2.11

Consider first the set: $$A = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < 0 \rbrace$$ It's clear that: $$A = \bigcup_{n=1}^\infty \lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace$$ Assume to the contrary that $m(A) > 0$. We have: $$m(A) \leq \sum_{n=1}^\infty m(\lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace)$$ Since $m(A) > 0$, for at least one $n$, $m(\lbrace x \hspace{0.25cm} | \hspace{0.25cm} f(x) < - \frac{1}{n} \rbrace)$ > 0. Call this set $E$. We have: $$\int_{E} f \leq \int_E -\frac{1}{n} = -\frac{1}{n} m(E) < 0$$ ...a contradiction. It's obvious that if you switch that conditions around such that for any measurable set $S \subset \mathbb{R}^d$, $$\int_S f \leq 0$$ that the same line of reasoning will result in $f \leq 0$ almost everywhere. Combining these conditions two conditions, i.e., for any measurable set $S \subset \mathbb{R}^d$, $$\int_S f = 0$$ will directly result in $0 \leq f \leq 0$ almost everywhere. I.e. $f = 0$ almost everywhere.