 Solution

If $a\ge 0$, 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 $\{f\le a\}$ 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 $f$ is a measurable function.