Let $f:[-1,1]$ be Riemann integrable and $\psi(x)=x\ sin(\frac{1}{x})$ for $0<x\leq1$ and $\psi(0)=0$. Show that $x\mapsto f(\psi(x))$ is Riemann integrable over $[0,1]$.

Also, just curious ... does the same conclusion hold if $\psi$ is changed to $\psi(x)=\sqrt{x}\ sin(\frac{1}{x})$ for $x>0$?

Appreciated for all the help!

  • $\begingroup$ The function $\psi$ is continuous on $[0,1]$ and so is the second $\psi$ you consider. $\endgroup$ – Julien Apr 11 '13 at 5:02
  • $\begingroup$ @julien: True, but the general fact here is (continuous) $\circ$ (Riemann integrable) = Riemann integrable. In general it doesn't work the other way (i.e., the way the OP is asking it) around: math.uga.edu/~pete/Lu99.pdf. $\endgroup$ – Pete L. Clark Apr 11 '13 at 5:34
  • $\begingroup$ @PeteL.Clark Thank you very much for the link! $\endgroup$ – Julien Apr 11 '13 at 5:41
  • $\begingroup$ Do you know Lebesgue's criterion for Riemann integrability? $\endgroup$ – 23rd Apr 15 '13 at 9:08

The answer to both your questions is YES, and richard’s comment shows the way. To expand on this comment : by Lebesgue’s criterion for Riemann integrability, the answer to your question will be YES whenever $\psi$ has the following property :

$(P_1) \ \psi^{-1}(N) \text{ stays null for any null set } N$

It is easy to see that $(P_1)$ holds whenever

$(P_2) \ \psi \ \text{is a diffeomorphism}$

(in that case, $\psi^{-1}$ is Lipschitz and it is well-known and easy that the image of a null set by a Lipschitz function is again a null set)

From this, one sees that $(P_1)$ will also hold whenever

$(P’_2) \text{There is a sequence } 1=a_1 \gt a_2 \gt a_3 \ldots \gt a_n \gt \ldots \gt 0 \ \text{such that}\ \psi \text{ is a diffeomorphism on each} \ [a_k,a_{k+1}] $

Both your functions $\psi$ have this last property.


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