Solution
= Solution
If $f$ is measurable, then for a threshold $a\leq0$, $\{f^+\geq a\}=E$, while for $a>0$, $\{f^+\geq a\}=\{f\geq a\}$. The half-line criterion makes $f^+$ measurable. By part (2), $-f$ is measurable, and $f^-=(-f)^+$ is therefore measurable as well.
Conversely, if $f^+$ and $f^-$ are measurable, part (2) and the permitted <closure> under sums show that $f=f^+-f^-$ is measurable. Thus \b[measurability is equivalent to measurability of both parts].