The Timpson Lyceum · Real Analysis (2020)

Lecture 19 (Real Analysis 2020)

Overview

This lecture establishes the basic algebraic and order properties of the Riemann (Darboux) integral. None of the results are conceptually surprising — they formalize facts that are intuitive when you picture the integral as signed area — but each is proved rigorously using Archimedean sequences of partitions and the squeeze between lower and upper Darboux sums. The five results covered are: additivity over an interval (splitting at an interior point), monotonicity (order is preserved), a bounding lemma for sums of functions, linearity (additivity over functions and scaling by a constant), and the triangle inequality for integrals. Together these are the "recipe" tools you reach for whenever you manipulate integrals.

Key Concepts

Setup and tools. A function $f:[a,b]\to\mathbb{R}$ is integrable iff there exists an Archimedean sequence of partitions $(P_n)$ — a sequence whose mesh (largest subinterval width) tends to $0$, equivalently for which $\lim_{n\to\infty}\big(U(f,P_n)-L(f,P_n)\big)=0$. When such a sequence exists, $\int_a^b f = \lim_{n\to\infty} U(f,P_n) = \lim_{n\to\infty} L(f,P_n).$ Two facts are used repeatedly: refining a partition (adding points) never worsens an approximation, and refining an Archimedean sequence keeps it Archimedean (the mesh still $\to 0$).

Additivity over the interval. If $f$ is integrable on $[a,b]$ and $a<c<b$, then $f$ is integrable on $[a,c]$ and $[c,b]$, and $\int_a^b f = \int_a^c f + \int_c^b f.$ Idea: take an Archimedean sequence on $[a,b]$, insert the point $c$ into every partition (a refinement, so still Archimedean), then let $P_n', P_n''$ be the partitions induced on $[a,c]$ and $[c,b]$. Each piece's mesh still $\to 0$, so each is Archimedean on its subinterval, and the upper sum splits as $U(f,P_n)=U(f,P_n')+U(f,P_n'')$. Taking limits gives the result.

Monotonicity (order preservation). If $f,g$ are integrable on $[a,b]$ and $f(x)\le g(x)$ for all $x$, then $\int_a^b f \le \int_a^b g.$ Idea: on each subinterval let $y_i$ be the point maximizing $f$. Then $U(f,P)=\sum f(y_i)\,\Delta x_i \le \sum g(y_i)\,\Delta x_i \le U(g,P)$, since $g\ge f$ everywhere (and the second $\le$ holds because the true max of $g$ can only be larger). Pass to the limit.

Bounding lemma for sums (no integrability needed). For bounded $f,g:[a,b]\to\mathbb{R}$ and any partition $P$, $L(f,P)+L(g,P)\ \le\ L(f+g,P)\ \le\ U(f+g,P)\ \le\ U(f,P)+U(g,P).$ The inequalities can be strict. Counterexample showing strictness: on $[0,1]$ take $f(x)=1$ on rationals, $-1$ on irrationals, and $g=-f$. Then $f+g\equiv 0$ so its sums are $0$, while $L(f,P)=-1$ and $U(f,P)=+1$ on every partition. The key engine is the supremum/infimum inequality: writing $M_i(h)=\sup_{I_i}h$ and $m_i(h)=\inf_{I_i}h$, $M_i(f+g)\le M_i(f)+M_i(g),\qquad m_i(f)+m_i(g)\le m_i(f+g),$ because $M_i(f)+M_i(g)$ is an upper bound for $f+g$ on $I_i$ and $M_i(f+g)$ is the least such bound.

Linearity. If $f,g$ are integrable on $[a,b]$ and $\alpha\in\mathbb{R}$:

• Additivity: $\displaystyle\int_a^b (f+g) = \int_a^b f + \int_a^b g.$ Apply the bounding lemma with one Archimedean sequence (the same sequence works for $f$, $g$, and $f+g$ since the mesh condition depends only on the interval; note $f+g$ is bounded). The lower-sum half squeezes $\int(f+g)\ge \int f+\int g$ and the upper-sum half gives $\le$, so equality.
• Homogeneity: $\displaystyle\int_a^b \alpha f = \alpha\int_a^b f.$ For $\alpha\ge 0$, $U(\alpha f,P)=\alpha\,U(f,P)$ (factor the constant out of the sum and the limit). For $\alpha<0$ multiplication flips extremes, so $U(\alpha f,P)=\alpha\,L(f,P)$; since the upper and lower limits agree, the value is still $\alpha\int_a^b f$.

Combining the two recovers the general statement $\int_a^b(\alpha f+\beta g)=\alpha\int_a^b f+\beta\int_a^b g$.

Triangle inequality for integrals. If $f$ and $|f|$ are integrable on $[a,b]$, then $\left|\int_a^b f(x)\,dx\right| \le \int_a^b |f(x)|\,dx.$ Idea: $-|f(x)|\le f(x)\le |f(x)|$ pointwise; integrating (monotonicity) gives $-\int|f|\le \int f\le \int|f|$, which is exactly the statement $|{\int f}|\le \int|f|$ via $|A|\le B \iff -B\le A\le B$. Equality can fail: a function $+1$ on part of $[-1,1]$ and $-1$ on the rest can have $\int f=0$ while $\int|f|=2$.

Worked Example

Claim. If $f$ is integrable on $[a,b]$ and $\alpha<0$, then $\displaystyle\int_a^b \alpha f = \alpha\int_a^b f$.

State the key sum identity first (negative scalar swaps the roles of sup and inf on each subinterval), then carry it through the limit. Let $(P_n)$ be an Archimedean sequence for $f$ on $[a,b]$ — it is also one for $\alpha f$, which stays bounded.

The last step uses that $\lim U(f,P_n)=\lim L(f,P_n)=\int_a^b f$ for an Archimedean sequence, so pulling the (negative) constant out of the limit lands back on the integral of $f$.

Summary

• Integrability provides an Archimedean sequence of partitions, and the integral equals the common limit of its upper and lower Darboux sums — the engine behind every proof here.
• Additivity over an interval: for $a<c<b$, $\int_a^b f=\int_a^c f+\int_c^b f$; proved by inserting $c$ into each partition (a refinement) and splitting the sums.
• Monotonicity: $f\le g$ on $[a,b]$ implies $\int_a^b f\le\int_a^b g$.
• Bounding lemma: for bounded $f,g$, $L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P)$ — driven by $\sup(f+g)\le\sup f+\sup g$ and the dual inf inequality; strict in general (the $\pm1$ rational/irrational example).
• Linearity: $\int_a^b(\alpha f+\beta g)=\alpha\int_a^b f+\beta\int_a^b g$; for $\alpha<0$ a negative scalar swaps upper and lower sums but the value is unchanged.
• Triangle inequality: $\big|\int_a^b f\big|\le\int_a^b |f|$, from $-|f|\le f\le|f|$ and monotonicity; equality can fail when $f$ changes sign.