# If $f(x)=o(g(x))$ at $0$ does $\int_0^tf(x)=o\left(\int_0^t g(x)dx\right)$ when $t\to 0$?

Suppose $$f(x)=o(g(x))$$ when $$x\to 0$$. Does $$\int_0^t f(x)dx=o\left(\int_0^t g(x)dx\right)$$ when $$t\to 0$$ ?

For me is true, but my teacher says that my proof is wrong without explanation. Can someone explain where I'm wrong ?

Let $$\varepsilon >0$$. There is $$\delta >0$$ s.t. $$|f(x)|\leq \varepsilon |g(x)|$$ when $$|x|\leq \delta$$. Let $$|t|\leq \delta$$, then $$\int_0^t|f(x)|dx\leq \varepsilon \int_0^t|g(x)|dx,$$ and thus $$\lim_{t\to 0}\frac{\int_0^t |f(x)|dx}{\int_0^t |g(x)|dx}\leq \varepsilon ,$$ therefore $$\int_0^t |f(x)|dx=o\left(\int_0^t|g(x)|\right).$$ So if $$f,g\geq 0$$, then the claim is true.

Question : So, I proved the claim for $$f,g\geq 0$$ only. But from here, isn't it obvious that the claim is true for integrable $$f$$ and $$g$$ by writting $$f=f^+-f^-$$ and $$g=g^+-g^-$$ (where $$f^+(x)=\max(f(x),0)$$ and $$f^-(x)=-\min(f(x),0)$$) ?

• No. Going from the case $f,g \geq 0$ to the general case is not obvious. – Kavi Rama Murthy Aug 14 at 11:17
• at least, the case $f,g\geq 0$ is correct ? If yes, how can you generalize it when $f$ and $g$ no positive ? @KaviRamaMurthy – user657324 Aug 14 at 11:26
• I don't see any simple way of going from positive to general case, so your approach does not work. – Kavi Rama Murthy Aug 14 at 11:31
• @KaviRamaMurthy: Maybe it's wrong in general ? (if $f$ and $g$ are not continuous) – user657324 Aug 14 at 11:36
• You can find $g$ such that $\int_0^{1/n} g(x)dx=0$ for every $n$ but there is a positive function $f$ with $f=o(g)$. In this case the conclusion fails. – Kavi Rama Murthy Aug 14 at 11:39

Assuming that $$f$$ and $$g$$ are continuous functions and $$\int_0^{t} g(x)dx$$ does not vanish for $$|t|$$ sufficiently small you get this result immediately from L'Hopital's Rule.