Proposition 1 (Inequality of the mean)

— Let f be a measurable numerical function that is bounded in measure. For every integrable numerical function $g \geqslant 0$, the function $fg$ (defined almost everywhere) is integrable and

Moreover, two of the three members of the inequality (3) cannot be equal unless, in the set of $x \in X$ such that $g(x) \neq 0$, f is equal to $M_\infty(f)$ almost everywhere or equal to $m_\infty(f)$ almost everywhere.

Indeed, $fg$ is measurable (§ 5, No. 3, Cor. 5 of Th. 1); moreover, the inequality $m_\infty(f)g(x) \leqslant f(x)g(x) \leqslant M_\infty(f)g(x)$ holds, not only locally almost everywhere, but even almost everywhere, because the set of points $x \in X$ where $g(x) \neq 0$ is a countable union of integrable sets (§ 5, No. 6, Lemma 1). It follows that $fg$ is integrable (§ 5, No. 6, Th. 5) and the inequality (3) holds. On the other hand, the function $M_\infty(f)g - fg$ is almost everywhere defined and equal to $(M_\infty(f) - f)g$; it is therefore $\geqslant 0$ almost everywhere in $X$; since the relation $M_\infty(f) \int g d|\mu| = \int fg d|\mu|$ is equivalent to $\int (M_\infty(f) - f)g d|\mu| = 0$, it can hold only if the function $(M_\infty(f) - f)g$ is negligible, which completes the proof.

Setting aside the trivial case that $\int g d|\mu| = 0$, the inequality (3) may be deduced from Th. 1 of No. 1 applied to the interval $D = [m_\infty(f), M_\infty(f)]$. One can bring to Th. 1 of No. 1 the complements analogous to those of Prop. 1, that specify the case in which the point $(\int fg d\mu)/(\int g d\mu)$ belongs to the boundary of D (Exer. 2).