 Solution

If $f$ is measurable, then for a threshold $a\le 0$, $\{f^{+}\ge a\}=E$, while for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. 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 measurability is equivalent to measurability of both parts.