 Solution

Let $A_n=\{x:f(x)\ge 1/n\}$. To avoid presupposing finite measure of $A_n$, first truncate it: $A_{n,k}=A_n\cap [-k,k]$ is measurable and has finite measure. Its indicator divided by $n$ is integrable and is bounded above by $f$. The allowed monotonicity property givesThus every $A_{n,k}$ is null. Their countable union $A_n$ is null by part (a), and so is $E={\bigcup }_n{A}_n$. Every positive value of $f$ is at least $1/n$ for some integer $n$, hence $\{f>0\}=E$. Consequently $f=0$ outside a null set.