We work on a probability space $(\Omega,\mathcal{F},P)$ and suppose that $Q$ is a probability measure s.t. $Q\ll P$, i.e. for $A\in\mathcal{F}$ with $P[A]=0$, also $Q[A]=0$.
Suppose $A\in\mathcal{F}$ with $Q[A]=1$. Then $P[A]>0$. Question: Does it even hold that $P[A]=1$?
Motivation: If we have that $P$ and $Q$ are equivalent, then a statement that is $P$-a.s. true is also $Q$-a.s. true and vice versa. If $Q$ is absolutely continuous with respect to $P$, then a statement that is $P$-a.s. true is also $Q$-a.s. true. I would like to understand if the converse holds.
My progress so far: For any $B∈\mathcal{F}$ we have: $$P[B]=1⇒P[B^c]=0⇒Q[B^c]=0⇒Q[B]=1$$ Thus, $Q[B]<1⇒P[B]<1$.
Another result that might be helpful (a proof is for example in Prop. 2.1 here): $Q\ll P$ is equivalent to: $\forall \epsilon>0, \exists\delta>0$ s.t. $P[B]<\delta \Rightarrow Q[B]<\epsilon$.
Any help on how to proceed with the proof, or, in case the statement is wrong, a (hint for a) counterexample is very much appreciated.
No; for this counterexample, we work in $\Omega = [0,1]$ equipped with its Borel structure. Let $\lambda$ denote the Lebesgue measure and $\delta_0$ the Dirac mass at zero. Set $P = \frac{1}{2}\lambda + \frac{1}{2} \delta_0$ and $Q = \lambda$.
To show absolute continuity, suppose $A \in \mathcal{F}$ has $P(A)=0$. Then this means that $0 \not\in A$ and $\lambda (A) = 0$, which implies that $Q(A) = 0$. Thus, $Q \ll P$. However, note that $Q((0,1]) = 1$, but $P((0,1]) = 0.5$, showing your claim is not true.