Example

— Let $X$ be a discrete space, $\mu$ a measure on $X$, and set $\alpha(x) = \mu(\varphi_{\{x\}})$ for every $x \in X$. The functions in $\mathscr{F}_F^1$ are then integrable, in other words $\mathscr{L}_F^1 = \mathscr{F}_F^1$; moreover, for every function $f \in \mathscr{L}_F^1$,

For, let $f \in \mathscr{F}_F^1$; we have $|\mu|^*(|f|) = \sum_{x \in X} |\alpha(x)| \cdot |f(x)| < +\infty$ (§ 1, No. 3, Example); for every $\varepsilon > 0$, there exists a finite subset $M$ of $X$ such that

The function $g$ equal to $f$ at the points $x \in M$ where $|f|$ is finite, and to 0 elsewhere, belongs to $\mathscr{H}(X; F)$ and, by the conventions that have been made,

which proves that $f \in \mathscr{L}_F^1$. On the other hand,

whence the second assertion.

In other words, the $\mu$-integrable functions $f$ are those for which the family $(\alpha(x)f(x))_{x \in X}$ is absolutely summable (GT, IX, §3, No. 6), and the integral $\int f\, d\mu$ is the sum of this family.

Since $\mu(f)$ is continuous on $\mathscr{L}_F^1$ by definition, and since it takes its values in a Hausdorff space, we have $\mu(f) = 0$ for every function that belongs to the closure of 0 in $\mathscr{L}_F^1$, that is, is negligible; if $f$ and $g$ are two equivalent integrable functions, then $\mu(f) = \mu(g)$. In other words, the value of $\mu(f)$ depends only on the class $\tilde{f}$ of the integrable function $f$; it is again denoted $\mu(\tilde{f})$, and the function $\tilde{f} \mapsto \mu(\tilde{f})$ is a continuous linear mapping of $L^1_F$ into $F$. If a function $f$, with values in $F$ and defined almost everywhere in $X$, is equivalent to an integrable function, we again say that $f$ is integrable and we write $\int f\, d\mu = \mu(\tilde{f})$; one defines similarly an integrable function with values in $\overline{\mathbf{R}}$, defined and finite almost everywhere, as well as its integral.