Lemma

If $f$ and $g$ are two functions in $\mathscr{L}^p$ such that $0 \leq f \leq g$, then

When $f$ and $g$ are continuous with compact support, the relation (9) may be written

and is then a consequence of the elementary inequality $(g - f)^p \leq g^p - f^p$ (No. 1, formula (2)). To pass from this to the general case, it suffices to observe that the two members of (9) are continuous functions on $\mathscr{L}^p \times \mathscr{L}^p$, and that every function $f \geq 0$ in $\mathscr{L}^p$ is the limit (for convergence in mean of order $p$) of a sequence of continuous functions $\geq 0$ with compact support, by the continuity of the mapping $g \mapsto |g|$ on $\mathscr{L}^p$ (Prop. 11).

The lemma having been established, for every $\varepsilon > 0$ there exists by hypothesis an $\tilde{f} \in \mathrm{H}$ such that, for every $\tilde{g} \geq \tilde{f}$ belonging to $\mathrm{H}$, we have $(\|\tilde{g}\|_p)^p - (\|\tilde{f}\|_p)^p \leq \varepsilon$; from this it follows that $(\|\tilde{g} - \tilde{f}\|_p)^p \leq \varepsilon$; thus, if $\tilde{g}_1$ and $\tilde{g}_2$ are two elements in $\mathrm{H}$ that are $\geq \tilde{f}$, then $\|\tilde{g}_1 - \tilde{g}_2\|_p \leq 2\varepsilon^{1/p}$, which proves that $\mathfrak{F}$ is a Cauchy filter base on $L^p$ and completes the proof of Prop. 14.