Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$?



Is a subset that contains no positive measurable subsets contained in a null measurable set?
You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali set is non-measurable, but has inner measure $0$. 


I claim that a ($\sigma$-finite) measure space has your property if and only if every set
is measurable with respect to its completion measure.


For the forward direction, suppose that $A$ is any subset
of the space $X$. Let $a$ be the inner measure of $A$, the
supremum of $\mu(A\_0)$ among all measurable $A\_0\subset A$ (and we may assume wlog this is finite).
By taking a union, it follows that the inner measure is
realized, so that there is some measurable $A\_0\subset A$
with $\mu(A\_0)=a$. It follows that $A-A\_0$ has inner
measure $0$. By the hypothesis, it follows that
$A-A\_0\subset B$ for some measurable set $B$ with
$\mu(B)=0$, and so $A-A\_0$ has measure $0$ with respect to
the completion. Consequently, $A$ differs from the
measurable set $A\_0$ on a completion-measure zero set
$A-A\_0$, and hence is measurable with respect to the
completion measure.


Conversely, suppose that every set is measurable with
respect to the completion of $\mu$. Suppose that $A$ has
inner measure $0$. By assumption, there is some measurable
set $A\_0$ such that the symmetric difference $A\triangle
A\_0\subset A\_1$ for some measurable $A\_1$ with
$\mu(A\_1)=0$. It follows that $A\_0-A\_1$ is a measurable
subset of $A$, and hence measure $0$, and so $A\subset
B=A\_0\cup A\_1=(A\_0-A\_1)\cup A\_1$ shows that $A$ is contained
in a measure $0$ set $B$, as desired.


In particular, a complete measure has the property if and
only if it measures every set. 


