Rightarrow P$.
Now we need to find a Borel set $A$ such that $P_{n}(A)\not\to P(A)$.
Consider the sets $\text{supp}(P_{n})$. $P(\text{supp}(P_{n}))=1/2^{n+1}$,
so
$$
P\big(\cap_{n=1}^{\infty}\text{supp}(P_{n})^{\complement}\big)
\geq 1-P\big(\cup_{n=1}^{\infty}\text{supp}(P_{n})\big)
\geq 1-\sum_{n=1}^{\infty}P(\text{supp}(P_{n}))
=1/2
$$
Denote this intersection $A$. As the countable intersection of closed sets,
$A$ is Borel-measurable. Furthermore, because $\text{supp}(P_{n})\cap A=\emptyset$
for all $n$, $P_{n}(A)=0$ for all $n$! Thus $P_{n}(A)\not\to P(A)$.
A: Let $X_n \sim Bin(n,p)$ and $Z \sim N(0,1)$. Then
$$Y_n=\frac{X_n - np}{\sqrt{npq}} \xrightarrow{d}Z \hspace{20pt} \text{(weakly convergence)}  $$
Let $A_n=\bigl\{\frac{k-np}{\sqrt{npq}}:k=0,1,2,...,n \bigr\} $ which is a finite set containing $n+1$ many points and $A=\cup A_n$.
Then $P[Y_n\in A]=1$ $\forall n$ but $P[Z\in A]=0$
(countable union of countable sets is countable)
A: I think there's an issue with the previous reply, and it's that the function proposed might not be countably additive: pick the interval $(0,2/n)$ and split it into $(0,1/n-\epsilon]$ and $(1/n-\epsilon,2/n)$, for some $\epsilon>0$ but small. A measure requires that the function evaluated in the original interval and the sum of the function evaluated in each of the subintervals is the same, however, if we use indicator functions, the bin $(\frac{1}{n}-\frac{1}{n2^{n+1}},\frac{1}{n})$ is double counted when evaluating for each subinterval. At least thats how I understand the previous construction. Now, there's a way around it and it's the following:
Let $F_n^i$ be the uniform CDF in $A_n^i=(\frac{i}{n}-\frac{1}{n2^{n+1}},\frac{i}{n})$ and define $\mu_n(A)=\frac{1}{n} \sum_{i=1}^n F_n^i(A_n^i \cap A)$, where A is any Borel Set in (0,1). $\mu_n(A)$ is for sure a measure because it is a linear combination of measures, and for the same reason, as uniform distribution is absolutely continuous with respect to lebesgue measure, $\mu_n$ is too. Now, if $x\in(0,1)$, $\mu_n((0,x])=\frac{1}{n}\lfloor xn \rfloor$. The reason is that before x, there are $\lfloor xn \rfloor$ full bins, each of them accounting for 1 unit in the measure (CDF $F_n^i$ of a full bin $A_n^i$ is 1). So $$\lim_{n\to\infty} \mu_n((0,x])=\lim_{n\to\infty} (\frac{1}{n}\lfloor xn \rfloor+\frac{F_{\lfloor xn \rfloor+1}^i (x)}{n})=\lim_{n\to\infty} (\frac{1}{n}xn-\frac{1}{n}(xn-\lfloor xn \rfloor)+\frac{F_{\lfloor xn \rfloor+1}^i (x)}{n})=x$$ as $(xn-\lfloor xn \rfloor))$ is uniformly bounded by 1, as well as any CDF. Note that $F_{\lfloor xn \rfloor+1}^i (x)$ is well defined as if $x \notin +A_{\lfloor xn \rfloor+1}^i$, $F_{\lfloor xn \rfloor+1}^i (x)=0$. This last term accounts for the fact that x might be included or not in the $\lfloor xn \rfloor + 1$ bin (draw a picture and you´ll see it).
So $\mu_n$ converges to a uniform distribution in (0,1). Then, the rest of the construction of the set including the points in the support, from the previous answer, works.
