Use the definition of the Riemann integral to prove that if \(f\) is bounded on \([a,b]\) and is continuous everywhere except for finitely many points in \((a,b)\), then \(f\in\mathscr{R}\) on \([a,b]\).
Solution
1. Setup: \(f\) is bounded, say \(|f| \le M\), and discontinuous only at finitely many points \(c_1 < \cdots < c_k\) in \((a,b)\).
hypothesis.
2. Given \(\eps > 0\), cover the discontinuity points by open intervals \(J_1, \ldots, J_k\) of total length \(< \frac{\eps}{4M}\).
possible since there are finitely many of them.
3. On the compact set \(K = [a,b] \setminus \bigcup_i J_i\), \(f\) is uniformly continuous.
\(K\) is closed and bounded, hence compact, and \(f\) is continuous on \(K\) (all discontinuities lie in the removed open intervals).
4. Choose \(\delta > 0\) such that \(|f(x) - f(y)| < \frac{\eps}{2(b-a)}\) whenever \(x, y \in K\) with \(|x - y| < \delta\).
uniform continuity on \(K\) (<1>3).
5. Build a partition \(P\) containing the endpoints of the \(J_i\) and with every sub-interval that lies inside \(K\) of length \(< \delta\); then \(U(P) - L(P) < \eps\).
5.1. The sub-intervals inside \(K\) contribute at most \(\frac{\eps}{2(b-a)}\sum \Delta x \le \eps/2\).
on each such sub-interval the oscillation is \(< \frac{\eps}{2(b-a)}\) by 4, and the total length of the good sub-intervals is \(\le b - a\).
5.2. The sub-intervals overlapping the bad set contribute at most \(2M \cdot \frac{\eps}{4M} = \eps/2\).
on each, the oscillation is \(\le 2M\) (\(f\) bounded by \(M\)), and their total length is the total length of the \(J_i\), \(< \eps/(4M)\).
5.3. Q.E.D.
1 and 2 give \(U(P) - L(P) < \eps/2 + \eps/2 = \eps\).
6. Q.E.D.
by the Riemann criterion, \(U(P) - L(P) \to 0\) over fine partitions (<1>5 for every \(\eps\)) implies \(f\) is Riemann integrable on \([a,b]\).