Improper integral of a non-increasing function Let $f: \mathbb{R} \to \mathbb{R}$ be a continous function and non increasing function and let $g$ be a solution of $g'=f$ on $\mathbb{R}_{+}$ and $g(0)=0$, exists $\displaystyle \lim_{t\to +\infty}\int_{0}^{g(t)}f(s)ds$?
I was thinking that in two cases if $\displaystyle \lim_{t\to +\infty}g(t)<\infty$ so, there is not problem and we can conclude that there exists that limit but if $\displaystyle \lim_{t\to+\infty} g(t)=+\infty$, so we have a improper integral $\displaystyle \int_{0}^{+\infty}f(s)ds$ and I don't know how to continue from there.
Also I have by hypothesis that the set $t\in \mathbb{R}$ such that $f(t)=0$ is nonempty set.
 A: I interpret non-increasing as: $t \leqslant t' \implies f(t') \leqslant f(t)$.
But to avoid unnecessary confusion with the "not"s, I shall use "decreasing" instead of "non-increasing".
By "eventually decreasing", I shall mean: there exists $M \in \Bbb R$ such that $M < t \leqslant t' \implies f(t') \leqslant f(t)$.
(Similar definitions apply for "increasing" replaced by "decreasing".)

Since $f$ is continuous, it has an antiderivative $F$ on all of $\Bbb R$. (That is, $F : \Bbb R \to \Bbb R$ satisfies $F' = f$ on all of $\Bbb R$.)
[The only reason I'm doing this is that the $g$ given only satisfies $g' = f$ on $\Bbb R_+$ but I want an antiderivative on $\Bbb R$.]
Let $$H(t) := \int_0^{g(t)}f(s) \ {\mathrm d}s$$
for $t > 0$.
Our job is to show that $\lim_{t \to \infty} H(t)$ exists.
Note $H(t) = F(g(t)) - F(0)$. (Here I'm using the Fundamental Theorem of Calculus. Note that $f$ is continuous.)
Since $g$ and $F$ are differentiable, we see that $H$ is differentiable on $\Bbb R_+$ and
$$H'(t) = F'(g(t))g'(t) = f(g(t))f(t) \tag{1}$$
for all $t > 0$.
[The above is where we utilised $F$. Note that $g(t)$ could very well be negative and so I really did need an antiderivative on $\Bbb R$.]
Now, the set $\{t : f(t) = 0\}$ is nonempty. Pick $t_0$ from that set. Then, we have that $f \leqslant 0$ on $[t_0, \infty)$ since $f$ is decreasing.
Since $g' = f$, we see that $g$ is eventually decreasing.
To summarise, there exists $M > 0$ such that the following hold on $(M, \infty)$:

*

*$f \leqslant 0$,

*$g$ is decreasing,

*(As a consequence,) $f \circ g$ is increasing.

Now, there are two cases:
Case 1. $f \circ g \leqslant 0$ on $(M, \infty)$.
In this case, we have $H' \geqslant 0$ on $(M, \infty)$.   (Use $(1)$.)
Case 2. $f \circ g$ is positive at some point $t_1 \in (M, \infty)$.
Since $f \circ g$ is increasing, we see that $f \circ g > 0$ on $[t_1, \infty)$. Again, by $(1)$, we see that $H' \leqslant 0$ on $[t_1, 0)$.
In either case, we see that $H$ is eventually monotonic. Now, it is easy to see that a monotonic function always has a limit at $\infty$ and thus, we are done.
