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 $Q(E)=0$ ?



Absolute continuity of probabilities on Polish spaces and open sets.
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. 


