# real analysis – Proving f Riemann integrable implies |f| Riemann integrable by contradiction.

I know that there exists a theorem telling us that if a real valued function $$f$$ defined over an interval $$(a,b)$$ is Riemann integrable, then |$$f$$| is too (moreover I believe the converse to be false). I’ve tried to look at the proofs, but they seem a bit beyond my ability at the moment. My first thought was to suppose there existed a Riemann integrable function, let us say $$g$$, such that |$$g$$| was not Riemann integrable, and show that this leads to some sort of contradiction.

I thought about this for a while and have not been able to think of such a function/proof. Would anybody know of such an example?