The Timpson Lyceum · Real Analysis (2020)

Lecture 20 (Real Analysis 2020)

Overview

This lecture establishes one of the central results of integration theory: every continuous function on a closed, bounded interval is integrable. We reach it in three steps. First a key lemma bounds the gap between the upper and lower Darboux sums of a partition by a single term $f(v) - f(u)$ times the length of the whole interval. Then we combine that lemma with uniform continuity to prove the main theorem. Finally we extend the result to functions that are merely bounded and continuous on the open interval, showing the integral is unaffected by the values at the two endpoints. The recurring strategy is to build a sequence of partitions whose mesh shrinks to zero and force the upper-minus-lower difference to vanish.

Key Concepts

Partition. A finite set $P = \{x_0, x_1, \dots, x_n\}$ with $a = x_0 < x_1 < \cdots < x_n = b$ , dividing $[a,b]$ into $n$ subintervals $[x_{i-1}, x_i]$ .
Per-interval extrema. On each subinterval define $m_i = \min_{[x_{i-1},x_i]} f$ and $M_i = \max_{[x_{i-1},x_i]} f$ . The Extreme Value Theorem guarantees these exist: a continuous function on a closed bounded interval attains a maximum and a minimum.
Darboux sums. The upper and lower sums are $U(f,P) = \sum_{i=1}^{n} M_i\,(x_i - x_{i-1}), \qquad L(f,P) = \sum_{i=1}^{n} m_i\,(x_i - x_{i-1}).$ The upper sum uses the tallest rectangles (above the curve), the lower sum the shortest (under the curve).
Gap / mesh of a partition. $\operatorname{gap}(P)$ is the length of the largest subinterval. A sequence of partitions $P_n$ is an Archimedean–Riemann sequence when $\lim_{n\to\infty}\operatorname{gap}(P_n) = 0$ .
Integrability criterion. $f$ is integrable on $[a,b]$ exactly when there is a sequence of partitions $P_n$ with $\lim_{n\to\infty}\big(U(f,P_n) - L(f,P_n)\big) = 0$ . The common limit of the sums is then $\int_a^b f$ .
Uniform continuity. Stronger than ordinary continuity: how much $f$ can change over an interval can be bounded uniformly, independent of where the interval sits in the domain. (Intuition: $f(x) = x^2$ on $\mathbb{R}$ is continuous but not uniformly continuous — over a fixed window of width 1 its growth is unbounded as you move right.) A continuous function on a closed bounded interval is automatically uniformly continuous, because its total variation is capped by $M - m$ . The sequential form used here: if $\lim (v_n - u_n) = 0$ , then $\lim\big(f(v_n) - f(u_n)\big) = 0$ .
Lemma (the gap bound). For continuous $f$ and any partition $P$ , there exist points $u, v$ lying in a single subinterval of $P$ such that $0 \le U(f,P) - L(f,P) \le f(v) - f(u).$ Here $v$ and $u$ are the max and min points of the subinterval where $M_i - m_i$ is largest; replacing every interval's height difference by this single largest one only enlarges the sum.
Main theorem. If $f:[a,b]\to\mathbb{R}$ is continuous, then $f$ is integrable on $[a,b]$ .
Extension to open-interval continuity. If $f:[a,b]\to\mathbb{R}$ is bounded on $[a,b]$ and continuous on the open interval $(a,b)$ , then $f$ is still integrable, and the value of $\int_a^b f$ does not depend on $f(a)$ or $f(b)$ . Intuitively, the two end rectangles have bounded height ( $\le 2M$ ) and shrinking width, so their contributions vanish.

Worked Example

Main theorem: a continuous $f$ on $[a,b]$ is integrable. Choose any partition sequence $P_n$ with $\operatorname{gap}(P_n) \to 0$ . By the lemma, for each $n$ there are points $u_n, v_n$ in one subinterval of $P_n$ with $0 \le U(f,P_n) - L(f,P_n) \le f(v_n) - f(u_n)$ . Since $u_n, v_n$ share a subinterval, $|v_n - u_n| \le \operatorname{gap}(P_n) \to 0$ , and uniform continuity gives $f(v_n) - f(u_n) \to 0$ . Squeezing the difference between $0$ and a quantity tending to $0$ forces it to $0$ , so $f$ is integrable.

The lemma's key algebraic step — factoring the common widths out of a difference of sums, then pulling the now-constant height factor out front — runs as follows:

\begin{aligned} U(f,P) - L(f,P) &= \sum_{i=1}^{n} M_i\,(x_i - x_{i-1}) - \sum_{i=1}^{n} m_i\,(x_i - x_{i-1}) \\ &= \sum_{i=1}^{n} (M_i - m_i)\,(x_i - x_{i-1}) \\ &\le \sum_{i=1}^{n} \big(f(v) - f(u)\big)\,(x_i - x_{i-1}) \\ &= \big(f(v) - f(u)\big) \sum_{i=1}^{n} (x_i - x_{i-1}) \\ &= \big(f(v) - f(u)\big)\,(b - a). \end{aligned}

The inequality uses that $f(v) - f(u) = M_j - m_j = \max_i (M_i - m_i) \ge M_i - m_i$ for every $i$ , where $j$ is the index of the worst subinterval. The final telescoping sum collapses to the total width $b - a$ .

Summary

A continuous function on a closed bounded interval is integrable — proved via a gap-bounding lemma plus uniform continuity.
The lemma bounds $U(f,P) - L(f,P)$ by $\big(f(v) - f(u)\big)(b-a)$ , where $u,v$ are the min/max points of the worst subinterval; the proof factors common widths and telescopes.
Integrability is shown by exhibiting partitions whose mesh $\to 0$ and squeezing the upper-minus-lower difference between $0$ and a vanishing bound.
Uniform continuity is the engine: shrinking mesh forces $v_n - u_n \to 0$ , hence $f(v_n) - f(u_n) \to 0$ .
The result extends to functions bounded on $[a,b]$ and continuous on $(a,b)$ : integrability survives, and $\int_a^b f$ is independent of the endpoint values, since the end rectangles' areas (height $\le 2M$ , width $\to 0$ ) vanish in the limit.
The Extreme Value Theorem underpins everything by guaranteeing the per-interval max $M_i$ and min $m_i$ exist.