If a measurable set $A$ is a subset of a measurable set $B$ then $B \setminus A$ is also measurable (by axioms of the $\sigma$-algebras) and since $A$ and $B\setminus A$ are disjoint, by additivity of measure $\mu(A) + \mu(B \setminus A) = \mu(B)$, so that $\mu(A) \leq \mu(B)$. In particular, when $B$ has measure $0$, so does $A$.
On the other hand, not every subset of a measurable set needs to be measurable. But for the subsets of measure zero there's no harm in introducing another measure on an extension of the original $\sigma$-algebra, a completion of $\mu$ which has the property that every subset of a measure $0$ set is measurable (and so has measure $0$) while it agrees with the original measure on the sets belonging to the original $\sigma$-algebra.
It is a matter of personal preference whether one works with complete measures or not. It makes some proofs cleaner but at the cost of introducing many new sets into the $\sigma$-algebra. E.g. completion of a Borel $\sigma$-algebra is the Lebesque $\sigma$-algebra which has strictly greater cardinality (at least for real line -- I am not sure in general).