Lemma

Let $J_0 = [\alpha, \beta]$ be a compact interval contained in $I$. In order that $\| \int_J f(t) \, dt - \int_{J'} f(t) \, dt \| \leq \varepsilon$ for every pair of compact intervals $J, J'$ contained in $I$ and containing $J'$ it is necessary that $\| \int_K f(t) \, dt \| \leq \varepsilon$, and it suffices that $\| \int_K f(t) \, dt \| \leq \varepsilon / 2$, for every compact interval $K$ contained in $I$ and having no interior point in common with $J_0$.

Indeed, if for $J_0 \subset J \subset I$ and $J_0 \subset J' \subset I$, one has

one sees in particular that, for $x \leq y \leq \alpha$, or for $\beta \leq x \leq y$ ($x$ and $y$ in $I$), one has $\| \int_x^y f(t) \, dt \| \leq \varepsilon$. Conversely, if $\| \int_K f(t) \, dt \| \leq \varepsilon / 2$ for every compact interval $K \subset I$ such that $K \cap J_0 = \emptyset$, and if $J = [x, y]$, $J' = [z, t]$ are two compact intervals contained in $I$ and containing $J_0$, one has

1 Recall (Set Theory, III, p. 144) that a set $\mathfrak{F}$ of subsets of $I$ is directed with respect to the relation $\subset$ if, for any $X \in \mathfrak{F}$, $Y \in \mathfrak{F}$, there exists $Z \in \mathfrak{F}$ such that $X \subset Z$ and $Y \subset Z$. If $S(X)$ denotes the subset of $\mathfrak{F}$ formed by the $U \in \mathfrak{F}$ such that $U \supset X$, then the $S(X)$ form a base for a filter on $\mathfrak{F}$, called the filter of sections of $\mathfrak{F}$; the limit (if it exists) of a map $f$ of $\mathfrak{F}$ into a topological space, with respect to the filter of sections of $\mathfrak{F}$, is called the limit of f with respect to the directed set $\mathfrak{F}$ (cf. Gen. Top., I, p. 70 and Gen. Top., IV, p. 348).

since