Absolutely Continuous Weakly Convergent Sequence Need Not Converge Strongly

Question

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

1. 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.

2. 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).

3. 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.

Answer 1

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 $P_{n}(A)\not\to P(A)$.

Answer 2

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)