TITLE: Show that an integral can be made as small as possible.
QUESTION [1 upvotes]: Consider a function $\mu(s)$ satisfying the following properties:

$\mu(s) \in C^0((0,+\infty))$,
$\mu(s) > 0$ and $\mu(s)$ is increasing in $s \in (0,+\infty)$,
$\displaystyle \int_0^1 {\dfrac{\mu(s)}{s} ds} < +\infty.$

Show that for every $\epsilon > 0$, there is a $\delta >0 $ such that
$$\displaystyle \int_0^\delta {\dfrac{\mu(s)}{s} ds} < \epsilon.$$
My idea: Set $\displaystyle F(\delta) := \int_0^\delta {\dfrac{\mu(s)}{s} ds}$. Then the conclusion follows if we can show that
$$ \lim_{\delta \to 0^+} {F(\delta)} = 0 $$
This, in turn, will follow if $F$ is continuous on $[0,\delta)$. There will be no problem if $\mu(s) /s$ were bounded on $[0,\delta)$ but this is not always the case; take $\mu(s) = s^a\ (0<a<1)$, for example.

REPLY [1 votes]: Well 
$$\int_{0}^\delta \frac{\mu(s)}{s} + \int_{\delta}^1 \frac{\mu(s)}{s} = \int_{0}^1 \frac{\mu(s)}{s} = l \in \mathbb{R} \ \ (\star)$$
Now since $f(\delta) = \int_\delta^1 \frac{\mu(s)}{s}$ is continuos, and since $f(0) = l$, $f(1) = 0$, it implies that $f(\delta)$ will eventually take any value between $0$ and $l$, and in particular $$\forall \epsilon > 0 \ \exists \delta : f(\delta) = l - \epsilon$$
And so substituting in $(\star)$
$$\int_0^\delta \frac{\mu(s)}{s} = l - f(\delta) = \epsilon$$
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