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Let $f$ be a function defined on the close interval $[a,b]$. Does the riemann stieltjes integrability of $f^3$ imply the riemann stieltjes integrability of $f$ ? The answer is trivially no in the case of $f^2$, but I am not able to find a counterexample for the case $f^3$.

AlmostSureUser
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    this may not be an "answer" but this should give you some idea....$\sum \frac{1}{n^3}$ is finite but $\frac{1}{n}$ is not finite.... SO? –  Dec 18 '13 at 13:51
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    @PraphullaKoushik But here we integrate on a closed interval $[a,b]$, the trivial $\int_1^{\infty} 1/x^3 \mathrm{d}x$ won't work here. – Jean-Claude Arbaut Dec 18 '13 at 13:52
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    @arbautjc : Valid point :D i missed that.... any how i will keep it as it is for time being until i think of a solution... –  Dec 18 '13 at 13:54
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    You put "Stieltjes" in your question. Are you doing integrals like $\int_a^b f(x)^3 d\phi(x)$ or something? – GEdgar Dec 18 '13 at 14:53
  • Yes, I mean an integral w.r.t. a function:

    http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral

     – AlmostSureUser Dec 18 '13 at 15:24

1 Answers1

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You could try proving the following theorem : If $f:[a,b]\to \mathbb{R}$ is Riemann-integrable, and $\varphi : [c,d]\to \mathbb{R}$ is continuous where $f([a,b])\subset [c,d]$, then $\varphi\circ f$ is Riemann integrable.

You need to begin with the Riemann condition for integrability (difference between Upper and Lower sum is small), and use the uniform continuity of $\varphi$. It takes a little more work than that, but that is the essence of the argument.

Now just take $\varphi(x) :=x^{1/3}$, to see that your statement is true.

Prahlad Vaidyanathan
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  • Wouldn't this argument be valid for $\varphi(x) :=x^{1/2}$ too? This is the trival counter-example from the OP. – gammatester Dec 18 '13 at 14:04
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    No, because $x \mapsto x^2$ is not injective. However, given $f^3$, $f$ is uniquely determined. – Prahlad Vaidyanathan Dec 18 '13 at 14:07
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    No. Let $\varphi (x)=x^2$, $\psi (t)=\sqrt t$. Then $\psi \circ \varphi = |\cdot |\neq \text{Id}$. – pppqqq Dec 18 '13 at 14:07
  • Sorry, I didnt understand. My hypothesis is that $f^3$ is integrable and I have (if it is true) to show that $f$ is integrable, why should I start with the hypotesis that $f$ is integrable? This is exactly what I want to prove/confute. – AlmostSureUser Dec 18 '13 at 15:33
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    You start with $f^3$ playing the role of $f$ in the above theorem, and $\varphi(x) =x^{1/3}$, so $f = \varphi\circ f^3$ – Prahlad Vaidyanathan Dec 18 '13 at 15:42
  • I suppose that if we only consider $[a,b]$ such that $0\le a \le b$, then $\phi(x) = x^{1/2}$ would work? – 3x89g2 Mar 02 '17 at 04:37
  • @Prahlad Vaidyanathan, but for the map $\phi$ if some entry $x$ is negetive, then what about $\phi(x)$. – MathBS May 17 '18 at 20:08