 Solution

For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\ge 1/n\}\cap [-m,m]$. This measurable set has finite measure, and $f\ge n^{-1}{1}_{E_{n,m}}$. Monotonicity of the Lebesgue integral givesso every $E_{n,m}$ is a Lebesgue null set. The set where $f>0$ is the countable union of these sets over $n,m$. Part (a) makes it null, henceThe bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.