ThmDex – An index of mathematical definitions, results, and conjectures.
Result R4331 on D5500: Almost sure event
Probability of binary intersection with an almost sure event
Formulation 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $E, F \in \mathcal{F}$ are each an D1716: Event in $P$
(ii) \begin{equation} \mathbb{P}(F) = 1 \end{equation}
Then \begin{equation} \mathbb{P}(E \cap F) = \mathbb{P}(E) \end{equation}
Proofs
Proof 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $E, F \in \mathcal{F}$ are each an D1716: Event in $P$
(ii) \begin{equation} \mathbb{P}(F) = 1 \end{equation}
Applying results
(i) R4145: Binary union is an upper bound to both sets in the union
(ii) R2090: Isotonicity of probability measure

we have \begin{equation} 1 = \mathbb{P}(F) \leq \mathbb{P}(E \cup F) \leq 1 \end{equation} Thus, $\mathbb{P}(E \cup F) = 1$. Next, using result R4445: Inclusion-exclusion principle for probability of binary union, we have \begin{equation} \mathbb{P}(E \cap F) = \mathbb{P}(E) + \mathbb{P}(F) - \mathbb{P}(E \cup F) = \mathbb{P}(E) + 1 - 1 = \mathbb{P}(E) \end{equation} This completes the proof. $\square$