TITLE: absolutely continuous measure -question
QUESTION [2 upvotes]: $\nu\ll\mu$ iff $|\nu|\ll\mu$ iff $\nu^{+}\ll\mu$ and $\nu^{-}\ll\mu$
Suppose $\nu\ll\mu$ and $\mu(E)=0$ (this if I understand follows from the hypothesis?) . 
Let $X=P\cup N$ be the Hahn decomposition w.r.t $\nu$. let $\nu^{+}(E)=\nu(E\cap P)$ and $\nu^{-}(E)=-\nu(E\cap N)$
$\mu(E)=0\Rightarrow \mu(E\cap P)=0\Rightarrow \nu(E\cap P)=0$
similarly 
$\mu(E)=0\Rightarrow \mu(E\cap N)=0\Rightarrow \nu(E\cap N)=0$
how do we know that?
We cannot use that $E\cap P\subseteq E$ and since $\mu(E)=0$ then $\mu(E\cap P)=0$ as we do not know if the measure is complete. What argument has been use in the above implications?

REPLY [2 votes]: Since $\mu \geqslant 0$, every measurable subset $A$ of a $\mu$-null set $B$ is a $\mu$-null set, since
$$0 =\mu(B) = \mu(A) + \mu(B\setminus A) \geqslant \mu(A) \geqslant 0.$$
Thus, since $P$ and $N$ belong to the $\sigma$-algebra $\mathcal{M}$, the two sets $E\cap P$ and $E\cap N$ are $\mu$-null sets whenever $E$ is one.
{"set_name": "stack_exchange", "score": 2, "question_id": 516512}