Solution
= Solution
Since $|f|=f^++f^-$, part (3) and <closure> under sums prove that $|f|$ is measurable whenever $f$ is.
\b[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.