Discussion
Suppose \(f(x)\) is defined on [0, 1] as follows :
\[ f(x) = \begin{cases} 0, & \text{if } x \text{ irrational} \\ 1/q, & \text{if } x \text{ rational} \end{cases} \]
and \(x = \dfrac{p}{q}\) with \((p, q) = 1\). Then \(f\) is :
Answer: A
Thomae's function is Riemann integrable (its integral is 0).
For any \(\varepsilon \gt 0\), only finitely many rationals \(p/q\) in [0, 1] have \(1/q \ge \varepsilon\). Cover them by intervals of tiny total length; on the rest of [0, 1] the function is below \(\varepsilon\). So upper sums can be made less than \(2\varepsilon\), while every lower sum is 0 because irrationals lie in every subinterval.
The function is continuous at every irrational point (values near an irrational have large denominators, so they are small) and discontinuous at every rational point (where \(f = 1/q \gt 0\) but irrationals arbitrarily close give 0). That rules out (B), (C) and (D).
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