([Apo74, 7.12] c.f. [Rud76, p. 138 #3]) Give an example of a bounded function \(f\) and an increasing function \(\alpha\) defined on \([a,b]\) such that \(|f|\in\mathcal R(\alpha)\) but \(f\notin\mathcal R(\alpha)\).
Solution
1. The example.Proof: take \([a,b] = [0,1]\), \(\alpha(x) = x\) (increasing), and \(f(x) = 1\) if \(x\) is rational, \(f(x) = -1\) if \(x\) is irrational.
2. \(|f| \in \mathcal{R}(\alpha)\).Proof: \(|f| \equiv 1\) on \([0,1]\), a constant, hence Riemann integrable with \(\int_0^1 |f|\,dx = 1\).
3. \(f \notin \mathcal{R}(\alpha)\).Proof: on any subinterval \([x_{i-1}, x_i]\), the interval contains both rationals and irrationals (density of both in \(\mathbb{R}\)), so \(\inf f = -1\) and \(\sup f = 1\) on every subinterval.Hence for every partition \(P\), \(L(f, P) = -1\) and \(U(f, P) = 1\), so \(U(f,P) - L(f,P) = 2\) never goes to \(0\); the Riemann criterion fails and \(f \notin \mathcal{R}\).
4. Q.E.D.