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FALSE: a bounded function on $[a, b]$ is Riemann integrable exactly when its set of discontinuities is nowhere dense

Statement

False claim: a bounded function $f : [a, b] \to R$ 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 $\int_{a}^{b} f$ ) if and only if its set of discontinuities is nowhere dense (Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

The claim replaces the correct smallness condition, 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), by the smallness condition of category. The two are independent, and the implication that fails below is the one from "nowhere dense" to "integrable": the indicator of the Smith-Volterra-Cantor set (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals) is discontinuous exactly on a closed nowhere dense set and is not integrable, because that set cannot be covered by intervals of small total length.

The other implication fails too, and more cheaply: Thomae's function is integrable and its discontinuity set is $Q \cap [0, 1]$ , which is dense in $[0, 1]$ and therefore not nowhere dense. One failing direction refutes the biconditional, and the harder one is worked out below.

Facts & Assumptions

Given: The Smith-Volterra-Cantor set $S \subseteq [0, 1]$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals) and its indicator $g : [0, 1] \to R$ , with $g (x) = 1$ for $x \in S$ and $g (x) = 0$ for $x \in [0, 1] ∖ S$ .

[A1]

The false claim, in the direction used here: if a bounded $f$ on $[a, b]$ has a nowhere dense set of discontinuities, then $f$ is Riemann integrable.

[L1]

$S$ is closed and bounded and nowhere dense, and if $(a_{k})$ , $(b_{k})$ are sequences of reals with $a_{k} \leq b_{k}$ , $S \subseteq ⋃_{k} [a_{k}, b_{k}]$ and $\sum_{k < i} (b_{k} - a_{k}) \leq M$ for every $i \in N$ , then $M \geq 2^{- 1}$ ; in particular $S$ does not have measure zero (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero, The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L2]

A set $A$ is nowhere dense when the interior of $\overline{A}$ is empty; for closed $A$ this says that $A$ contains no nonempty open set, equivalently that every neighbourhood of every point contains a point outside $A$ (Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , 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, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

A set is closed exactly when its complement is open, that is, when every point outside it has a neighbourhood missing it (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L4]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $Δ_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1$ , $I_{i} = [t_{i}, t_{i + 1}]$ , and $[0, 1] = ⋃_{i < n} I_{i}$ (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).

[L5]

$L (g, P) = \sum_{i < n} m_{i} Δ_{i}$ , $U (g, P) = \sum_{i < n} M_{i} Δ_{i}$ with $m_{i} = \inf g [I_{i}]$ and $M_{i} = \sup g [I_{i}]$ ; $\underline{\int_{0}^{1}} g$ 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 $\int_{a}^{b} f$ ).

[L6]

A set with a least element has it as its infimum and one with a greatest element has it as its supremum (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L7]

Finite sums: scaling, splitting, monotonicity in the terms, and $\sum_{i < n} 0 = 0$ ; a finite list may be extended to a sequence by degenerate intervals of length $0$ without changing any partial total (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L8]

Ordered-field arithmetic: the order is total, so any two reals have a maximum and a minimum; adding a constant preserves an inequality. For $0 \leq x \leq 1$ and a real $ρ > 0$ , the reals $u : = \max {0, x - ρ}$ and $v : = \min {1, x + ρ}$ satisfy $u < v$ , by checking the four cases of which member each of the two attains, and $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ , since $z \in (u, v)$ gives $x - ρ \leq u < z < v \leq x + ρ$ and $0 \leq u < z < v \leq 1$ (Maximum and minimum of a set, 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 $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). 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

1.1

$g$ is bounded, taking only the values $0$ and $1$ , so its Darboux sums and integrals are defined by [L5].

givenL5

1.2

$g$ is discontinuous at every point of $S$ . Let $x \in S$ , so $g (x) = 1$ , and let a real $ρ > 0$ be given. Put $u : = \max {0, x - ρ}$ and $v : = \min {1, x + ρ}$ ; then $u < v$ because $0 \leq x \leq 1$ and $ρ > 0$ , and $(u, v) \subseteq N_{ρ} (x) \cap [0, 1]$ by [L8]. Since $S$ is closed and nowhere dense it contains no nonempty open set by [L1] and [L2], so there is $y \in (u, v)$ with $y \notin S$ . Then $y \in [0, 1]$ , $∣ y - x ∣ < ρ$ and $∣ g (x) - g (y) ∣ = 1$ , so the continuity condition fails at $x$ for $ε : = 1$ .

givenL1L2L6L8

1.3

$g$ is continuous at every point of $[0, 1] ∖ S$ . Let $x \in [0, 1]$ with $x \notin S$ . Since $S$ is closed, [L3] gives a real $ρ > 0$ with $N_{ρ} (x) \cap S = \emptyset$ , so $g$ vanishes identically on $N_{ρ} (x) \cap [0, 1]$ and $∣ g (y) - g (x) ∣ = 0 < ε$ there, for every $ε > 0$ .

givenL1L3

2.1

So the set of discontinuities of $g$ in $[0, 1]$ is exactly $S$ , which is nowhere dense by [L1].

step 1.2step 1.3L1

2.2

Every upper Darboux sum of $g$ is at least $2^{- 1}$ . Let $P = (n, t)$ be a partition of $[0, 1]$ and let $B : = {i < n : I_{i} \cap S \neq \emptyset}$ . For $i \in B$ the set $g [I_{i}]$ contains $1$ , so $M_{i} = 1$ by [L6] and $g \leq 1$ ; for $i \notin B$ one has $g [I_{i}] = {0}$ and $M_{i} = 0$ . Hence $U (g, P) = \sum_{i < n} M_{i} Δ_{i}$ is the sum of the $Δ_{i}$ with $i \in B$ , by [L5] and [L7].

step 1.1L5L6L7

2.3

Every lower Darboux sum of $g$ is $0$ . With $P$ as above and $i < n$ : $t_{i} < t_{i + 1}$ by [L4], so $(t_{i}, t_{i + 1})$ is a nonempty open subset of $[0, 1]$ , and by [L1] and [L2] it is not contained in $S$ ; a point of it outside $S$ lies in $I_{i} \cap [0, 1]$ and has $g$ -value $0$ , so $g [I_{i}]$ contains $0$ and $m_{i} = 0$ by [L6], $g$ being nonnegative. Hence $L (g, P) = 0$ by [L5] and [L7].

step 1.1L1L2L4L5L6L7

3.1

The intervals $I_{i}$ with $i \in B$ cover $S$ : a point of $S$ lies in $[0, 1] = ⋃_{i < n} I_{i}$ by [L4], hence in some $I_{i}$ , and that $i$ is in $B$ . Extending this finite list of closed intervals to a sequence by degenerate intervals $[0, 0]$ ([L7]) gives a cover of $S$ all of whose partial total lengths are at most $U (g, P)$ , so [L1] gives $U (g, P) \geq 2^{- 1}$ .

step 2.2L1L4L7

4.1

Therefore $\underline{\int_{0}^{1}} g = 0$ by [L6] and step 2.3, while $\overline{\int_{0}^{1}} g \geq 2^{- 1}$ by step 3.1, since every upper sum is at least $2^{- 1}$ and the infimum of such a set is at least $2^{- 1}$ . The two differ, so $g$ is not Riemann integrable by [L5].

step 3.1step 2.3L5L6

5.1

So $g$ is a bounded function on $[0, 1]$ whose set of discontinuities is nowhere dense and which is not Riemann integrable; [A1] fails at $g$ , and with it the claimed equivalence.

step 2.1step 4.1A1∎

Remarks

Where the false claim comes from. For a closed discontinuity set, being nowhere dense and being null are both ways of saying "small", and for the Cantor set they agree. They come apart exactly because a nowhere dense closed set may still swallow a fixed fraction of the length of every interval it meets, which is what the Smith-Volterra-Cantor construction arranges: it removes a middle interval of length $4^{- n}$ at stage $n$ rather than a fixed proportion (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals).
The correct statement is measure zero, in both directions. That is 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, and it explains both failures at once: $S$ is nowhere dense and not null, so $g$ is not integrable; $Q \cap [0, 1]$ is dense and null, so Thomae's function is integrable (FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero).
Nothing here uses any choice principle. The non-integrability of $g$ is proved directly from claim 4 of The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero, which is a statement about interval covers, rather than through the forward half of 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, which spends countable choice. See What this page costs in choice: Riemann's criterion, the Darboux-Riemann equivalence and integrability of a monotone function are theorems of ZF; integrability of a continuous function inherits the single use of countable choice inside Heine-Cantor; and only the forward half of the Lebesgue criterion spends countable choice, once, at the countable union of null sets.

Depends on

Used by

The indicator of the Smith-Volterra-Cantor set is discontinuous exactly on a nowhere dense set, and is not Riemann integrable, because that set does not have measure zero Counterexample

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Sources

Smith-Volterra-Cantor set (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
MAT425 Lecture Notes (Princeton University) (standard reference, not scraped)
J. Hunter, Chapter 11: The Riemann Integral (standard reference, not scraped)

FALSE: a bounded function on [a,b][a,b] is Riemann integrable exactly when its set of discontinuities is nowhere dense