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On a polish space $\mathcal{X}$ i consider two Borel probabilities $P$ and $Q$ such that for any open set $E$ of $\mathcal{X}$ we have : $P(E) =0$ implies $Q(E)=0$. Does this imply that $Q$ is absolutely continuous with respect to $P$ that is, that for any Borel set $E$ of $\mathcal{X}$ we have $P(E)=0$ implies $Q(E)=0$ ?

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No: If $P$ is Lebesgue measure on $[0,1]$, then $P(E) > 0$ for every open set $E$, and every probability measure $Q$ on $[0,1]$ vacuously satisfies your assumption. –  Dan Apr 12 '13 at 12:45
Sorry i don't understand your answer, how does this prove that the claim is wrong ? –  Polite Apr 12 '13 at 13:10
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closed as off-topic by Nate Eldredge, Andrey Rekalo, Stefan Kohl, Ramiro de la Vega, Noah Stein Feb 19 at 19:40

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1 Answer

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Your condition means precisely that the support of the measure $Q$ (the minimal closed set of full measure) is contained in the support of the measure $P$. Obviously it has nothing to do with absolute continuity.

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Why is this obvious ? Do you have any counter exemple ? –  Polite Apr 12 '13 at 13:20
There is a lot of pairwise singular measures with the same support. For instance, look at the distribution of the sum $$ \sum_{k=1}^\infty x_k/2^k \;, $$ where $x_k$ are independent and take values 1 with probability $p$ and 0 with probability $1−p$. The resulting measures $m_p$ are pairwise singular (as it follows, for instance, from the law of large numbers) and have the same support $[0,1]$ (for $p=1/2$ this is the Lebesgue measure). –  R W Apr 12 '13 at 13:47
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