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Let $A_1, A_2, \ldots$ and $B_1, B_2, \ldots$ be two sequences of events in some probability triple $(\Omega, \mathcal{F}, \mathbf{P})$. Now, it is true that $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supseteq \limsup_n (A_n \cap B_n)$, and I can think of many different cases where this relation holds true for equality (e.g., let $A_n = B_n$, or $A_n = B_n^c$, or $B_n = \emptyset$, or $B_n = \Omega$). However, I cannot think of any case where $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supset \limsup_n (A_n \cap B_n)$ holds for strict inclusion.

My proof for the relation is simple. Since $A_n \cap B_n \subseteq A_n$ and $A_n \cap B_n \subseteq B_n$, then we can perform some set algebra as follows:

$$\bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcup_{k=n}^\infty A_k$$ $$\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$$

Then we can use the same logic for $B_n$, and come to the conclusion as stated in the beginning.

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up vote 2 down vote accepted

Just take \begin{align} \emptyset &\not= A_1 = B_2 = A_3 = B_4 = ...\\ \emptyset &= B_1 = A_2 = B_3 = A_4 = ... \end{align}

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Well, that was too easy! Thanks! – Matt Nov 6 '12 at 17:15

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