prove that $\frac{1}{h}\int^{h}_{0}F(y)\,dy\to F(+\infty)$ as $h\to +\infty$ If F is a non-decreasing, right continuous and a bounded function show that 
$$\frac{1}{h}\int^{h}_{0}F(y)\,dy\to F(+\infty)\text{ as }h\to +\infty.$$
 A: As your assumption, we can conclude that $g(h)=\int_0^h F(y)dy$ goes to $\infty$ when $h\rightarrow+\infty$. Otherwise, $\lim_{h\rightarrow+\infty}F(h)=0$ is valid from the convergence of improper integral which is contradict to assumption. So that 
$$
\lim_{h\rightarrow+\infty} \frac{\int_0^h F(y)dy}{h}=\lim_{h\rightarrow+\infty} F(h)=F(+\infty)
$$
In the first equality L'Hospital principle have been used.
A: This should better be shown from first principles without using L'Hopital rule. 
Define $G(h) = \frac1h \int_0^h F(y) \, dy$.  Since $F$ is bounded and not decreasing, the limit of $F$ when $h\rightarrow \infty$ exists (and is finite, written as $F(+\infty)$.  Now choose some $\epsilon > 0$, and fine large enough $y_\epsilon$ such that $F(y_\epsilon) > F(+\infty) - \epsilon$. Write for $h > y_\epsilon$
\begin{align}
   G(h) &= \frac1h \left( \int_0^{y_\epsilon} F(y)\, dy + \int_{y_\epsilon}^h F(y)\, dy \right) \\
      &\ge \frac{h-y_\epsilon}{h}\cdot \frac{\int_{y_\epsilon}^h F(y)\, dy}{h-y_\epsilon}  \\
   &\ge \frac{h-y_\epsilon}{h} (F(+\infty)-\epsilon)
\end{align} and now letting $h$ go to infinity this approaches $F(+\infty)-\epsilon$. Since this can be done foe any positive $\epsilon$, we are almost done. And, since it is clear the limit cannot be any larger, we are done. (As written, this proof assumes the integrand is never negative. But that problem can be solved in like manner).
A: Use L'hopitals rule in the limit for h->infinity
