A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere

Assume $\int fd\mu =0$. For $n\ge 1$ let $E_n:=\{f\ge 1\mathrm{/}n\}$. Then $(1\mathrm{/}n){\chi }_{E_n}\le f$, so [L1] and [L4] give $\frac{1}{n}\mu (E_n)=\int (1\mathrm{/}n){\chi }_{E_n}d\mu \le \int fd\mu =0$. Hence $\mu (E_n)=0$ for every $n$. Since $\{f>0\}={\bigcup }_n{E}_n$, the exceptional set where $f\mathrm{\ne }0$ is null, so $f=0$ almost everywhere by [L2].

Assume $f=0$ almost everywhere, and let $N$ be a measurable null set containing $\{f>0\}$. For any disjoint simple representation $s={\sum }_j{c}_j{\chi }_{E_j}\le f$, every $E_j$ with $c_j>0$ lies inside $N$ and has measure zero. Every zero-coefficient term contributes zero, including an omitted complement cell of infinite measure, by the simple integral's $0\cdot (+\mathrm{\infty })=0$ convention. Thus $\int sd\mu =0$. The now-complete finite refinement proof of The simple integral is independent of the chosen representation makes this value independent of representation. Taking the supremum over all simple minorants in [L3] gives $\int fd\mu =0$.

Step 1.1 proves the forward implication and step 1.2 proves the reverse implication.