34

Suppose $f,g$ are two Riemann Integrable functions .Is it true that $f\circ g$ is also Riemann Integrable?

Trying this for a long time but not getting the answer

Amzoti
  • 56,093
Learnmore
  • 31,062

1 Answers1

83

A function on a bounded interval is Riemann-integrable iff it is bounded and almost everywhere continuous. So the functions $$ f(x) = \begin{cases} 1 & \text{ for }x \ne 0 \\ 0 & \text{ for } x = 0 \end{cases} \quad \text{ and } \quad g(x) = \begin{cases} 1/q & \text{ for }x=p/q \\ 0 & \text{ for } x \notin \mathbb{Q} \end{cases} $$ are Riemann-integrable over any bounded interval, since $f$ is continuous everywhere except at $0$, and $g$ is continuous at every irrational $x$. (In the definition $x=p/q$ is the unique representation of rational $x$ with $p$ and $q$ relatively prime integers and $q>0$.)

The composition of these functions is $$ f(g(x)) = \begin{cases} 1 & \text{ for }x \in \mathbb{Q} \\ 0 & \text{ for } x \notin \mathbb{Q} \end{cases} $$ which is nowhere continuous, so not Riemann-integrable over any interval.

Lukas Geyer
  • 18,259
  • Thanks for your effort sir but my problem has been solved by David Mitra – Learnmore Dec 10 '14 at 15:44
  • 10
    That's fine there are lots of other people who benefit from this answer – timur Dec 10 '18 at 03:02
  • Does anyone have a reference for why g(x) above is continuous on IR/IQ ? Does it have to do with, if you want to approximate an irrational with rationals, the denominator of those irrationals has to go to infinity (if you bound the denominator, you don't find enough numbers)? – dasWesen Nov 12 '20 at 13:40
  • 1
    @dasWesen: Basically yes, and this function even has its own Wikipedia page, with a formal proof of continuity/discontinuity: https://en.wikipedia.org/wiki/Thomae%27s_function – Lukas Geyer Nov 14 '20 at 02:39
  • @LukasGeyer Thank you! – dasWesen Nov 14 '20 at 16:27
  • @LukasGeyer The composition we get here is not a Reimann Integrable function but is a Lebesgue measurable function, Is it true in Lebesgue case? – Kalas678 Mar 12 '22 at 13:34
  • @Sushant; Compositions of Lebesgue-measurable functions need not be Lebesgue measurable. Even the composition of Lebesgue measurable and homeomorphism need not be, because a homeomorphism can map a non-measurable set into a Lebesgue null set. However, compositions of Borel-measurable functions are again Borel-measurable. – Lukas Geyer Mar 13 '22 at 21:04