Lebesgue measurable set that is not a Borel measurable set

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?

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Why not just ask for clarification in a comment there? –  The Chaz 2.0 May 4 '12 at 17:36
Prove: there are $\mathfrak{c} = \#\mathbb{R}$ Borel measurable sets. Every subset of the Cantor set is Lebesgue measurable. There are $2^\mathfrak{c} = \#P(\mathbb{R})$ subsets of the Cantor set. –  t.b. May 4 '12 at 17:52
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 $S \subseteq \psi(C)$. It's 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.
This is. ${}{}{}{}$ –  leo May 5 '12 at 0:36