UPSC Maths 2025 Paper 2 Q4b — Solution

15 marks · Section A

Question

Prove that every continuous function is Riemann integrable.

Technique

Use Riemann’s criterion (upper sum − lower sum $<\varepsilon$ ), powered by uniform continuity of a continuous function on a closed bounded interval (the Heine–Cantor theorem).

Solution

Claim. If $f:[a,b]\to\mathbb{R}$ is continuous, then $f$ is Riemann integrable on $[a,b]$ .

Preliminaries. For a partition $P=\{a=x_0<x_1<\cdots<x_n=b\}$ , with $\Delta x_i = x_i-x_{i-1}$ , set on each subinterval $M_i = \sup_{[x_{i-1},x_i]} f,\qquad m_i = \inf_{[x_{i-1},x_i]} f,$ and the upper/lower Darboux sums $U(P,f) = \sum_{i=1}^n M_i\,\Delta x_i,\qquad L(P,f) = \sum_{i=1}^n m_i\,\Delta x_i.$ Riemann’s criterion: $f$ is integrable iff for every $\varepsilon>0$ there is a partition $P$ with $U(P,f) - L(P,f) < \varepsilon.$

Step 1 — $f$ is bounded. A continuous function on the closed bounded interval $[a,b]$ is bounded (extreme value theorem). Hence $M_i,m_i$ are finite and are actually attained (max/min on each compact subinterval).

Step 2 — $f$ is uniformly continuous (Heine–Cantor). Since $[a,b]$ is compact and $f$ is continuous, $f$ is uniformly continuous: for every $\varepsilon>0$ there exists $\delta>0$ such that $|s-t|<\delta \;\Longrightarrow\; |f(s)-f(t)| < \frac{\varepsilon}{b-a}\qquad \text{for all } s,t\in[a,b].$

Step 3 — Choose a fine partition. Let $\varepsilon>0$ . Take $\delta$ from Step 2, and choose any partition $P$ with mesh $\|P\| = \max_i \Delta x_i < \delta$ . On each subinterval $[x_{i-1},x_i]$ of length $<\delta$ , $f$ attains its max at some $\xi_i$ and min at some $\eta_i$ , with $|\xi_i-\eta_i|\le\Delta x_i<\delta$ , so by uniform continuity $M_i - m_i = f(\xi_i) - f(\eta_i) < \frac{\varepsilon}{b-a}.$

Step 4 — Bound $U-L$ . $U(P,f) - L(P,f) = \sum_{i=1}^n (M_i - m_i)\,\Delta x_i < \frac{\varepsilon}{b-a}\sum_{i=1}^n \Delta x_i = \frac{\varepsilon}{b-a}\,(b-a) = \varepsilon.$

Step 5 — Conclusion. For every $\varepsilon>0$ we produced a partition $P$ with $U(P,f)-L(P,f)<\varepsilon$ . By Riemann’s criterion, $f$ is Riemann integrable on $[a,b]$ . $\qquad\blacksquare$

Key point. The crux is uniform continuity: ordinary pointwise continuity gives a $\delta$ depending on the point, which is insufficient to bound all oscillations $M_i-m_i$ simultaneously. Compactness of $[a,b]$ upgrades continuity to uniform continuity.

Answer

Every continuous $f:[a,b]\to\mathbb{R}$ is Riemann integrable. Proof outline: $f$ is bounded and (by Heine–Cantor) uniformly continuous; choosing a partition of mesh $<\delta$ makes each oscillation $M_i-m_i<\varepsilon/(b-a)$ , so $U(P,f)-L(P,f)<\varepsilon$ , satisfying Riemann’s criterion.