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Let $E$ an event and $P(E) = 1$,

does this mean that:

$\forall A$ from the sample space : $P(E|A)=P(E)$ ?

( I need to prove that $P(B)=1 \rightarrow \forall C : P(B\cap C)=P(C)$ )

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    $\begingroup$ Yes. Try to prove it yourself! $\endgroup$
    – GEdgar
    Dec 27, 2015 at 20:59
  • $\begingroup$ If the answer to the question is "YES" then I'd say i've proven it. The only issue was with that. $\endgroup$
    – reaffer
    Dec 27, 2015 at 21:01
  • $\begingroup$ Kolmogorov's factorization definition of independence should not apply to sets of measure 0 or 1 since it leads to such absurb statements as $A$ being independent of $A$ if $P(A)=0,1$ when its obviously not the case. So in fact no, event $E$ is NOT independent of any other event since it's not independent of any event contained fully either inside $E$ or $E^c$. $\endgroup$
    – A.S.
    Dec 27, 2015 at 22:18
  • $\begingroup$ @A.S. why is it "obviously not the case" though? My intuitions about independence is weak for events of probability 0,1. And the consequence of the kolmogorov definition, $P(A)=P(A)^2$ implies $A$ trivial, leads to intuitive results e.g., Kolmogorovs 0-1 law. $\endgroup$
    – snarfblaat
    Dec 27, 2015 at 22:37
  • $\begingroup$ @snarf Because even if $P(A)=0$, you know that probability of $A$ happening give that $A$ happened is $1$ (since $A$ implies $A$). More generally $P(A|B)=1$ for $A^c\cap B=\emptyset$ and $P(A|B)=0$ for $A\cap B=\emptyset$. I don't see how you last remark touches on independence. $\endgroup$
    – A.S.
    Dec 27, 2015 at 22:57

1 Answer 1

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Here's one argument:

\begin{align} P(C) &= P(\{ B \cup B^c \} \cap C) \\ &= P(B \cap C) + P(B^c \cap C) \\ &= P(B \cap C) \end{align}

since $P(B^c \cap C) \leq P(B^c) = 0$.

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