I found myself stuck with an "elementary" claim in some article. A simplified version of the problem is  :
Let $p : [0,1]\to [0,1]$ be a continuous and non decreasing function such that $p(0)=0$ and $\int_0^1 (\log \frac{1}{u})^{1/2} \, dp(u) < + \infty$. Show that $$ \lim_{h\to 0} \frac{\int_0^h (\log \frac{1}{u})^{1/2} \, dp(u)}{p(h) (\log \frac{1}{h})^{1/2}} $$ exists and is finite.

I think that a good idea is to use integration by parts formula which gives $$ \int_0^h (\log \frac{1}{u})^{1/2}\, dp(u) = p(h) (\log \frac{1}{h})^{1/2} + \int_0^h \frac{p(u)}{2u(\log \frac{1}{u})^{1/2}} du, $$ but I don't see how to conclude...

I found myself stuck with an "elementary" claim in some article. A simplified version of the problem is: 
Let $p : [0,1]\to [0,1]$ be a continuous and non decreasing function such that $p(0)=0$ and $$ \int_0^1 \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u) < + \infty. $$ Show that $$ \lim_{h\to 0} \frac{\displaystyle\int_0^h \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u)}{p(h) \left(\log \dfrac{1}{h}\right)^{1/2}} $$ exists and is finite.

I think that a good idea is to use integration by parts formula which gives $$ \int_0^h \left(\log \frac{1}{u}\right)^{\!1/2}\, dp(u) = p(h) \left(\log \frac{1}{h}\right)^{\!1/2} + \int_0^h \frac{p(u)}{2u\left(\log \frac{1}{u}\right)^{\!1/2}} du, $$ but I don't see how to conclude...