I am not sure I am using the standard definitions so I will open by defining what I need:
Let $X$ be a set, $\nu:\, \mathscr{P}(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any $\{A_{i}\}_{i=1}^{\infty}\subseteq\mathscr{P}(X)$ (not neccaseraly disjoint) it holds that $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$
Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$
I have an exersice that asks me to prove that if $\nu(A)=0$ then $A$ is $\nu$- measurable.
Note: I have already proved that the set of all $\nu$-measurable sets, denoted by $M$, is a $\sigma$-algebra.
What I tried:
For any $E\subseteq X$:
From containment: $\nu(E\cap A^{c})\leq\nu(E)$
But $E=(E\cap A)\cup(E\cap A^{c})$ thus $\nu(E)=\nu((E\cap A)\cup(E\cap A^{c}))\leq\nu(E\cap A)+\nu(E\cap A^{c})$
This is the part I want to say that since $E\cap A\subseteq A$ and $\nu(A)=0$ then $\nu(E\cap A)=0$ and so $\nu(E)=\nu(E\cap A^{c}) $
thus
$\nu(E)=0+\nu(E\cap A^{c})=\nu(E\cap A)+\nu(E\cap A^{c})$
but the problem is that I do not know that $\nu$ is monotone (I can argue that its monotone on sets in $M$ but $E\cap A^{c}$and the other sets here need not be in $M$).
Can someone please help me to prove that $\nu$ is monotone, or suggest another approach ?