I found an example in this journal paper. Let $\beta$ be a Bernstein set, i.e. a subset of $\mathbb{R}$ such that both it and its complement intersects every uncountable closed subset of $\mathbb{R}$. (This post describes how to construct such a set using the axiom of choice.) And let $\gamma$ be a dense measure-$0$ $G_\delta$ subset of the fat Cantor set. (This answer describes how to construct such a set.)
Then $\beta\cap\gamma$ is a Lebesgue measurable set which cannot be written as a union of a Borel set and a subset of a measure $0$ $F_\sigma$ set. This is proven in the linked paper.
Example of a set $S\subseteq\mathbb R$ of Lebesgue measure zero which is not the union of a Borel set and a meager set.
Let $G$ be a dense $G_\delta$ set of measure zero.
Let $B$ be a Bernstein set, i.e., $B$ meets every uncountable closed set but contains no uncountable closed set. (Such sets exists assuming the axiom of choice.)
Let $S=B\cap G$.
$S$ has measure zero, since $S\subseteq G$.
Assume for a contradiction that $S=A\cup M$ where $A$ is a Borel set and $M$ is meager.
Since $B$ is a Bernstein set, $B$ contains no uncountable closed set; since every uncountable Borel set contains an uncountable closed set, $B$ contains no uncountable Borel set; since $A\subseteq S\subseteq B$ and $A$ is a Borel set, $A$ is countable.
Since $S=A\cup M$, where $A$ is countable and $M$ is meager, $S$ is meager.
Since $S$ is meager, there is a dense $G_\delta$ set $H$ such that $S\cap H=\emptyset$.
Since $G$ and $H$ are dense $G_\delta$ sets, $G\cap H$ is a dense $G_\delta$ set.
Since $G\cap H$ is a dense $G_\delta$ set, there is an uncountable closed set $F\subseteq G\cap H$.
Now $B\cap F\subseteq B\cap G\cap H=S\cap H=\emptyset$, so $B\cap F=\emptyset$. Since $F$ is an uncountable closed set, this contradicts the fact that $B$ is a Bernstein set.
Best Answer
If $A$ and $C$ are each Lebesgue measurable, then so is $C\setminus A$ and so by finite additivity it has measure zero.
But then $B\setminus A$ also has measure zero and hence is measurable, so $$B=A\cup(B\setminus A)$$ is the union of two measurable sets, hence is measurable.