Suppose I have two positive Borel measures on $[0,\infty]$, say $\mu$ and $\nu$ and I know that $\nu([0,x]) = 0$ implies $\mu([0,x]) = 0$ for any $x$ and that $\nu((a,b]) = 0$ implies $\mu((a,b]) = 0$ for any pair $a,b$.
Does this imply that $\mu < < \nu$? I would think so but I have trouble proving it, I thought that the monotone class theorem was the answer but I don't see how the sets with the property $\nu(E) = 0 \Longrightarrow \mu(E) = 0$ are closed under monotone intersections.
We may assume that the measures are finite, if that helps anything.
The situation is even worse than my comment. Take any measure $\nu$ that's never $0$ on a nondegenerate interval (the Lebesgue measure works, as do many finite measures). If $\nu$ has sets of measure $0$, then take one such set $E$, let $\lambda$ be a probability measure on $E$, and let $\mu = \nu + \lambda$. Then $\mu$ and $\nu$ satisfy the condition on intervals, but $\mu$ is not absolutely continuous with respect to $\nu$.