Let $\beta$ be the $\sigma$-algebra of Borel sets on $\mathbb{R}^n$ and let $A, B \subset \mathbb{R}^n$ such that $A \subset B$, $B\in \beta$ and such that measure of $B$ is less than $1$. Is $A$ Lebesgue measurable?
So my first guess was not. I know that it isn't true without the restriction on the measure of $B$. Just take $[0,1]$ and construct the Vitali set. But somehow the thing with the measure being less than one keeps bugging me. Can I imitate the construction of a Vitali set in, say, $[0, 1/2]$? Thanks in advance and sorry if the question is somewhat stupid.