How to complete this simple example of the Vitali-Hahn-Saks Theorem?

Question

I'm studying the Vitali-Hahn-Saks Theorem and I need some help with an example in which the conclusion of the theorem fails.

Theorem. Let $(\Omega,\mathcal{F}, P)$ be a finite measure space, and let $(P_n)$ a sequence of finite measures on $(\Omega, \mathcal{F})$ each of which is absolutely continuous with respect to $P$. Suppose the sequence $(P_n(\Omega))$ is bounded. If $(P_n)$ converges setwise to the set function $P_\infty$, then $(P_n)$ is uniformly absolutely continuous with respect to $P$ ($\forall \epsilon > 0, \exists \delta>0, \forall n, \forall A \in \mathcal{F}, P(A) < \delta$ implies $P_n(A) < \epsilon$). Moreover, $P_\infty$ is a finite measure that is absolutely continuous with respect to $P$.

I thought I'd cook up a simple example where uniform absolute continuity fails in order to see how setwise convergence fails. Start with a countable probability space given by $P(\omega_n) = 2^{-n}$. Let the events $E_n = \Omega - \cup_{i=1}^n \{\omega_i\}$. And let $P_n = P(\cdot \mid E_n)$. Note that $\inf_n\{P(E_n) \}=0$.

The sequence $(P_n)$ is not uniformly absolutely continuous with respect to $P$. To see that let $\epsilon = 1/2$, and let $\delta > 0$ be given. For large enough $n$, $P(E_n) < \delta$ but $P_n(E_n) = 1 > \epsilon.$

Question. How to see that $(P_n)$ does not converge setwise? I've played around with a few example events $A$, but $(P_n(A))$ always converges. Hints are appreciated.

Answer 1

Here are some hints about what kind of set $A$ you are looking for:

If $A$ is finite, then eventually $A\cap E_n = \emptyset$, so $$ P_n(A)=\frac{P(A\cap E_n)}{P(E_n)} \longrightarrow 0 $$

Thus we are looking for infinite sets. Moreover, if we eventually have $E_n \subset A$ then $A \cap E_n = E_n$ so $$ P_n(A) = \frac{P(A \cap E_n)}{P(E_n)} = \frac{P(E_n)}{P(E_n)} \longrightarrow 1 $$

Thus we are looking for a set $A$ that has infinitely many "holes" (more precisely, for any $N$ we can find a $n >N$ such that $\omega_n$ is not in $A$).

Can you think of a very simple set $A$ with these properties?

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Added later:

Elaborating on the hint given above, one of the simplest set to satisfy the properties we want is the set of $\omega_i$ with even index. That is, we take $$ A = \{ \omega_{2 i} : i \geq 1 \} $$

In this case, after some calculation we find that $P_n(A) = 1/3$ when $n$ is even and $P_n(A) = 2/3$ when $n$ is odd. Consequently, the sequence $P_n(A)$ does not converge.