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$F'(x)$ is positive and integrable on $[a,b]$. Does that imply $F$ is strictly increasing?

Intuitively I think it's true, but don't know how to prove.

Hongyan
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  • @AndrewLi how to prove it? – Hongyan Mar 04 '18 at 05:48
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    @Hongyan mean value theorem. – IntegrateThis Mar 04 '18 at 05:50
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    not sure why you need the integrability of $F'$, but yes, $F$ is strictly increasing on $[a,b]$ ... – D.Hutchinson Mar 04 '18 at 05:54
  • @Hongyan The derivative exists so differentiability implies continuity – Andrew Li Mar 04 '18 at 05:57
  • @D.Hutchinson Sorry, you are right. – Hongyan Mar 04 '18 at 06:08
  • What I really mean is this question :https://math.stackexchange.com/questions/2675833/fx-is-positive-and-integrable-on-a-b-does-that-imply-int-ax-ftdt – Hongyan Mar 04 '18 at 06:15
  • You can't use the current question to solve the question linked in your previous comment. Why? Because if $F(x) =\int_{a} ^{x} f(t) , dt$ then you don't know if $F$ is differentiable or even if it is differentiable tlthen the equality $F'=f$ holds or not. The linked question is significantly difficult than the current question which is a mere application of mean value theorem. – Paramanand Singh Mar 04 '18 at 09:34

2 Answers2

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By the fact that $F'(x)$ exists we can conclude that $F$ is continuous.

Hence by the mean value theorem for all $a_0,b_0$ such that $b\ge b_0>a_0\ge a$ exists $c$ such that $a_0<c<b_0$ and $F'(c)=\frac{F(b_0)-F(a_0)}{b_0-a_0}$. Because $b_0-a_0$ and $F'(c)$ are positive from assumption $F(b_0)-F(a_0)>0\implies F(b_0)>F(a_0)$ for all $b_0>a_0$

ℋolo
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Note: $$\int_a^x f(t)dt=\lim_\limits{n\to\infty} \sum_{k=0}^n \frac1n \cdot f\left(a+k\frac{x-a}{n}\right)>0,$$ because $f(x)>0, x\in (a,b]$.

farruhota
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  • How does that help? Limit of a positive function is not necessarily positive. The fact that (Riemann /Lebesgue) integral of a positive function is positive is a non-trivial fact. – Paramanand Singh Mar 04 '18 at 09:36
  • @ParamanandSingh, I answered the version $3$ of OP. Now I see it is $4$. – farruhota Mar 04 '18 at 09:42
  • I understand your point. But still a positive $f$ means a positive Riemann sum, but it does not follow immediately that limit of Riemann sum is also positive. The conclusion is true, but the proof is reasonably difficult. See https://math.stackexchange.com/a/519921/72031 – Paramanand Singh Mar 04 '18 at 10:37