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The Dirichlet function on $[0, 1]$ has lower Darboux integral $0$ and upper Darboux integral $1$ , so it is bounded and not Riemann integrable

Statement refuted

Refuted: that every bounded function on a closed bounded interval with distinct endpoints is Riemann integrable (FALSE: every bounded function on $[a, b]$ is Riemann integrable, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

The witness is the Dirichlet function $1_{Q}$ restricted to $[0, 1]$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ). It takes only the values $0$ and $1$ , so it is bounded (Lower bound, bounded below, bounded set); every lower Darboux sum is $0$ and every upper Darboux sum is $1$ ; hence

$\underline{\int_{0}^{1}} 1_{Q} = 0 \neq 1 = \overline{\int_{0}^{1}} 1_{Q},$

and the function is not Riemann integrable. The two Darboux integrals are as far apart as the range of the function allows.

Facts & Assumptions

Given: $g : [0, 1] \to R$ with $g (x) = 1$ for rational $x$ and $g (x) = 0$ for irrational $x$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ).

[A1]

The refuted claim: every bounded function on such an interval is Riemann integrable (FALSE: every bounded function on $[a, b]$ is Riemann integrable).

[L1]

Both $Q$ and $R ∖ Q$ are dense in $R$ , so every nonempty open interval contains a rational and an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L2]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1$ , $I_{i} = [t_{i}, t_{i + 1}]$ , and $(t_{i}, t_{i + 1})$ is a nonempty open interval contained in $I_{i}$ (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

$m_{i} = \inf g [I_{i}]$ , $M_{i} = \sup g [I_{i}]$ , $L (g, P) = \sum_{i < n} m_{i} Δ_{i}$ , $U (g, P) = \sum_{i < n} M_{i} Δ_{i}$ ; $\underline{\int_{0}^{1}} g$ is the supremum of the lower sums and $\overline{\int_{0}^{1}} g$ the infimum of the upper sums; $g$ is integrable exactly when they agree (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L4]

A set with a least element has it as its infimum and one with a greatest element has it as its supremum; the supremum and infimum of ${c}$ are both $c$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L5]

Finite sums: scaling and $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: $0 \neq 1$ , and the order is total and transitive (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

$g$ is bounded, with $0 \leq g (x) \leq 1$ for every $x \in [0, 1]$ , so its Darboux sums and integrals are defined by [L3].

givenL3

1.2

Let $P = (n, t)$ be any partition of $[0, 1]$ and $i < n$ . By [L2] the interval $(t_{i}, t_{i + 1})$ is nonempty and open, so by [L1] it contains a rational and an irrational, both lying in $I_{i}$ . Hence $g [I_{i}] = {0, 1}$ and, by [L4], $m_{i} = 0$ and $M_{i} = 1$ .

givenL1L2L4

2.1

Therefore $L (g, P) = \sum_{i < n} 0 \cdot Δ_{i} = 0$ and $U (g, P) = \sum_{i < n} 1 \cdot Δ_{i} = \sum_{i < n} Δ_{i} = 1$ , for every partition $P$ of $[0, 1]$ , by [L3], [L5] and [L2].

step 1.2L2L3L5

3.1

The set of lower sums is ${0}$ and the set of upper sums is ${1}$ , so $\underline{\int_{0}^{1}} g = 0$ and $\overline{\int_{0}^{1}} g = 1$ by [L4] and [L3]. Since $0 \neq 1$ by [L6], $g$ is not Riemann integrable on $[0, 1]$ .

step 2.1L3L4L6

4.1

So $g$ is bounded on $[0, 1]$ , an interval with $0 < 1$ , and is not Riemann integrable: [A1] is refuted.

step 1.1step 3.1A1∎

Remarks

The failure is uniform over partitions. No partition does better than any other here: $U (g, P) - L (g, P) = 1$ for every $P$ , so Riemann's criterion (Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ) fails as badly as it can. Contrast the Cantor-set indicator, where the gap does go to $0$ (The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral $0$ even though it is discontinuous at uncountably many points).
The Lebesgue criterion gives the same verdict for the same reason. $g$ is discontinuous at every point of $[0, 1]$ (The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ ), and $[0, 1]$ is not null (A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero), so Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero refuses it too. The direct computation above is kept because it is elementary and costs no choice principle.
The Riemann sums are not even a warning. Every uniform partition with rational tags gives Riemann sum $1$ , so along that one sequence of tagged partitions the sums converge; this is precisely why the Riemann definition quantifies over all tagged partitions of small mesh (For the Dirichlet function every uniform partition with rational tags gives Riemann sum $1$ , so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence).

Depends on

Used by

For the Dirichlet function every uniform partition with rational tags gives Riemann sum 1, so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence Counterexample

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Sources

Dirichlet function (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
J. Hunter, Chapter 11: The Riemann Integral (standard reference, not scraped)
MIT 18.013A, Nonintegrable Functions (standard reference, not scraped)

The Dirichlet function on [0,1][0,1] has lower Darboux integral 00 and upper Darboux integral 11, so it is bounded and not Riemann integrable