{c}

If $a\geq0$, the set $\{f>a\}$ is contained in $E$, since $f=0$ outside $E$. It is therefore a <null set> and is <Lebesgue measurable>. If $a<0$, its complement $\{f\leq a\}$ is contained in $E$ and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving \b[$f$ is a measurable function].

Solution

Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / D / 1 / e / Solution

 Solution (source code)