Solution
= Solution
{c}
Since $|f|=f^++f^-$, the previous part and closure under addition prove measurability of the <absolute value>.
The converse fails. Choose a nonmeasurable subset $N$ of $[0,1]$ and define $f=1$ on $N$, $f=-1$ on its complement. Then $|f|\equiv1$ is measurable, whereas $f^{-1}([0,\infty))=N$ is not. Thus \b[measurability of $|f|$ does not imply measurability of $f$].