If $\frac{P(E_n)}{P(F_n)} \to 1$, does $\frac{P(A \cap F_n)}{P(A \cap E_n)} \to 1$? Let $(\Omega, \mathcal F, P)$ be a probability space. Let $E_1\supset E_2\supset ...$ and $F_1 \supset F_2 \supset...$ be two decreasing sequences of events in $\mathcal F$ with the following properties.

*

*$E_n \subset F_n$.


*$\bigcap_n E_n  = \bigcap_n F_n \neq \emptyset$.


*$P(E_n) > 0$.
Let $A \in \mathcal F$. Let's assume in addition that


*$P(A \cap E_n)>0$.

Question. If $P(E_n)/P(F_n) \to 1$, does $P(A \cap F_n)/P(A \cap E_n) \to 1$ as well?
The answer is obviously yes if $P(A \cap \bigcap_nE_n)>0$, so we can assume this isn't the case.
Heuristically, it seems to me that the answer should be yes: $P(F_n)$ and $P(E_n)$ are decreasing to the same limit at the same rate and $P(A \cap E_n)$ and $P(A \cap F_n)$ are decreasing to the same limit. I can't see why intersecting with a single set $A$ would change the rates of convergence, so it seems like the result should hold.
I think I'm just missing an easy algebraic trick or something like that, however, because I haven't been able to prove this.
 A: If
$$P(E_n \cap A) = 2^{-(n+2)}$$
$$P(F_n \cap A) = 2^{-(n+1)}$$
$$P(E_n \cap A^c)=P(F_n \cap A^c) = \frac{1}{2}$$
Then
$$P(E_n \cap A)/P(F_n \cap A) \to \frac{1}{2}$$
So you can construct a counterexample like this.
For example, consider the uniform distribution on $[0,1]$ and let $E_n$ be the event that $X>\frac{1}{2}$ or the first $n+2$ digits of $X$ in binary are 0. Let $F_n$ be the event that $X>\frac{1}{2}$ or the first $n+1$ digits of $X$ in binary are 0. Where $A$ is the event $X \le \frac{1}{2}$
A: Choose a convergent increasing positive sequence, say $S_j=\sum_{1}^{j}a_i$. Then $P=m/2S_{\infty}$ where $m$ is the Lebesgue measure on $\mathbb{R}$ is a probability measure on $[-S_{\infty},S_{\infty}]$. Define $E_n=\{0\}\cup\left(\bigcup_{n}^{\infty}[S_j,S_{j+1}]\right)$ and $F_n=E_n\cup\left(\bigcup_{n}^{\infty}[-\sum_{1}^{j+1}a_i/i,-\sum_{1}^{j}a_i/i]\right)$. Obviously $1\le P(F_n)/P(E_n)\le1+1/(n+1)\rightarrow1$. Now define $A=[-S_{\infty},0]\cup\left(\bigcup_{1}^{\infty}[S_j,S_j+a_{j+1}/(j+1)]\right)$ and we have $2S_{\infty}\cdot P(A\cap E_n)=P(F_n)-P(E_n)$ and $2S_{\infty}\cdot P(A\cap F_n)=2\left(P(F_n)-P(E_n)\right)$. Hence the ratio is 2.
