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FALSE: every bounded function on $[a, b]$ is Riemann integrable

Statement

False claim: every bounded function $f : [a, b] \to R$ (Lower bound, bounded below, bounded set) is Riemann integrable on $[a, b]$ (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 $\int_{a}^{b} f$ ).

Boundedness is exactly what is needed for the Darboux sums to exist (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ ), and the claim above confuses that with their two extrema agreeing. The witness below is bounded, takes only the values $0$ and $1$ , and has lower Darboux integral $0$ and upper Darboux integral $1$ : as far apart as the values allow.

Facts & Assumptions

Given: The Dirichlet function $1_{Q}$ restricted to $[0, 1]$ , that is $g : [0, 1] \to R$ 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 false claim: every bounded function on a closed bounded interval with distinct endpoints is Riemann integrable.

[L1]

Both $Q$ and $R ∖ Q$ are dense in $R$ , and a set is dense exactly when every neighbourhood of every real meets it (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 $[0, 1]$ : $n \geq 1$ , $t_{i} < t_{i + 1}$ , $Δ_{i} = t_{i + 1} - t_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1 - 0 = 1$ , and $I_{i} = [t_{i}, t_{i + 1}]$ (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 these 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: the midpoint $(u + v) \cdot 2^{- 1}$ of $u < v$ satisfies $u < (u + v) \cdot 2^{- 1} < v$ , and $N_{ρ} (c) \subseteq (u, v)$ for $c$ the midpoint and $ρ : = (v - u) \cdot 2^{- 1}$ (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), The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ). 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.

Refutation

technique · direct

1.1givenL3

$g$ is bounded: $0 \leq g (x) \leq 1$ for every $x \in [0, 1]$ , since $g$ takes only the values $0$ and $1$ . So its Darboux sums and integrals are defined by [L3].

1.2L1L2L6

Every subinterval of every partition of $[0, 1]$ contains both a rational and an irrational. Let $P = (n, t)$ be a partition and $i < n$ ; then $t_{i} < t_{i + 1}$ by [L2], so with $c : = (t_{i} + t_{i + 1}) \cdot 2^{- 1}$ and $ρ : = (t_{i + 1} - t_{i}) \cdot 2^{- 1}$ one has $N_{ρ} (c) \subseteq (t_{i}, t_{i + 1}) \subseteq I_{i}$ by [L6], and $N_{ρ} (c)$ meets $Q$ and meets $R ∖ Q$ by [L1].

2.1step 1.2givenL4

Hence $g [I_{i}] = {0, 1}$ for every $i < n$ , so $m_{i} = 0$ and $M_{i} = 1$ by [L4].

3.1step 2.1L2L3L5

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].

4.1step 3.1L3L4

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 $0 \neq 1$ , $g$ is not Riemann integrable on $[0, 1]$ .

5.1step 1.1step 4.1A1∎

$g$ is a bounded function on $[0, 1]$ , an interval with $0 < 1$ , and it is not Riemann integrable; so [A1] fails at $g$ and the claim is false.

Remarks

What the true statement is. Boundedness is necessary and not sufficient; the exact condition is that the discontinuity set have measure zero (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). The Dirichlet function 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 the criterion refuses it for the strongest possible reason.
The gap is maximal for the values available. A function with values in ${0, 1}$ can have upper minus lower integral at most $1$ , and this one attains that. Halving the function halves the gap and changes nothing else, so no bound on the size of the failure can be extracted from boundedness alone.
The support is not what decides. Thomae's function is nonzero at exactly the same points as the Dirichlet function, namely the rationals, and differs from it only in the values it takes there; yet it is integrable, with integral $0$ (FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero). What matters is not where a function is nonzero but where it is discontinuous.

FALSE: every bounded function on $[a, b]$ is Riemann integrable

Statement

Facts & Assumptions

Refutation

Remarks

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Dependency tree · 2 levels

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FALSE: every bounded function on [a,b][a,b] is Riemann integrable

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: every bounded function on $[a, b]$ is Riemann integrable