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Suppose we have an event $A$ with $P(A) = 0$, and let $B$ be any other event. Prove that $A$ and $B$ are independent.

It seems rather obvious that since $P(A \cap B) = P(A)P(B)$ if they're independent, and $P(A)P(B) = 0$, so $A$ and $B$ will be independent regardless since $A \cap B$ must have Lebesgue measure 0. However, I'm not satisfied with this "proof". Is it possible that it's not always true?

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What do you find unsatisfactory about it? I find it unsatisfactory from the following point-of-view: If A and B are independent, I excpect P(B|A) = P(B). However, this only makes sense when P(A) is not 0. – echoone Dec 4 '12 at 2:21
Example: Suppose X is uniformly distributed on [0,1]. Let A be the event X = 0.6 and let B be the event X > 0.5 Then P(B) = 0.5 and P(B|A) = 1, event though P(A) = 0. – echoone Dec 4 '12 at 2:23
up vote 2 down vote accepted

It is always true. If $\Pr(A)=0$, then $\Pr(A\cap B)=0$. So $\Pr(A\cap B)=\Pr(A)\Pr(B)$. It is hardly ever Lebesgue measure, it is whatever probability measure in involved.

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