1

Give an example of 2 integrable functions $f$ and $g$ such that $g$ o $f$ is not integrable

Here is my example but I'm not so sure about it

$f(x)=-1$ is integrable and so is $g(x)=\sqrt{x}$. But $g$ o $f= \sqrt{-1}$ doesn't exist so it isn't integrable.

I wonder if my example is correct. If not Can you tell me why and give a better example?

Diane Vanderwaif
  • 4,470

1 Answers1

2

For a non-trivial example consider the integrals over $[0,1]$ of the functions $f$ and $g$ defined by: $f(x)=0$ if $x$ irrational, $f(x)=1/q$ if $x=p/q$ with $p$, $q $ integers with no common factors other than $1$ and $q>0$; $g(x)=1$ if $x\ne0$, $g(0)=0$.

This is Example 4.9 in Gelbaum and Olmsted's Counterexamples in Analysis.

David Mitra
  • 76,313
  • is your $g(x)$ integrable? I;m not sure if I'm correct but from your example I see that $U(g,P)=1$ and $L(g,P)=0$ for all partition $P$. If I let $\epsilon=1/4$ then $U(g,P)-L(g,P)=1>1/4=\epsilon$ which show that $g(x)$ is not integrable. Is it? – Diane Vanderwaif Dec 01 '13 at 13:21
  • @DianeVanderwaif Yes, $g$ is integrable over $[0,1]$. The lower sums are not always $0$, $g$ takes the value $0$ only at $x=0$. (See this post to see that $f$ is integrable.) – David Mitra Dec 01 '13 at 13:32