(1) Suppose the function $g(x) = \sqrt{|f(x)|}$ is measurable on $[0,1]$ with respect to classical Lebesgue measure. Does it follow that the function $f(x)$ is measurable on $[0,1]$?
(2) And if $f$ is integrable on Lebesgue in $[0,1]$, then what can you say about the function $g$ ?
I don't know answer to (1), but I think that in (2) can be a counterexample to the fact that this is wrong.
I would be very grateful if you helped me.
Consider a non-measurable function $f$ such that $f(x)\in\{-1,1\}$ for all $x$ (take $S$ non measurable and $f=\chi_S-\chi_{S^c}$).
Notice that $\sqrt{|f(x)|}\leqslant |f(x)|+1$.