Can the identity Lebesgue function be almost surely Borel?

Question

I am trying to construct a counterexample to a homework question, and my intuition says my example is good, but I couldn't prove it. Here is the problem:

Let $f : \left[0, 1\right] -> \left[0, 1\right]$ be the identity map $f(x) = x$. When both spaces are equipped with the Lebesgue $\sigma$-algebra, this map is obviously measurable.

Is it possible (I think not) to find an equivalent (i.e. equal "Lebesgue-almost-everywhere") function $g$ that satisfies the following: For every Lebesgue set $L \subset \left[0, 1\right]$, $g^{-1}\left(L\right)$ is a Borel set.

(In less precise words, can you weaken the domain's $\sigma$-algebra from Lebesgue to Borel without weakening the range, in a way that the function is still equivalent to a measurable map?)

Answer 1

No. There are contimuum many ($\mathfrak{c}$) Borel sets, but $2^\mathfrak{c}$ Lebesgue measurable sets. And this holds for the restriction to every subset with positive measure. So on any full measure set, one can generate $2^\frak{c}$ many sets that are preimages of Lebesgue measurable sets under the identity and they cannot be all Borel sets.