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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.

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    A variant: [Measurability of the composition of a measurable map with a surjective map satisfying an expansion condition](http://math.stackexchange.com/q/33742). That $f$ is assumed to be surjective in that thread isn't used in a crucial way and integrability is easy as soon as you have measurability.2012-10-24
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    A possible duplicate of [this](https://math.stackexchange.com/questions/2463714/fx-fy-ge-cx-y-with-c0-then-for-gb-to-mathbbr-integrable-the) question.2017-10-14

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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...

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    Sorry for the typo. I have just edited it.2012-10-24
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    This example doesn't satisfy the "expansion" condition.2017-10-12
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    @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.2017-10-12
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    Don't we have $f(0,0,1/2) = f(0,0,0) = (0,0)?$2017-10-13