Lebesgue measurable set with non-zero measure

Question

every Lebesgue measurable set with non-zero measure contains a non-measurable set.what is the proof?

Answer 1

There is the Vitali set, $E \subset [0,1]$ such that all of its translates by rational numbers are disjoint, and cover the reals. Now if we have $X\subseteq E$ and $X$ is measurable then $$\bigcup\limits_{r \in \mathbb{Q}\cap[0,1]} X+r\subseteq [-1,2]$$ and all $X+r$ are disjoint thus we must have $m(X)=0$. Thus any measurable subset of the Vitali set has measure zero.

Now if $A$ is your given measurable set of positive measure,

$$A=\bigcup\limits_{r \in \mathbb{Q}} (E+r)\cap A$$ if all the sets $(E+r)\cap A$ are measureable they have measure zero and thus $A$ has measure zero, thus for some $r$, $(E+r)\cap A$ is a non measureable set.