Theorem 3

— *For every increasing sequence $(f_n)$ of numerical functions $\geqslant 0$ defined on X,

Since each of the functions $f_n$ is less than or equal to $\sup_n f_n$, everything comes down to proving that $\mu^*(\sup_n f_n) \leqslant \sup_n \mu^*(f_n)$; this is obvious if the second member of the inequality is $+\infty$. In the contrary case, $\mu^*(f_n) < +\infty$ for all $n$; we are going to show that, for every $\varepsilon > 0$, there exists an increasing sequence $(g_n)$ of functions in $\mathscr{I}_+$ such that $f_n \leqslant g_n$ and $\mu^*(g_n) \leqslant \mu^*(f_n) + \varepsilon$. If $g$ is the upper envelope of the sequence $(g_n)$, we will then have $\mu^*(g) = \sup_n \mu^*(g_n)$ (No. 1, Th. 1), whence $\mu^*(g) \leqslant \sup \mu^*(f_n) + \varepsilon$; since $\sup f_n \leqslant g$ and $\varepsilon$ is arbitrary, the theorem will then have been proved.

By hypothesis, there exists a function $h_n \in \mathscr{J}_+$ such that $f_n \leq h_n$ and $\mu^*(f_n) \leq \mu^*(h_n) \leq \mu^*(f_n) + \varepsilon/2^n$; let us show that the functions $g_n = \sup(h_1, h_2, \ldots, h_n)$ meet the requirements. They belong to $\mathscr{J}_+$, form an increasing sequence, and satisfy $f_n \leq g_n$ for all $n$; we shall prove that

Let us argue by induction on $n$; the case $n = 1$ is trivial. On the other hand $g_{n+1} = \sup(g_n, h_{n+1})$, $g_n \geq f_n$ and $h_{n+1} \geq f_{n+1} \geq f_n$, whence $\inf(g_n, h_{n+1}) \geq f_n$; since

it follows from Th. 2 of No. 1 that

Q.E.D.