TIFR 2013 problem 13 | Riemann integrable functions

Try this true/false problem from TIFR 2013 Problem 13 based on composite of Riemann integrable functions.

Question: TIFR 2013 problem 13

True/False?

Let \(f:[a,b]\to[c,d]\) and \(g:[c,d]\to\mathbb{R}\) be Riemann integrable  functions. Then the composite \(gof\) is also Riemann integrable.

Hint: Can the Thomae function(which is known to be integrable) be composed with some function to give the characteristic function of rational numbers (which is known to be not integrable in the Riemann sense)?

Discussion:

Let \(f:[0,1]\to[0,1]\) be the Thomae function, \(f(x)=\begin{cases} 1/q & \text{ for }x=p/q & \text{where}& gcd(p,q)=1 \\ 0 & \text{ for } x \notin \mathbb{Q} \end{cases}\).

\(f\) is known to be Riemann integrable over [0,1] (it is continuous at all irrationals and therefore set of discontinuity is countable).

We want a function \(g:[0,1]\to\mathbb{R}\) such that, for \(1/q\) it gives value \(1\) and for \(0\) it gives value \(0\). Then we will end up with the \(h(x)=\begin{cases} 0 & \text{ for }x\in\mathbb{Q}  \\ 1 & \text{ for } x \notin \mathbb{Q} \end{cases}\). Moreover, g should be Riemann integrable.

Let \(g(x) = \begin{cases} 1 & \text{ for }x \ne 0 \\ 0 & \text{ for } x = 0 \end{cases}\)

Then g is Riemann integrable since it is discontinuous at only 1 point. And as desired, \(gof=h\). \(h\) is well known to be not Riemann integrable. This proves the fact that the statement is False.

Some Useful Links:

• TIFR 2014 Problem 22
• Fundamental Theorem of Abelian Groups – Video