Let $m^*$ be the Lebesgue outer measure. If $A\subset R$ and $m^*(A)=0$ then there exists Borel measurable sets $B$ and $C$ such that $A=B- C$.
Can I say:
Since $m^*(A)=0$, $A$ must be a union of a bunch of singletons, which makes $A$ Borel measurable. Also, since $m^*(A)=0$ there exists some Borel measurable $B$ such that $A \subset B$. Lastly, since the collection of Borel-measurable sets is a sigma-algebra, $C=B \cap A^c$ is also Borel-mesurable.
NO. Any set is the union of a "bunch" of singletons. Perhaps you suppose (incorrectly) that if $m^*(A)=0$ then the bunch must be countable. The set-difference $B\setminus C$ of two Borel sets $B,C$ is also a Borel set. But not every $A$ such that $m^*(A)=0$ is a Borel set.
The cardinal of the family of all Borel sets is $c=2^{\aleph_0},$ which is the cardinal of $\Bbb R.$ The Cantor set $D$ has cardinal $c$ and is a Borel set, and $m^*(D)=0,$ so $m^*(E)=0$ for any $E\subset D .$ But the cardinal of $\{E: E\subset D\}$ is $2^c,$ which is larger than $c.$ So there are non-Borel subsets of $D$ that have outer measure $0.$
There seems to be a teaching style that avoids mentioning $inner$ measure. The inner measure $m^i(A)$ of $A\subset \Bbb R$ can be defined as $\sup \{m^*(F): F=\overline F\subset A\}.$ Then $A$ is Lebesgue-measurable iff $m^*(A)=m^i(A)$.... iff there exist Borel sets $A',A''$ such that $A'\subset A\subset A''$ and $m^*(A')=m^*(A'')$ ....iff there are $G,H$ such that $A=G\cup H$ where $G$ is Borel and $m^*(H)=0$.
The family $\Bbb B$ of Borel sets is the smallest $\sigma$-algebra containing the open sets. The family $\Bbb L$ of Lebesgue sets is the smallest $\sigma$-algebra that has $\Bbb B$ as a subset and which contains every set that has Lebesgue outer measure $0.$.... The idea was to extend Borel measure to a countably-additive, translation-invariant measure for which any subset of a measure-$0$ set is also measurable (with measure $0$ of course.)