Let \(f\) be a Riemann integrable function on a closed interval \([a, b]\). For \(x \in [a, b]\), let \(g(x) = \int_a^x f(t)\,dt\). Prove that \(g\) is continuous.
Solution
1. Since \(f\) is Riemann integrable on \([a,b]\), it is bounded: there is \(M\) with \(|f(t)| \le M\) for all \(t \in [a,b]\).
Riemann integrable functions are bounded.
2. For \(x, y \in [a,b]\),\[|g(x) - g(y)| = \left| \int_a^x f - \int_a^y f \right| = \left| \int_y^x f(t)\,dt \right| \le M |x - y|.\]
1 and the triangle inequality for integrals.
3. Hence \(g\) is Lipschitz continuous (with constant \(M\)).
2.
4. Therefore \(g\) is continuous.
3 (Lipschitz implies continuous).
5. Q.E.D.
4.