Past exam of the options I course of the University of Oxford / 2010 / D / D1 / 4 / Solution

Since $|f|=f^{+}+f^{-}$, part (3) and closure under sums prove that $|f|$ is measurable whenever $f$ is.

The converse is false. Let $A\subseteq [0,1]$ be a non-Lebesgue-measurable set and define $f=1$ on $A$ and $f=-1$ on its complement in $[0,1]$. Then $|f|=1$ is measurable, but $f^{-1}([0,\infty ))=A$ is not. Taking absolute values can conceal a nonmeasurable sign choice.