02.04.03 · analysis / integration

Integrability of continuous functions on [a,b]

shipped3 tiersLean: none

Anchor (Master): Heine 1872 J. Reine Angew. Math. 74 (Heine-Cantor uniform continuity); Lebesgue 1901 C. R. Acad. Sci. Paris 132 (measure-zero criterion); Rudin Real and Complex Analysis Ch. 2

Intuition [Beginner]

A continuous curve has no jumps and no breaks. When you draw it without lifting your pen, every intermediate height gets visited. This smoothness means that the tallest and shortest rectangles in any strip stay close to each other, because the function cannot leap across a gap.

When you slice the interval into thin strips and build upper and lower rectangle stacks, the gap between overestimate and underestimate depends on how much the function wiggles inside each strip. For a continuous function, the wiggle shrinks as the strips get thinner.

This concept exists because continuity is the most natural condition that guarantees the area is well-defined. Any function you can draw as an unbroken curve has a definite area under it.

Visual [Beginner]

The picture shows a smooth curve drawn without lifting the pen over the interval from $0$ to $3$ . Two sets of rectangles sit on this interval. The dark rectangles (upper) touch the highest point of the curve in each strip. The light rectangles (lower) touch the lowest point. Because the curve has no jumps, the dark and light rectangles are nearly the same height in every thin strip.

As the strips become thinner, the tallest and shortest rectangles in each strip converge to the same height, squeezing both estimates toward the true area.

Worked example [Beginner]

Consider $f (x) = x^{2} + 1$ on the interval from $0$ to $2$ . Divide into $4$ equal strips of width $1/2$ .

Step 1. The largest value in each strip: $f (1/2) = 5/4$ , $f (1) = 2$ , $f (3/2) = 13/4$ , $f (2) = 5$ . Upper total is $(5/4 + 2 + 13/4 + 5) \times 1/2 = 55/8$ .

Step 2. The smallest value in each strip: $f (0) = 1$ , $f (1/2) = 5/4$ , $f (1) = 2$ , $f (3/2) = 13/4$ . Lower total is $(1 + 5/4 + 2 + 13/4) \times 1/2 = 35/8$ .

Step 3. The gap is $55/8 - 35/8 = 20/8 = 2.5$ . With $8$ strips (width $1/4$ ) the gap halves to approximately $1.25$ . The exact area is $14/3 \approx 4.667$ .

What this tells us: the continuous function has no jumps, so the gap between upper and lower estimates shrinks predictably as strips get thinner.

Check your understanding [Beginner]

Exercise (easy, multiple choice).

Why does a continuous function on a closed interval always have a well-defined area?

A. Continuous functions are always increasing B. The gap between upper and lower rectangle estimates shrinks to zero as strips get thinner C. Continuous functions are always positive D. The area is always zero for continuous functions

Hint

Think about what happens to the tallest and shortest rectangles in each strip when the function has no jumps.

Answer

B. Continuity prevents jumps, so as strips get thinner the tallest and shortest rectangles in each strip converge to the same height. This forces the gap between upper and lower estimates to zero. Feedback-correct: this is the geometric content of the theorem. Feedback-wrong: continuous functions need not be monotone, can be negative, and generally have non-zero area.

Formal definition [Intermediate+]

A function $f : [a, b] \to R$ is uniformly continuous on $[a, b]$ if for every $ϵ > 0$ there exists $δ > 0$ such that $∣ f (x) - f (y) ∣ < ϵ$ whenever $∣ x - y ∣ < δ$ and $x, y \in [a, b]$ .

Uniform continuity differs from pointwise continuity: the same $δ$ works simultaneously for every point in the domain, rather than depending on the point.

Definition (Measure zero). A subset $E \subset R$ has Lebesgue measure zero if for every $ϵ > 0$ there exists a countable collection of open intervals ${(a_{k}, b_{k})}$ such that $E \subset ⋃_{k} (a_{k}, b_{k})$ and $\sum_{k} (b_{k} - a_{k}) < ϵ$ .

Counterexamples to common slips [Intermediate+]

Confusing pointwise and uniform continuity. The function $f (x) = 1/ x$ is continuous at every point of $(0, 1)$ but not uniformly continuous there. Uniform continuity fails because points near $0$ require increasingly small $δ$ values.
Assuming all bounded functions are integrable. The Dirichlet function $f = 1_{Q}$ is bounded on $[0, 1]$ but not integrable. Boundedness alone does not suffice; control over oscillation is required.
Conflating "measure zero" with "countable." Every countable set has measure zero, but some uncountable sets (such as the Cantor set) also have measure zero. Countability is sufficient but not necessary.

Key theorem with proof [Intermediate+]

Theorem (Continuous on $[a, b]$ implies integrable). If $f : [a, b] \to R$ is continuous, then $f$ is Riemann-Darboux integrable on $[a, b]$ .

Proof. Since $[a, b]$ is a closed bounded interval and $f$ is continuous, the Heine-Cantor theorem 02.01.02 guarantees that $f$ is uniformly continuous on $[a, b]$ .

Given $ϵ > 0$ , uniform continuity provides $δ > 0$ such that $∣ f (x) - f (y) ∣ < ϵ / (b - a)$ whenever $∣ x - y ∣ < δ$ .

Choose a partition $P = {x_{0}, x_{1}, \dots, x_{n}}$ of $[a, b]$ with mesh $∥ P ∥ < δ$ . On each subinterval $[x_{i - 1}, x_{i}]$ , the extreme value theorem gives points $c_{i}, d_{i} \in [x_{i - 1}, x_{i}]$ with $f (c_{i}) = M_{i} = sup_{[x_{i - 1}, x_{i}]} f$ and $f (d_{i}) = m_{i} = in f_{[x_{i - 1}, x_{i}]} f$ . Since $∣ c_{i} - d_{i} ∣ \leq ∥ P ∥ < δ$ , uniform continuity gives:

$M_{i} - m_{i} = f (c_{i}) - f (d_{i}) < \frac{ϵ}{b - a} .$

Therefore:

$U (f, P) - L (f, P) = i = 1 \sum n (M_{i} - m_{i}) Δ x_{i} < \frac{ϵ}{b - a} i = 1 \sum n Δ x_{i} = ϵ .$

By the Darboux integrability criterion 02.04.01, $f$ is integrable. $□$

Bridge. The Heine-Cantor theorem is the foundational reason that continuity on a compact interval upgrades to uniform continuity, and this is exactly the mechanism that makes the gap $U (f, P) - L (f, P)$ shrink. The central insight is that compactness transforms a pointwise property (continuity) into a global one (uniform continuity), and the bridge is from the topology of the domain to the existence of the integral. This result builds toward 02.04.04 where the fundamental theorems of calculus use integrability of continuous functions as their starting hypothesis, and appears again in the Lebesgue criterion below, which identifies the precise class of discontinuities that integrability tolerates.

Exercises [Intermediate+]

Exercise 4 (medium, multiple choice).

Which step in the proof that continuous functions on $[a, b]$ are integrable uses the compactness of $[a, b]$ ?

A. The extreme value theorem (to guarantee $M_{i}$ and $m_{i}$ are attained) B. The Heine-Cantor theorem (to upgrade continuity to uniform continuity) C. The construction of the partition $P$ D. Both A and B

Hint

Compactness of $[a, b]$ is used in two places: the extreme value theorem requires a closed bounded domain, and the Heine-Cantor theorem requires a compact domain.

Answer

D. The extreme value theorem requires $[a, b]$ to be closed and bounded (compact), so that $M_{i}$ and $m_{i}$ are actually attained. The Heine-Cantor theorem requires the domain to be compact to upgrade pointwise continuity to uniform continuity. Both results fail on non-compact domains.

Exercise 7 (hard, symbolic).

Let $f : [a, b] \to R$ be integrable with $\int_{a}^{b} f (x) d x = 0$ and $f (x) \geq 0$ for all $x \in [a, b]$ . Prove that $f (x) = 0$ at every point of continuity of $f$ .

Hint

Suppose $f (c) > 0$ at a point $c$ where $f$ is continuous. Use continuity to find a small interval around $c$ where $f$ stays positive, and show that the integral over that interval is positive.

Answer

Suppose for contradiction that $f$ is continuous at some $c \in [a, b]$ with $f (c) > 0$ . Set $ϵ = f (c) /2 > 0$ . By continuity at $c$ , there exists $δ > 0$ such that $∣ f (x) - f (c) ∣ < ϵ$ for $∣ x - c ∣ < δ$ in $[a, b]$ . On the interval $I = [c - δ, c + δ] \cap [a, b]$ , we have $f (x) > f (c) - ϵ = f (c) /2 > 0$ . Since $f \geq 0$ on all of $[a, b]$ and $f > f (c) /2$ on $I$ , the integral satisfies:

$\int_{a}^{b} f (x) d x \geq \int_{I} f (x) d x \geq \frac{f ( c )}{2} \cdot ∣ I ∣ > 0.$

This contradicts the hypothesis $\int_{a}^{b} f = 0$ . Therefore $f (c) = 0$ at every point of continuity.

Advanced results [Master]

Theorem 1 (Lebesgue's criterion for Riemann integrability). A bounded function $f : [a, b] \to R$ is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero.

The forward direction proceeds by covering the oscillation set with small intervals. The reverse direction applies the Darboux criterion on the continuity set and bounds the contribution from the discontinuity set.

Theorem 2 (Composition preserves integrability). If $f : [a, b] \to [c, d]$ is integrable and $g : [c, d] \to R$ is continuous, then $g \circ f : [a, b] \to R$ is integrable.

The proof uses that $g$ is uniformly continuous on $[c, d]$ , so the oscillation of $g \circ f$ on any subinterval is controlled by the oscillation of $f$ . In particular, $∣ f ∣^{p}$ , $e^{f}$ , and $sin f$ are all integrable when $f$ is.

Theorem 3 (Product of integrable functions). If $f$ and $g$ are integrable on $[a, b]$ , then $f \cdot g$ is integrable.

This follows from the identity $f g = \frac{1}{4} [(f + g)^{2} - (f - g)^{2}]$ and the fact that the square of an integrable function is integrable (by Theorem 2 with $g (x) = x^{2}$ ).

Theorem 4 (Oscillation characterisation). The oscillation of $f$ at a point $x_{0}$ is $ω_{f} (x_{0}) = lim_{δ \to 0} [sup_{∣ x - x_{0} ∣ < δ} f (x) - in f_{∣ x - x_{0} ∣ < δ} f (x)]$ . Then $f$ is continuous at $x_{0}$ if and only if $ω_{f} (x_{0}) = 0$ , and $f$ is integrable on $[a, b]$ if and only if for every $ϵ > 0$ the set ${x \in [a, b] : ω_{f} (x) \geq ϵ}$ has measure zero.

This reframes Lebesgue's criterion in terms of the oscillation function, making the connection to the Darboux criterion direct: the partition construction controls the total oscillation by separating small-oscillation regions from large-oscillation regions.

Theorem 5 (Interchange of limit and integral). If $f_{n} : [a, b] \to R$ is a sequence of integrable functions converging uniformly to $f$ , then $f$ is integrable and $\int_{a}^{b} f = lim_{n \to \infty} \int_{a}^{b} f_{n}$ .

Uniform convergence guarantees that the gap $U (f_{n}, P) - L (f_{n}, P)$ carries over to $f$ in the limit. This is the first of the limit-interchange theorems in integration theory; the Lebesgue dominated convergence theorem 02.11.04 generalises it to pointwise convergence with an integrable bound.

Theorem 6 (Integrable functions are dense in $L^{1}$ ). The set of Riemann integrable functions on $[a, b]$ is dense in $L^{1} ([a, b])$ : for every $f \in L^{1}$ and every $ϵ > 0$ there exists a Riemann integrable function $g$ with $\int_{a}^{b} ∣ f - g ∣ < ϵ$ .

Step functions approximate from below, and continuous functions approximate step functions by the integrability result above.

Synthesis. The foundational reason that continuous functions are integrable is that compactness upgrades continuity to uniform continuity, which in turn controls oscillation. The central insight of the Lebesgue criterion is that integrability depends not on continuity everywhere but on the size of the discontinuity set, and this is exactly the condition that identifies the Riemann integrable functions as the bounded functions whose discontinuities are sparse in the measure-theoretic sense. Putting these together, the oscillation characterisation (Theorem 4) makes the bridge between the Darboux partition condition and the Lebesgue measure condition explicit, and the pattern generalises to the Lebesgue integral, where every measurable bounded function on a finite-measure set is integrable. The composition and product theorems build toward the algebraic structure of the space of integrable functions, and this space appears again in 02.11.04 as the Banach space $L^{1} ([a, b])$ .

Full proof set [Master]

Proposition 1 (Lebesgue's criterion, reverse direction). If $f : [a, b] \to R$ is bounded and the set $D$ of discontinuities of $f$ has Lebesgue measure zero, then $f$ is Riemann integrable.

Proof. Let $∣ f (x) ∣ \leq M$ and let $ϵ > 0$ . For each $n \in N$ , set $D_{n} = {x \in [a, b] : ω_{f} (x) \geq 1/ n}$ . Then $D = ⋃_{n = 1}^{\infty} D_{n}$ , and since $D$ has measure zero, each $D_{n}$ has measure zero.

Since $D_{n_{0}}$ has measure zero (where $n_{0}$ is chosen so that $1/ n_{0} < ϵ / (2 (b - a))$ ), cover it with countably many open intervals of total length less than $ϵ / (4 M)$ . By compactness of $D_{n_{0}} \cap [a, b]$ , a finite subcover suffices; call these $J_{1}, \dots, J_{k}$ .

On $[a, b] ∖ ⋃ J_{i}$ , the oscillation $ω_{f} (x) < 1/ n_{0} < ϵ / (2 (b - a))$ . By a compactness argument, there exists $δ > 0$ such that any subinterval of $[a, b] ∖ ⋃ J_{i}$ of length less than $δ$ has oscillation less than $ϵ / (2 (b - a))$ .

Construct a partition $P$ that includes all endpoints of $J_{1}, \dots, J_{k}$ and has mesh less than $δ$ on the complement. Then:

$U (f, P) - L (f, P) = bad \sum (M_{i} - m_{i}) Δ x_{i} + good \sum (M_{i} - m_{i}) Δ x_{i} .$

The bad sum (subintervals inside $⋃ J_{i}$ ) is at most $2 M \cdot ϵ / (4 M) = ϵ /2$ . The good sum is at most $(ϵ / (2 (b - a))) \cdot (b - a) = ϵ /2$ . Total: $U - L < ϵ$ . By the Darboux criterion, $f$ is integrable. $□$

Proposition 2 (Composition preserves integrability). If $f : [a, b] \to [c, d]$ is integrable and $g : [c, d] \to R$ is continuous, then $g \circ f$ is integrable on $[a, b]$ .

Proof. Since $g$ is continuous on the compact interval $[c, d]$ , $g$ is uniformly continuous: for every $η > 0$ there exists $δ > 0$ with $∣ g (s) - g (t) ∣ < η$ whenever $∣ s - t ∣ < δ$ .

Let $ϵ > 0$ . On any subinterval $[x_{i - 1}, x_{i}]$ where $ω_{f} < δ$ , the oscillation of $g \circ f$ satisfies $ω_{g \circ f} \leq η$ , because any two values $f (x), f (y)$ in that subinterval are within $δ$ , and $g$ changes by less than $η$ over a $δ$ -range.

The set where $ω_{f} \geq δ$ has measure zero (since $f$ is integrable, by Lebesgue's criterion). On the remaining set, $ω_{g \circ f} \leq η$ . Since $g \circ f$ is bounded (by the extreme value theorem for $g$ ), the set where $g \circ f$ has oscillation $\geq η$ is contained in ${ω_{f} \geq δ}$ , which has measure zero. By Lebesgue's criterion, $g \circ f$ is integrable. $□$

Connections [Master]

Step-function integral and the Darboux integral 02.04.01. The Darboux criterion is the prerequisite framework: continuous functions are integrable because the gap $U (f, P) - L (f, P)$ vanishes under partition refinement. The integrability of continuous functions is the first substantial application of the Darboux machinery after its introduction.
Fundamental theorems of calculus 02.04.04. FTC1 requires $f$ to be continuous on $[a, b]$ to conclude that $F^{'} (x) = f (x)$ . The integrability of continuous functions is the prerequisite that guarantees $F (x) = \int_{a}^{x} f (t) d t$ is well-defined, without which FTC1 cannot be stated.
Banach spaces and bounded linear operators 02.11.04. The space $L^{1} ([a, b])$ is a Banach space, and the Riemann integrable functions form a dense subspace by Theorem 6. The completion of the Riemann integrable functions under the $L^{1}$ -norm yields the Lebesgue integrable functions, and this completion is the bridge from Riemann to Lebesgue integration.

Historical & philosophical context [Master]

Heine 1872, in a paper on the foundations of function theory ^[Heine1872], proved that a continuous function on a closed bounded interval is uniformly continuous, building on ideas from Cantor's 1870 work on trigonometric series uniqueness ^[Cantor1870]. The application of this result to prove integrability of continuous functions appears in standard form in the late nineteenth-century analysis textbooks.

Lebesgue 1901, in C. R. Acad. Sci. Paris 132 ^{[Lebesgue1901]}, established the definitive characterisation: a bounded function is Riemann integrable if and only if its discontinuity set has measure zero. This criterion explained why continuous functions are integrable (the empty set has measure zero), why monotone functions are integrable (at most countably many discontinuities, hence measure zero), and why the Dirichlet function is not (the full interval has positive measure). The measure-zero concept introduced here became the foundation for the Lebesgue integral and, through it, for modern measure theory.

Bibliography [Master]

@article{Heine1872,
  author  = {Heine, Eduard},
  title   = {Die Elemente der Functionenlehre},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {74},
  pages   = {172--188},
  year    = {1872}
}

@article{Cantor1870,
  author  = {Cantor, Georg},
  title   = {Beweis, dass eine f\"ur jeden reellen Werth von $x$ durch eine trigonometrische Reihe gegebene Funktion sich nur auf einzige Weise darstellen l\"asst},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {72},
  pages   = {139--142},
  year    = {1870}
}

@article{Lebesgue1901,
  author  = {Lebesgue, Henri},
  title   = {Sur une g\'en\'eralisation de l'int\'egrale d\'efinie},
  journal = {Comptes rendus de l'Acad\'emie des Sciences Paris},
  volume  = {132},
  pages   = {1025--1028},
  year    = {1901}
}

@book{Apostol1967,
  author    = {Apostol, Tom M.},
  title     = {Calculus, Vol. 1: One-Variable Calculus, with an Introduction to Linear Algebra},
  publisher = {Wiley},
  year      = {1967}
}

@book{Rudin1976,
  author    = {Rudin, Walter},
  title     = {Principles of Mathematical Analysis},
  publisher = {McGraw-Hill},
  edition   = {3},
  year      = {1976}
}

Prerequisites

02.04.01

Tier anchors

beginner: Strogatz Infinite Powers (smooth curves have well-defined area); 3Blue1Brown integration visualisation
intermediate: Apostol Calculus Vol. 1 Ch. 7; Rudin Principles of Mathematical Analysis Ch. 6
master: Heine 1872 J. Reine Angew. Math. 74 (Heine-Cantor uniform continuity); Lebesgue 1901 C. R. Acad. Sci. Paris 132 (measure-zero criterion); Rudin Real and Complex Analysis Ch. 2

References

TODO_REF
Apostol — Calculus Vol. 1 · Ch. 7, integrability of continuous functions
TODO_REF
Rudin — Principles of Mathematical Analysis · Ch. 6, Riemann integrability criteria
TODO_REF
Heine 1872 — Die Elemente der Functionenlehre · J. Reine Angew. Math. 74, 172-188, originator of the uniform-continuity proof for compact domains
TODO_REF
Lebesgue 1901 — Sur une generalisation de l'integrale definie · C. R. Acad. Sci. Paris 132, 1025-1028, originator of the measure-zero integrability criterion
TODO_REF
Cantor 1870 — Beweis, dass eine fur jeden reellen Wert von x durch eine trigonometrische Reihe gegebene Funktion sich nur auf einzige Weise darstellen lasst · J. Reine Angew. Math. 72, the Heine-Cantor theorem

Reviewer

TBD

Estimated time

beginner: 15m
intermediate: 35m
master: 70m