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Any idea about this problem:

Let $f:A\subset\mathbb{R}^m \longrightarrow B\subset\mathbb{R}^n$ continuous such that:

$\|f(x)-f(y)\|\ge \alpha\cdot\|x-y\|,\forall x,y\in A$ ($\alpha >0$ is a constant)

If $g:B \longrightarrow \mathbb{R}$ is an Riemann integrable function, prove that $g\circ f:A \longrightarrow \mathbb{R}$ is an Riemann integrable function.

Any hints would be appreciated.

felipeuni
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1 Answers1

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I think the problem is wrong as stated. $A \subset \mathbb R^3$ the unit ball, $B \subset \mathbb R^2$ the disk of radius 1 billion, and $f: A \to B$ defined as

$$f(x_1,x_2,x_3)=(x_1,x_2) \, \mbox{ if $x_1$ is rational} $$ $$f(x_1,x_2,x_3)=(x_1+10, x_2) \, \mbox{ if $x_1$ is irrational}$$

It is easy to check taht this function satisfies the above relation with $\alpha=1$..

If $g$ is continuous, typically $g \circ f$ is discontinuous at all points...

P.S. Are you talking about Riemann or Lesbegue integrability... My example only shows it is not Riemann integrable...

N. S.
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  • Sorry for the typo. I have just edited it. – felipeuni Oct 24 '12 at 22:53
  • This example doesn't satisfy the "expansion" condition. – zhw. Oct 12 '17 at 18:31
  • @zhw. If both $x_1's$ are rational or irrational you get equality in the "expansion" condition. And if one is rational and the other irrational, the fact that $10$ is much larger than the diameter of $A$ gives immediately the expansion condition. – N. S. Oct 12 '17 at 22:27
  • Don't we have $f(0,0,1/2) = f(0,0,0) = (0,0)?$ – zhw. Oct 13 '17 at 15:30