= Solution
Let $A_n=\{x:f(x)\geq1/n\}$. 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 gives
$$
0\leq\frac1n m(A_{n,k})\leq\int_{\mathbb R}f=0.
$$
Thus every $A_{n,k}$ is null. Their countable <union> $A_n$ is null by part (a), and so is $E=\bigcup_nA_n$. Every positive value of $f$ is at least $1/n$ for some <integer> $n$, hence $\{f>0\}=E$. Consequently \b[$f=0$ outside a <null set>].
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