Indeed, since $E$ is complete the Cauchy criterion shows that for the integral $\int_I f(t) \, dt$ to be convergent it is necessary and sufficient that for any $\varepsilon > 0$ there exists a compact interval $J_0 = [\alpha, \beta]$ with for every compact interval $J$ such that $J_0 \subset J \subset I$ one has $\| \int_J f(t) \, dt - \int_{J_0} f(t) \, dt \| \leq \varepsilon$. The proposition will follow from the following lemma:

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).

Indeed, since $f \geq 0$, the relation $J \subset J'$ implies that

When the map $J \mapsto \int_J f(t)\,dt$ is not bounded it has limit $+\infty$ with respect to the directed set $\mathscr{K}(I)$; then one says, by abuse of language, that the integral $\int_a^b f(t)\,dt$ is equal to $+\infty$. The properties of integrals established in n 1 extend (when dealing with functions $\geq 0$) to the case where certain of the integrals concerned are infinite, provided that the relations in which they feature make sense.

PROPOSITION 3 (comparison principle). Let $f$ and $g$ be two real regulated functions on an interval $I \subset \mathbf{R}$, such that $0 \leq f(x) \leq g(x)$ at each point where $f$ and $g$ are continuous (cf.* II, p. 61, prop. 6). If the integral of $g$ over $I$ is convergent, so also is the integral of $f$, and one has $\int_I f(t)\,dt \leq \int_I g(t)\,dt$. Further, the two integrals cannot be equal unless $f(x) = g(x)$ at every point of $I$ where $f$ and $g$ are continuous.*

Now for every compact interval $J \subset I$ one has

since $\int_J g(t)\,dt \leq \int_I g(t)\,dt$, the integrals $\int_J f\,dt$ are bounded above, so the integral $\int_I f(t)\,dt$ is convergent; further, on passing to the limit, one has $\int_I f(t)\,dt \leq \int_I g(t)\,dt$. Suppose further that $f(x) < g(x)$ at a point $x \in I$ at which $f$ and $g$ are continuous; there exists a compact interval $[c,\ d]$ contained in $I$, not reducing to a (single) point, and such that $x \in [c,\ d]$; one has $\int_c^d f(t)\,dt < \int_c^d g(t)\,dt$ (II, p. 61, cor. 1), and since on the other hand $\int_a^c f(t)\,dt \leq \int_a^c g(t)\,dt$ and $\int_d^b f(t)\,dt \leq \int_d^b g(t)\,dt$ from the above, one sees, on adding term-by-term, that $\int_a^b f(t)\,dt < \int_a^b g(t)\,dt$.

This proposition provides the most frequently used means for deciding if the integral of a function $f \geq 0$ is or is not convergent: namely, comparing $f$ to a simpler function $g \geq 0$ whose integral one already knows to be, or not to be, convergent; we shall see in chap. V how to search for comparator functions, in the most usual cases; and we shall deduce everyday criteria for the convergence of integrals and of series.

Indeed, for every compact interval $J \subset I$ one has (II, p. 61, formula (16))

If the integral of the positive function $\| \mathbf{f}(x) \|$ is convergent, then for every $\varepsilon > 0$ there exists a compact interval $[\alpha,\ \beta]$ contained in $I$, such that, for every compact interval $[\lambda,\ \gamma]$ contained in $I$ and having no interior point in common with $[\alpha,\ \beta]$, one has $\int_{\lambda}^{\gamma} \| \mathbf{f}(t) \, dt \| \leq \varepsilon$ (II, p. 65, prop. 1); one deduces that $\left| \int_{\lambda}^{\gamma} \mathbf{f}(t)\,dt \right| \leq \varepsilon$, which shows convergence of the integral over $I$ (II, p. 16, prop. 1); on passing to the limit in (3) one deduces the inequality (2).

Indeed, there exists a number h > 0 such that one has $\| [x.y] \| \leq h \| x \| . \| y \|$ identically (Gen. Top., IX, p. 173, th. 1); if one puts $k = \sup_{x \in I} \| f(x) \|$, then one has $\| [f(x).g(x)] \| \leq h k \| g(x) \|$ on I; the comparison principle now shows that the integral of [f.g] is absolutely convergent, and, from (2),