If $A\subseteq\mathbb{R}$ and $\lambda^*(A)>0$ then $A$ contains a measurable subset that is not Borel. If $A\subseteq\mathbb{R}$ and $\lambda^*(A)>0$ then it contains a subset which is not Borel but Lebesgue measurable.
I don't know if the statement is true. Can someone provide some hint?
 A: A similar question has been asked on MO. Here I shall try to be more explicit, but most of what I write below was already said in the answers to the linked question.
If the set is measurable, the result is true: by regularity, $A$ contains a compact uncountable subset $C:0<\mu(C)<+\infty$. By Cantor-Bendixson, $C=P\cup Q$ with $P$ perfect, $Q$ countable. Mimicking the usual construction of a Cantor set, we get a subset $R$ of $P$ of measure $0$ and cardinality of the continuum.Now, every subset of $R$ is Lebesgue measurable. Since there are $2^\mathfrak c$ subsets of $R$ and only $\mathfrak c$ Borel sets, it's clear that there is a Lebesgue measurable subset of $A$ which is not Borel.
In general, the answer is independent from ZFC:
For the answer to be no, there must exist a subset $S$ of $\mathbb{R}$, such that it does not contain any uncountable null subset (otherwise, the same cardinality reasoning as before yields a Lebesgue measurable non Borel subset EDIT: as @Dave L. Renfro noted in the comments this requires an additional hypotesis, the requirement that $2^{\aleph_1}>2^{\aleph_0}$, which is known as Luzin's hypotesis. The claim is still valid since LH is independent from ZFC). Such a set $S$ is called a Sierpiński set, and its existence is independent from ZFC.
We have proved that, if there is no Sierpiński set, the answer is positive. It remains to prove that if a Sierpiński set exists, the answer is no:
let $B\subset S$ be a Lebesgue measurable set.
It's easy to see that it can be written as $C\cup N$, where $C$ is a Borel set and $N$ is a null set. Since $S$ is Sierpinski and $N\subset B\subset S$, $N$ is countable. Thus $B=C\cup N$ is Borel measurable.
As a final note, it is not hard to prove (using transfinite induction) that the continuum hypotesis implies the existence of a Sierpinski set:
Let $\mathcal{B}$ be the set of Borel sets of measure $0$. It is easy to see that $|\mathcal{B}|=\mathfrak c$. Let us well order $\mathcal{B}$ as $\{b_\lambda\}_{\{\lambda<\omega_1\}}$ ($\omega_1$ is the first uncountable ordinal).
For every $\lambda<\omega_1$, let $x_\lambda$ be so that $x_\lambda\not\in \cup_{\alpha<\lambda} b_\alpha$ (this is possible since the union has measure $0$). If we define $S=\{x_\lambda\}$, it is easy to see that $S$ has at most countable intersection with Borel sets of measure $0$. Since every null set is contained in a null Borel set (by outer regularity), the result follows.
