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.

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    $\begingroup$ No it's not necessarily measurable. Yes, you can imitate the construction of the Vitali set, or just scale it down. $\endgroup$ – Lorenzo Aug 17 '17 at 19:24
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    $\begingroup$ If $B$ is a non-measurable subset of $[0,1]$ then at least one of the sets $B\cap[0,\frac12],\ B\cap[\frac12,1]$ is non-measurable, although $[0,\frac12]$ and $[\frac12,1]$ are Borel sets of measure less than $1.$ More generally, every set of positive Lebesgue measure contains a non-measurable subset. $\endgroup$ – bof Aug 17 '17 at 19:33
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    $\begingroup$ It might be nitpicking but i'll just add that we need the axiom of choice to assert the existence of non-measurable sets. $\endgroup$ – tristan Aug 17 '17 at 20:01

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