The Timpson Lyceum · Real Analysis (2020)

Lecture 17 (Real Analysis 2020)

Overview

This lecture begins the theory of integration by building the Darboux (Riemann) integral from the ground up. Just as the derivative captures the slope of a function, the integral captures the area under a curve on a closed interval. The strategy is to approximate that area with rectangles, sandwich the true area between an under-estimate and an over-estimate, and then squeeze the two together. To make this precise we define partitions, lower and upper Darboux sums, refinements, and finally the lower and upper integrals — culminating in the definition of what it means for a function to be integrable.

Key Concepts

Partition. A partition $P$ of a closed interval $[a,b]$ is a finite set of points $P = \{x_0, x_1, \dots, x_n\}, \qquad a = x_0 < x_1 < \cdots < x_n = b.$ Each $x_i$ is a partition point, and each $[x_{i-1}, x_i]$ (for $1 \le i \le n$ ) is a partition interval. Dividing $[a,b]$ into $n$ subintervals requires $n+1$ points — always one more point than there are intervals.

Infimum and supremum on a subinterval. For a bounded function $f$ on $[a,b]$ and each partition interval, define

$m_i = \inf\{ f(x) : x \in [x_{i-1}, x_i] \}$ — the greatest lower bound (smallest height) on that interval,
$M_i = \sup\{ f(x) : x \in [x_{i-1}, x_i] \}$ — the least upper bound (largest height) on that interval.

By definition $m_i \le M_i$ on every interval. These need not occur at the endpoints — they can be attained anywhere inside the subinterval.

Lower and upper Darboux sums. Stacking rectangles whose heights are the min (resp. max) on each subinterval: $L(f, P) = \sum_{i=1}^{n} m_i\,(x_i - x_{i-1}), \qquad U(f, P) = \sum_{i=1}^{n} M_i\,(x_i - x_{i-1}).$ The lower sum uses rectangles that "barely reach" the curve; the upper sum uses rectangles that "barely clear" it. Since $m_i \le M_i$ , we always have $L(f,P) \le U(f,P)$ , with equality precisely for a constant function. Intuitively the true area lies between them.

Telescoping of widths. For any partition, the subinterval widths add up to the whole interval: $\sum_{i=1}^{n} (x_i - x_{i-1}) = b - a.$

Coarse bounds (a lemma). If $m \le f(x) \le M$ for all $x \in [a,b]$ (global bounds, no subscript), then for every partition $P$ : $m(b-a) \le L(f,P) \le U(f,P) \le M(b-a).$ Every lower sum is at least the single under-estimating rectangle, and every upper sum is at most the single over-estimating rectangle.

Refinement. A partition $P^*$ is a refinement of $P$ if $P \subseteq P^*$ — it keeps all of $P$ 's points and adds more. Refining can only sharpen the estimates: $L(f,P) \le L(f,P^*), \qquad U(f,P^*) \le U(f,P).$ Adding points pushes the lower sum up toward the true area and the upper sum down toward it.

Any lower sum ≤ any upper sum. For any two partitions $P_1, P_2$ (related or not), $L(f, P_1) \le U(f, P_2).$ Proof idea: let $P^* = P_1 \cup P_2$ , a common refinement of both. Then $L(f,P_1) \le L(f,P^*) \le U(f,P^*) \le U(f,P_2)$ .

Lower and upper integrals. Take the best possible estimates over all partitions: $\underline{\int_a^b} f = \sup_P L(f,P), \qquad \overline{\int_a^b} f = \inf_P U(f,P).$ The lower integral is the supremum (least upper bound) of all lower sums; the upper integral is the infimum (greatest lower bound) of all upper sums. Because every lower sum is below every upper sum, it always follows that $\underline{\int_a^b} f \;\le\; \overline{\int_a^b} f.$

Integrability. A bounded $f$ is integrable on $[a,b]$ exactly when these two numbers agree, and then the integral is their common value: $\int_a^b f = \underline{\int_a^b} f = \overline{\int_a^b} f.$ A constant function $f(x) = c$ is integrable with $\int_a^b c = c(b-a)$ . The Dirichlet function on $[0,1]$ — $f(x) = 0$ for rational $x$ , $f(x) = 1$ for irrational $x$ — is not integrable: since rationals and irrationals are both dense, every subinterval has $m_i = 0$ and $M_i = 1$ , so $\underline{\int_0^1} f = 0$ while $\overline{\int_0^1} f = 1$ .

Worked Example

Compute $\displaystyle \int_0^2 x\,dx$ from the definition by splitting $[0,2]$ into $n$ equal pieces and taking the upper sum, then letting $n \to \infty$ . Each subinterval has width $\frac{2}{n}$ , the right endpoint of the $i$ -th rectangle is $i \cdot \frac{2}{n}$ , and its height is $f\!\left(i\cdot\tfrac{2}{n}\right) = \tfrac{2i}{n}$ . Using $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ :

\begin{aligned} \int_0^2 x\,dx &= \lim_{n\to\infty}\; \sum_{i=1}^{n} f\!\left(\tfrac{2i}{n}\right)\cdot \tfrac{2}{n} = \lim_{n\to\infty}\; \frac{2}{n}\sum_{i=1}^{n} \frac{2i}{n} \\ &= \lim_{n\to\infty}\; \frac{4}{n^2}\sum_{i=1}^{n} i = \lim_{n\to\infty}\; \frac{4}{n^2}\cdot \frac{n(n+1)}{2} \\ &= \lim_{n\to\infty}\; \frac{4n^2 + 4n}{2n^2} = \lim_{n\to\infty}\left( 2 + \frac{2}{n} \right) = 2. \end{aligned}

This matches the geometric area of the triangle under $y = x$ from $0$ to $2$ : $\tfrac{1}{2}(\text{base})(\text{height}) = \tfrac{1}{2}(2)(2) = 2$ .

Summary

The integral is the area under a curve, approximated by rectangles and obtained as a limit — the integration analogue of the derivative-as-slope idea.
A partition chops $[a,b]$ into subintervals; on each, $m_i$ and $M_i$ are the inf and sup of $f$ , giving the lower sum $L(f,P)$ and upper sum $U(f,P)$ , with $L \le U$ always.
Refining a partition raises lower sums and lowers upper sums, and any lower sum is $\le$ any upper sum (compare via a common refinement).
The lower integral is $\sup_P L(f,P)$ and the upper integral is $\inf_P U(f,P)$ ; the lower integral never exceeds the upper integral.
A function is integrable iff the lower and upper integrals coincide; their common value is $\int_a^b f$ . Constant functions are integrable; the Dirichlet function is the classic non-integrable example.
Computing an integral directly means evaluating $\lim_{n\to\infty}$ of a Darboux sum, which relies on summation formulas such as $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ .