Proposition 6

Let $\mathbf{f}$ be a regulated vector function, and $g$ a regulated function $\geq 0$ on $[a, +\infty[$.

1. *If the integral $\int_{a}^{+\infty} g(t) \, dt$ converges then the relation $\mathbf{f} \ll g$ (resp. $\mathbf{f} \sim c g$, where $c$ is constant) implies that $\int_{x}^{+\infty} \mathbf{f}(t) \, dt \ll \int_{x}^{+\infty} g(t) \, dt$ (resp. $\int_{x}^{+\infty} \mathbf{f}(t) \, dt \sim c \int_{x}^{+\infty} g(t) \, dt$).

2. If the integral $\int_{a}^{+\infty} g(t) \, dt$ is infinite then the relation $\mathbf{f} \ll g$ (resp. $\mathbf{f} \sim c g$) implies that

for any $\alpha$ and $\beta$ in $[a, +\infty[$.

It is enough to prove the proposition for the relation $\mathbf{f} \ll g$ since, if $c \neq 0$, the relation $\mathbf{f} \sim c g$ is equivalent to $\mathbf{f} - c g \ll g$.

The first part is an immediate consequence of the mean value theorem, since if $\| \mathbf{f}(x) \| \leq \varepsilon g(x)$ for $x \geq x_0$ one deduces that

In the second place, suppose that $\int_{a}^{+\infty} g(t) \, dt = +\infty$. If $\| \mathbf{f}(x) \| \leq \varepsilon g(x)$ for $x \geq x_0 \geq \max(\alpha, \beta)$, one has

Now there exists an $x_1 \geq x_0$ such that for every $x \geq x_1$

whence, for $x \geq x_1$

which completes the proof, since $\varepsilon > 0$ is arbitrary.

In other words, one can integrate the two terms of a strong relation $\mathbf{f} \ll g$, $\mathbf{f} \sim \mathbf{a}g$, when $g$ is positive on an interval $[a, +\infty[$, and the relation persists between the primitives of the two terms, provided one takes care to integrate from $x$ to $+\infty$ if $\int_{a}^{+\infty} g(t) \, dt$ converges, and from $\alpha$ to $x$ (for any $\alpha$ in $[a, +\infty[$) in the opposite case.

Note that props. 1 (V, p. 228) and 6 (V, p. 230) remain valid when $\mathfrak{F}$ is a base for the trace filter of the intervals $[t, +\infty[$ (where $t > a$) on the complement of a countable set (cf. I, p. 15, th. 2).