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 $Q(E)=0$ ?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

closed as offtopic by Nate Eldredge, Andrey Rekalo, Stefan Kohl, Ramiro de la Vega, Noah Stein Feb 19 at 19:40
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Nate Eldredge, Andrey Rekalo, Stefan Kohl, Ramiro de la Vega, Noah Stein
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. 

