I like this example, which is as natural as can be an example with sets that are not Lebesgue measurable. Start from the Cantor function $f:[0,1]\rightarrow \mathbb{R}$, and consider $h(x):= x+f(x)$, which is a homeomorphism $[0,1]\rightarrow[0,2]$. On each interval on the complement of the Cantor set $C$ this functions is a translation. Therefore $|h([0,1]\setminus C)|=|[0,1]\setminus C|=1$. Thus $|h(C)|=|[0,2]\setminus h([0,1]\setminus C)|=1$. So there exists a non measurable subset $V$ of $h(C)$; let $W$ be $h^{-1}(V)\subset C$. Finally, the homeomorphism $h^{-1}$ h$maps this Lebesgue measurable set$W$into the non-measurable set$V$. Also note that any Lebesgue, non Borel set in$h(C)$is mapped by the homeomorphism$h^{-1}$into a Lebesgue, non Borel subset of$C$. 1 I like this example, which is as natural as can be an example with sets that are not Lebesgue measurable. Start from the Cantor function$f:[0,1]\rightarrow \mathbb{R}$, and consider$h(x):= x+f(x)$, which is a homeomorphism$[0,1]\rightarrow[0,2]$. On each interval on the complement of the Cantor set$C$this functions is a translation. Therefore$|h([0,1]\setminus C)|=|[0,1]\setminus C|=1$. Thus$|h(C)|=|[0,2]\setminus h([0,1]\setminus C)|=1$. So there exists a non measurable subset$V$of$ h(C)$; let$W$be$h^{-1}(V)\subset C$. Finally, the homeomorphism$h^{-1}$maps this Lebesgue measurable set$W$into the non-measurable set$V\$.