Being curious I'm wondering:
Let $V$ be a Vitali set defined as usually as a choice of $v\in[r]$ with $0\leq v\leq 1$ for every $[r]\in\mathbb{R}/\mathbb{Q}$. Since the countable disjoint union of translates $V+q$ by $q\in[-1,1]\cap\mathbb{Q}$ lie by construction in $[-1,2]$ any measurable subset $E\subseteq V$ has to have Lebesgue mass zero!
So my hypothesis, given a measure space does every nonmeasurable set $A$ split into a measurable subset $E$ of finite length and a nonmeasurable subset $A_0$ whose measurable subsets have all zero mass?
For a set $E$, let $\mu_*(E) $ denote the supremum of $\mu(A)$ over all measurable subsets $A\subset E$.
Given $E$ with $\mu_*(E)<\infty$, pick a measurable subset $A_1\subset E$ such that $\mu(A_1)\ge \frac12 \mu_*(E)$, and let $E_1=E\setminus A_1$. Observe that $\mu_*(E_1)\le \frac12\mu_*(E)$. This process yields sequences $A_n$ and $E_n$ such that $\mu_*(E_n)\to 0$.
Let $A=\bigcup_{n=1}^\infty A_n$; this is a measurable subset of $E$. The set $E\setminus A$ is contained in $E\setminus A_n$ for every $n$, hence $\mu_*(E\setminus A)=0$. The decomposition $E=A\cup (E\setminus A)$ has the desired property.