= Solution
{c}
If $f$ is measurable, then for $a\leq0$, $\{f^+\geq a\}=E$, and for $a>0$, $\{f^+\geq a\}=\{f\geq a\}$. Thus the given superlevel criterion proves $f^+$ measurable. Part 2 makes $-f$ measurable, and the same argument applied to $(-f)^+$ proves $f^-$ measurable.
Conversely $f=f^+-f^-$. If both parts are measurable, part 2 and the permitted closure under addition prove $f$ measurable. Therefore \b[$f$ is measurable exactly when its positive and negative parts are measurable].
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