If $A$ has a positive Lebesgue measure then there exist subsets which are not measurable I was thinking if there is a solution to this problem without trying to explicitly create Vitali sets in $A$. Does anyone have any ideas? 
 A: We do not explicitly need to produce a Vitali subset of $A$, but the existence of Vitali sets is all that we need:

Theorem. Any set $A\subseteq\mathbb R$ that is not of Lebesgue measure zero has a subset that is not measurable.

To see this, note first that if $A$ itself is nonmeasurable, we are done. Assume now that $A$ is measurable. Let $V$ be a Vitali set contained in $[0,1]$, that is, $V\subset[0,1]$ and, for any real $x$, $(x+\mathbb Q)\cap V$ is a singleton. Note that 
 $$ A=\bigcup\{A\cap(q+V)\mid q\in\mathbb Q\}. $$
Now, either one of the sets $A\cap(q+V)$ is nonmeasurable (and we are done), or else they are all measurable, but then all of them have measure zero, and therefore so does $A$: 
The point is that $A\cap(q+V)$ is measurable iff $(-q+A)\cap V$ is measurable, in which case both have the same measure, but any measurable subset of a Vitali set must be null. This is the usual argument: If $\{r_i\mid i\in\mathbb N\}$ enumerates $\mathbb Q\cap[0,1]$, and $B\subseteq V$ is measurable, then (letting $\lambda$ denote Lebesgue measure) we have that $\lambda(B+r)=\lambda(B)$ for any $r$ but also
 $$ \sum_i\lambda(B+r_i)=\lambda\bigl(\bigcup_i(B+r_i)\bigr)\le\lambda([0,2])=2. $$
A: I don't think the Vitali construction is useful here. Instead, I'd use the other (Bernstein's?) method of constructing a nonmeasurable set. We need two facts:


*

*There are only $2^{\aleph_0}$ closed sets of real numbers.

*Every closed set of positive measure (in fact every uncountable Borel set but we don't need that) has cardinality $2^{\aleph_0}$.
Let $A$ be a set of positive measure. Using the above facts, a straightforward transfinite induction will serve to construct two disjoint sets $B,C\subseteq A$, each of which has nonempty intersection with every closed subset of $A$ which has positive measure. (If you want to, you can just as easily construct a pairwise disjoint family of $2^{\aleph_0}$ such sets instead of only two.) It is easy to see that $B$ and $C$ are nonmeasurable. ($B$ can't have positive measure because it doesn't contain a closed set of positive measure; $A\setminus B$ can't have positive measure for the same reason. Since $A$ has positive measure, this means that $B$ is nonmeasurable.)
A: If $A$ is non-measurable, we are done. Suppose that $A$ is measurable. Then there exists $r>0$
such that $(-r, r)$ is a subset of $A-A$. We can construct a subset $N$ of $(-r, r)$ which is non-measurable. Therefore, for any $a\in A$ we have, $N+a$ is non-measurable subset of $A.$
