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Let $t^*\in(0,1)$ and $(\Omega,\mathcal{F},P)$ be a probability space. Suppose I have set $A\in\mathcal{F}$ and an uncountable family of sets $(B_t : t\in[0,1])\subset\mathcal{F}$ with the following properties:

  1. $P(A)>0$ and $P(B_t)=p$ where $p>0$ is a constant.
  2. For each $\omega\in A$ there exists a $\delta>0$ such that $\omega\notin B_t$ for all $t\in (t^*-\delta, t^*+\delta)$.

Is it true that $\limsup_{t\rightarrow t^*}P(A\cap B_t) = 0$? Is the limit well defined?

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1 Answer 1

up vote 2 down vote accepted

Let $t_n$ be any sequence converging to $t^*$, and let $C_n = \bigcup_{i=n}^\infty B_{t_n}$. Then $C_n$ is decreasing, so we have $$\lim_{n\to\infty}P(A\cap C_n)= P(A\cap\bigcap_{n=1}^\infty C_n)$$ because finite measures are continuous from above. But $\bigcap_n C_n$ is disjoint from $A$: for any $\omega\in A$, there exists $\delta$ such that $\omega\notin B_t$ if $|t-t^*|<\delta$, so there exists $N$ such that $\omega\notin B_{t_n}$ if $n> N$, so there exists $N$ such that $\omega\notin C_N$. Therefore $$\limsup_{n\to\infty}P(A\cap B_{t_n})\leq\lim_{n\to\infty}P(A\cap C_n)=P(A\cap\bigcap_{n=1}^\infty C_n)=0.$$

So $\limsup_{t\to t^*}P(A\cap B_t)=0$.

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Just for clarity, shouldn't we say that $\limsup_{n\rightarrow\infty} P(A\cap B_{t_n}) \leq \lim_{n\rightarrow\infty} P(A\cap C_n)$, since, a priori, we do not know that the limit of $P(A\cap B_{t_n})$ exists? –  Carl Morris Sep 20 '12 at 17:33
1  
That's true, I'll edit to reflect that. Thanks! –  Owen Biesel Sep 21 '12 at 2:16

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