Exercise: Prove inequality using Riemann lower sum and integral 
Let $\;F:\mathbb R^m \rightarrow \mathbb R\;$ be a continuous,
  non-negative function and $\;g:\mathbb R \rightarrow \mathbb R^m\;$ a
  continuous map. Consider $\;x_n \in \mathbb R\;$ a divergent sequence
  and $\;\varepsilon \gt 0\;$ such that:
$\;F(g(x))\ge \frac {\varepsilon}{2}\;\;\forall x\in
 I_n=(x_n-\frac{l^2}{2s},x_n-\frac{l^2}{2s})\;$ for some $\;l,s \gt
 0\;$. Show that: For every $\;L\gt 0\;$
$\;\sum_{x_n \le L} {\frac{\varepsilon l^2}{s} \le
 \int_{0}^{x
_n+\frac{l^2}{2s}} F(g(x)) \;dx }\;$.

My attempt:
By passing to a subsequence , I can assume that the intervals $\;I_n\;$ are disjoint. Then for every $\;I_n\;$ it holds, $\;\frac {\varepsilon}{2}(x_n+ \frac{l^2}{2s}-x_n+\frac{l^2}{2s})  \le \int_{x_n-\frac{l^2}{2s}}^{x_n+\frac{l^2}{2s}} F(g(x)) \;dx \Rightarrow \frac {\varepsilon l^2}{2s}\le \int_{x_n-\frac{l^2}{2s}}^{x_n+\frac{l^2}{2s}} F(g(x)) \;dx\;$.
At this point I've been stuck although I feel I'm pretty close to the solution.My questions are below:


*

*Can $\;I_n\;$ be considered as partition? I can't see how...

*Why does such $\;L\;$ exist?

*How can I conclude to $\;0\;$ instead of $\;x_n-\frac{l^2}{2s}\;$ in the integral?(I tried to change variable but then the whole integral changed)


I would appreciate any help. Hints are welcome too.
Thnaks in advance!
 A: I assume by divergent you mean $x_n \to +\infty$.
If you are, in fact, considering Riemann sums, the inequality probably should be 
$$\sum_{x_n \leqslant L}\frac{\epsilon}{2} \frac{l^2}{s} \leqslant \int_0^{x_2 + \frac{l^2}{2s}}F(g(x)) \, dx $$
Consider $L$ such that $0 < x_1 \leqslant x_2 \leqslant L.$  If the intervals $I_1$ and $I_2$ do not overlap then
$$\int_0^{x_2 + \frac{l^2}{2s}}F(g(x)) \, dx \geqslant  \int_{x_1 - \frac{l^2}{2s}}^{x_1 + \frac{l^2}{2s}}F(g(x)) \, dx +\int_{x_2 - \frac{l^2}{2s}}^{x_2 + \frac{l^2}{2s}}F(g(x)) \, dx \\ \geqslant \frac{\epsilon}{2}2\frac{l^2}{2s} +   \frac{\epsilon}{2}2\frac{l^2}{2s} \\ = \epsilon \frac{l^2}{s} \\ =  \sum_{x_n \leqslant L}\frac{\epsilon}{2} \frac{l^2}{s}.$$
This argument can be extended by induction to prove the inequality for $n > 2$.
If the intervals overlap then $x_2 - \frac{l^2}{2s} < x_1 + \frac{l^2}{2s}$ which implies  $x_2 - x_1 < \frac{l^2}{s}$ and, 
$$\int_0^{x_2 + \frac{l^2}{2s}}F(g(x)) \, dx =  \int_{x_1 - \frac{l^2}{2s}}^{x_1 + \frac{l^2}{2s}}F(g(x)) \, dx +\int_{x_1 + \frac{l^2}{2s}}^{x_2 + \frac{l^2}{2s}}F(g(x)) \, dx \\ \geqslant \frac{\epsilon}{2}2\frac{l^2}{2s} +   \frac{\epsilon}{2}\left(x_2 + \frac{l^2}{2s} - \left(x_1 + \frac{l^2}{2s}\right)\right) \\ =  \epsilon\frac{l^2}{2s} +   \frac{\epsilon}{2}(x_2 - x_1) \\ <  \epsilon\frac{l^2}{s} \\ = \sum_{x_n \leqslant L}\frac{\epsilon}{2} \frac{l^2}{s}.$$
In this case the inequality could be violated. This case could be precluded if $l$ and $s$ were chosen such that intervals do not overlap.
