Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / D / 1 / 3 / Solution

If $f$ is measurable, then for $a\le 0$, $\{f^{+}\ge a\}=E$, and for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. 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 $f$ is measurable exactly when its positive and negative parts are measurable.