Proposition 8

— Let D be a countable dense subset of $\mathbf{R}$. In order that a numerical function $f$ (finite or not) be measurable, it is sufficient (and necessary) that for every $a \in D$, the set of $x \in X$ such that $f(x) \geq a$ be measurable.

For, if this is so, for every $b \in \overline{\mathbf{R}}$ the set of $x$ such that $f(x) \geqslant b$ is measurable, being the intersection of the sets (which form a countable family) of the $x$ such that $f(x) \geqslant a$, as $a$ runs over the set of points of D that are $\leqslant b$. The set of $x$ such that $f(x) < b$ is measurable, being the complement of a measurable set. Next, if $b$ is finite then the set of $x$ such that $f(x) \leqslant b$ is measurable, being the intersection of the sets of the $x$ such that $f(x) < b + 1/n$; and $f(-\infty)^{-1}$ is measurable, being the intersection of the sets of the $x$ such that $f(x) < n$, as $n$ runs over $\mathbf{Z}$. Finally, the inverse image under $f$ of every closed interval of $\overline{\mathbf{R}}$ is measurable, being the intersection of two measurable sets, and Th. 4 may be applied.

One could show similarly that it suffices that for every $a \in \mathrm{D}$, the set of $x$ such that $f(x) > a$ be measurable.