Definition 1

— Let X be a locally compact space, $\mu$ a measure on X. Given a numerical function f (finite or not), defined locally almost everywhere in X, we call maximum in measure, or $\mu$-maximum (resp. minimum in measure, or $\mu$-minimum) of the function f, and denote by $M_\infty(f)$ (resp. $m_\infty(f)$), the infimum (resp. supremum) of the set of numbers $\alpha$ such that $f(x) \leq \alpha$ (resp. $f(x) \geq \alpha$) locally almost everywhere (for $\mu$).

It follows at once from the definition that $m_\infty(f) = -M_\infty(-f)$, thus from every property of the maximum in measure one deduces a corresponding property of the minimum in measure.

For every $\alpha > M_\infty(f)$, the set of $x \in X$ such that $f(x) > \alpha$ is locally negligible; now, the set of $x \in X$ such that $f(x) > M_\infty(f)$ is the union of the sets where $f(x) > r_n$, with $r_n$ running over the set of rational numbers $> M_\infty(f)$; therefore $f(x) \leq M_\infty(f)$ locally almost everywhere (§ 5, No. 2). Similarly $f(x) \geq m_\infty(f)$ locally almost everywhere; it follows that $m_\infty(f) \leq M_\infty(f)$ if the measure $\mu$ is nonzero; moreover, the relation $m_\infty(f) = M_\infty(f)$ is equivalent to saying that $f$ is equal to a constant locally almost everywhere. It is clear that if the measure $\mu$ is nonzero, then

If two functions $f, g$ are equal locally almost everywhere, then $m_\infty(f) = m_\infty(g)$ and $M_\infty(f) = M_\infty(g)$.

Finally, if $f$ and $g$ are two functions such that $f + g$ is defined locally almost everywhere, then

(1)

provided the second member is defined, as follows at once from Def. 1; similarly, if $f$ and $g$ are both $\geq 0$ and are such that $fg$ is defined locally almost everywhere, then

(2)

provided the second member is defined.

If $M_\infty(f) < +\infty$, then $f(x) < +\infty$ locally almost everywhere, but not necessarily almost everywhere. A numerical function $f$ is said to be bounded in measure (for the measure $\mu$) if it is defined and finite almost everywhere and if, moreover, the numbers $m_\infty(f)$ and $M_\infty(f)$ are both finite (the latter condition amounts to saying that $M_\infty(|f|) < +\infty$).