Absolutely continous probability measures and evens with full probability

Question

If $P,Q$ are probability measure s.t. $P$ is absolutely continuous w.r.t $Q$, this means that $Q(A) = 0$ implies $P(A) = 0$. Now assume that $Q(B) = 1$. Then $Q(B^c) = 0$ and this implies $P(B^c)= 0$, hence $P(B) = 1$. This seems strange, because I interpret $P \ll Q$ as "$P \leq Q$". But if $Q(B) = 1$ imples $P(B) = 1$, this interpretation makes no sense. What is wrong here?

Answer 1

Well, the obvious thing that is wrong is that you shouldn't interpret $P\ll Q$ as $P\leq Q$. The symbol $P\ll Q$ means that $P$ is absolutely continuous with respect to $Q$. This means in some sense that $P$ isn't "large" relative to $Q$, but it does not imply that $P\leq Q$ in any obvious sense. For instance, it's perfectly possible that $P\ll Q$ but $P(A)>Q(A)$ for some $A$. (You would have a valid point if you protested that the symbol $\ll$ is at least a little misleading, but it is the standard notation.)