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Suppose that $\{B_n\}_{n\in\mathbb N}$ are almost sure events, i.e. $\mathbb P(B_n)=1, \forall n$. Then how do I show that their intersection is also a sure event, i.e. $\mathbb P\left(\cap_{n=1}^\infty B_n\right)=1$. Thanks!

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Let's consider complements $B_n^c$ to $B_n$. For any $n$ it holds that $\mathbb{P}(B_n^c) = 0$.

Using countable additivity for measures we get: $$ \mathbb{P} \left(\bigcap_{n = 1}^{\infty} B_n \right) = 1 - \mathbb{P}\left(\bigcup_{n = 1}^{\infty} B_n^c \right) \geq 1 - \sum_{n = 1}^{\infty} \mathbb{P}(B_n^c) = 1. $$ So $$ 1 \geq \mathbb{P}(\cap_{n = 1}^{\infty} B_n) \geq 1. $$ Consequently $$ \mathbb{P}(\cap_{n = 1}^{\infty} B_n) = 1. $$

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