Proposition 2

Let $f$ be a real regulated function $\geq 0$ on an interval $I \subset \mathbf{R}$ with endpoints $a$ and $b$ ($a < b$). For the integral $\int_a^b f(t)\,dt$ to exist it is necessary and sufficient that the set of numbers $\int_J f(t)\,dt$ be bounded above when J runs through the set of compact intervals contained in I; the integral $\int_a^b f(t)\,dt$ is then the least upper bound of the set of $\int_J f(t)\,dt$.

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

the map $J \mapsto \int_J f\,dt$ is thus increasing, and the proposition follows from the monotone limit theorem (Gen. Top., IV, p. 349, th. 2).

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.