# If $\|f(x)-f(y)\|\ge \alpha\cdot\|x-y\|$ and $g$ integrable $\Longrightarrow$ $g\circ f:A \longrightarrow \mathbb{R}$ is an integrable function

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.

-
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

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...

-
Sorry for the typo. I have just edited it. –  felipeuni Oct 24 '12 at 22:53