Discussion
Let \((\Omega, F, P)\) be a probability space. Let \(\{A_{n}\}\) be a sequence of events such that \(P(A_{n}) = 1\) for each \(n \ge 1\). Then \(P\left(\bigcap_{n=1}^{\infty} A_{n}\right)\) =
Answer: B
The probability is One.
Pass to complements. Each \(A_{n}^{c}\) has probability 0, and by countable subadditivity
\[ \begin{aligned} P\left(\bigcup_{n} A_{n}^{c}\right) &\le \sum_{n} P(A_{n}^{c}) \\ &= 0 \end{aligned} \]
By De Morgan's law, \(\bigcap_{n} A_{n}\) is the complement of \(\bigcup_{n} A_{n}^{c}\), so its probability is \(1 - 0 = 1\). A countable intersection of almost sure events is almost sure.
The intersection is an event in F because F is a \(\sigma\)-field, so its probability is defined; it cannot be infinity, since probabilities lie in [0, 1]; and zero contradicts the calculation above.
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