Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
A variant: Measurability of the composition of a measurable map with a surjective map satisfying an expansion condition. 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. –  user45928 Oct 24 '12 at 22:46

1 Answer 1

up vote 0 down vote accepted

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...

share|improve this answer
    
Sorry for the typo. I have just edited it. –  felipeuni Oct 24 '12 at 22:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.