Lemma 1

Let g be a function with values in F, such that Np(g) < +\infty* (in other words, a function in $\mathscr{F}_F^p$). The set A of points x \in X such that g(x) \neq 0 is contained in the union of a negligible set and a sequence of compact sets.

Let $A_n$ be the set of points $x \in X$ such that $|g(x)| \geq 1/n$; A is the union of the $A_n$, and $\varphi_{A_n} \leq n|g|$, whence $|\mu|^*(A_n) \leq (n N_p(g))^p$; it follows that $A_n$ is contained in the union of a negligible set and a sequence of compact sets (§ 4, No. 6, Cor. 3 of Th. 4), therefore so is A.

The lemma proved, consider first the case that f has compact support K. By Cor. 1 of Th. 3 of No. 5, there exists a sequence $(g_n)$ of measurable step functions such that $|g_n(x)| \leq |f(x)|$ at every point $x \in X$ and such that $g_n(x)$ tends to $f(x)$ almost everywhere. Now, $g_n$ is a linear combination of characteristic functions of measurable sets contained in K; since these sets are integrable by Prop. 3 of No. 1, $g_n$ belongs to $\mathscr{L}_F^p$. Since $N_p(f) < +\infty$, Lebesgue’s theorem (§ 3, No. 7, Th. 6) shows that f belongs to $\mathscr{L}_F^p$.

In the general case, it follows from Lemma 1 that there exists an increasing sequence $(K_n)$ of compact sets such that $f(x)$ is zero almost everywhere in the complement of the union of the $K_n$. Let $f_n$ be the function equal to $f(x)$ on $K_n$ and to 0 elsewhere; $f_n$ is measurable by No. 3, Cor. 5 of Th. 1; since $|f_n| \leq |f|$, $f_n$ belongs to $\mathscr{L}_F^p$ by the first part of the argument. Since $f(x)$ is equal almost everywhere to the limit of the sequence $f_n(x)$, Lebesgue’s theorem again proves that $f \in \mathscr{L}_F^p$, which completes the proof.

One should take care to note that a function that is locally negligible but not negligible is not integrable; thus, a function equal locally almost everywhere to an integrable function is not necessarily integrable.