exact duplicate of Lebesgue measurable but not Borel measurable
BUT! can you please translate Miguel's answer and expand it with a formal proof? I'm totally stuck…
In short: Is there a Lebesgue measurable set that is not Borel measurable?
They are an order of magnitude apart so there should be plenty examples, but all I can find is "add a Lebesgue-zero measure set to a Borel measurable set such that it becomes non-Borel-measurable". But what kind of zero measure set fulfills such a property?
Best Answer
Let $\phi(x)$ be the Cantor function, which is non-decreasing continuous function on the unit interval $\mathcal{U}_{(0,1)}$. Define $\psi(x) = x + \phi(x)$, which is an increasing continuous function $\psi: [0,1] \to [0,2]$, and hence for every $y \in [0,2]$, there exists a unique $x \in [0,1]$, such that $y = \psi(x)$. Thus $\psi$ and $\psi^{-1}$ maps Borel sets into Borel sets.
Now choose a non Borel subset $S \subseteq \psi(C)$. Its preimage $\psi^{-1}(S)$ must be Lebesgue measurable, as a subset of Cantor set, but it is not Borel measurable, as a topological mapping of a non-Borel subset.