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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 $f^-(x)=-\min(f(x),0)$) ?

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    $\begingroup$ No. Going from the case $f,g \geq 0$ to the general case is not obvious. $\endgroup$ – Kavi Rama Murthy Aug 14 at 11:17
  • $\begingroup$ at least, the case $f,g\geq 0$ is correct ? If yes, how can you generalize it when $f$ and $g$ no positive ? @KaviRamaMurthy $\endgroup$ – user657324 Aug 14 at 11:26
  • $\begingroup$ I don't see any simple way of going from positive to general case, so your approach does not work. $\endgroup$ – Kavi Rama Murthy Aug 14 at 11:31
  • $\begingroup$ @KaviRamaMurthy: Maybe it's wrong in general ? (if $f$ and $g$ are not continuous) $\endgroup$ – user657324 Aug 14 at 11:36
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Aug 14 at 11:39
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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.

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