Theorem 4

— For a set $A$ to be integrable, it is necessary and sufficient that, for every $\varepsilon > 0$, there exist an integrable open set $G$ and a compact set $K$, such that $K \subset A \subset G$ and

a) The condition is sufficient, because it means that $\varphi_K \leq \varphi_A \leq \varphi_G$ and $\int (\varphi_G - \varphi_K) d|\mu| \leq \varepsilon$; since $\varphi_G$ and $\varphi_K$ are integrable, so is $\varphi_A$ (§ 3, No. 4, Prop. 8).

b) The condition is necessary. If $A$ is integrable, there exists an open set $G \supset A$ such that $|\mu|^*(G)$ is arbitrarily close to $|\mu|^*(A) = |\mu|(A)$ (§ 1, No. 4, Prop. 19); thus, it all comes down to proving that for every $\varepsilon > 0$, there exists a compact set $K \subset A$ such that $|\mu|(A) - |\mu|(K) \leq \varepsilon$. Since $\varphi_A$ is integrable, there exists a function $f \geq 0$, upper semi-continuous and with compact support $S$, such that $f \leq \varphi_A$ and $\int (\varphi_A - f) d|\mu| \leq \varepsilon/2$ (No. 4, Th. 3). Let $\delta > 0$ be an arbitrary number and let $K$ be the set of points $x \in X$ such that $f(x) \geq \delta$; $K$ is closed and is contained in $S$, hence is compact, and since $f \leq \varphi_A$ we have $K \subset A$. The set $B = A - K$ is integrable, and $f \leq \varphi_K + \delta \varphi_B$, whence

and finally which completes the proof, since $\delta$ is arbitrary.