= Solution
{c}
For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\geq1/n\}\cap[-m,m]$. This measurable set has finite measure, and $f\geq n^{-1}1_{E_{n,m}}$. Monotonicity of the <Lebesgue integral> gives
$$
0=\int_{\mathbb R}f\geq\frac1n\,\operatorname{meas}(E_{n,m}),
$$
so 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, hence
$$
\boxed{f=0\quad\text{almost everywhere}.}
$$
The bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.
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