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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

Statement refuted

Refuted: that a bounded function on $[a,b]$ is Riemann integrable whenever its set of discontinuities is nowhere dense (FALSE: a bounded function on $[a,b]$ is Riemann integrable exactly when its set of discontinuities is nowhere dense, Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ ).

The witness is the indicator $g$ of the Smith-Volterra-Cantor set $S \subseteq [0,1]$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$ , an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals). Its set of discontinuities is exactly $S$ , which is closed, nowhere dense and not of measure zero (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero); and $g$ is not Riemann integrable, with

$\underline{\int_0^1} g \;=\; 0, \qquad \overline{\int_0^1} g \;\ge\; \tfrac12 .$

The contrast with the Cantor set is the whole point. The Cantor set is also closed and nowhere dense, and its indicator is integrable, with integral $0$ (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 two sets differ only in measure, and that is what decides (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 proof below is direct, from claim 4 of The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero; it does not go 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 and so uses no choice principle.

Facts & Assumptions

Given: The Smith-Volterra-Cantor set $S \subseteq [0,1]$ and its indicator $g : [0,1] \to \mathbb{R}$ , with $g(x) = 1$ for $x \in S$ and $g(x) = 0$ otherwise.

[A1]

The refuted claim: a bounded function on a closed bounded interval with distinct endpoints whose set of discontinuities is nowhere dense is Riemann integrable.

[L1]

$S$ is closed, bounded and nowhere dense, so $S$ contains no nonempty open set; and if $(a_k)$ , $(b_k)$ are sequences of reals with $a_k \le b_k$ , $S \subseteq \bigcup_k[a_k,b_k]$ and $\sum_{k<i}(b_k-a_k) \le M$ for every $i \in \mathbb{N}$ , then $M \ge 2^{-1}$ (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 $n \ge 1$ , an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals, Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ , Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

[L2]

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

[L3]

For a partition $P = (n,t)$ of $[0,1]$ : $n \ge 1$ , $t_i < t_{i+1}$ , $\Delta_i > 0$ , $I_i = [t_i,t_{i+1}]$ , $[0,1] = \bigcup_{i<n}I_i$ , and $(t_i,t_{i+1})$ is a nonempty open subset of $[0,1]$ (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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L4]

$m_i = \inf g[I_i]$ , $M_i = \sup g[I_i]$ , $L(g,P) = \sum_{i<n}m_i\Delta_i$ , $U(g,P) = \sum_{i<n}M_i\Delta_i$ ; $\underline{\int_0^1}g$ 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$ , 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 $\int_a^b f$ , Lower bound, bounded below, bounded set).

[L5]

A set with a least element has it as its infimum and one with a greatest element has it as its supremum; an infimum of a set all of whose members are $\ge c$ is $\ge c$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L6]

Finite sums: scaling, splitting, monotonicity in the terms, $\sum_{i<n}0 = 0$ ; and a finite list of closed intervals extends 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 $\varepsilon$ ) and content zero (a finite such cover)).

[L7]

Ordered-field arithmetic: for $0 \le x \le 1$ and a real $\rho > 0$ the reals $u := \max\{0,x-\rho\}$ and $v := \min\{1,x+\rho\}$ satisfy $u < v$ and $(u,v) \subseteq N_\rho(x)\cap[0,1]$ ; and $0 < 2^{-1} < 1$ (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 $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ , Intervals of $\mathbb{R}$ : 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.

Counterexample

technique · direct

1.1

$g$ takes only the values $0$ and $1$ , so it is bounded on $[0,1]$ and its Darboux sums and integrals are defined by [L4].

givenL4

1.2

Discontinuity on $S$ . Let $x \in S$ , so $g(x) = 1$ , and let a real $\rho > 0$ be given. By [L7] the interval $(u,v) \subseteq N_\rho(x)\cap[0,1]$ is nonempty and open, so by [L1] it contains a point $y \notin S$ ; then $|y-x| < \rho$ , $y \in [0,1]$ and $|g(x)-g(y)| = 1$ , so continuity fails at $x$ for $\varepsilon := 1$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind).

givenL1L7

1.3

Continuity off $S$ . Let $x \in [0,1]$ with $x \notin S$ . Since $S$ is closed, [L2] gives a real $\rho > 0$ with $N_\rho(x)\cap S = \varnothing$ , so $g$ vanishes on $N_\rho(x)\cap[0,1]$ and continuity holds at $x$ .

givenL1L2

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 lower sum is $0$ . Let $P = (n,t)$ be a partition of $[0,1]$ and $i < n$ . By [L3] the interval $(t_i,t_{i+1})$ is a nonempty open subset of $[0,1]$ , so by [L1] it contains a point outside $S$ , at which $g$ takes the value $0$ ; as $g \ge 0$ , that value is the least element of $g[I_i]$ and $m_i = 0$ by [L5]. Hence $L(g,P) = 0$ by [L4] and [L6].

step 1.1L1L3L4L5L6

2.3

Every upper sum is at least $2^{-1}$ . With $P$ as above put $B := \{\, i < n : I_i \cap S \ne \varnothing \,\}$ . For $i \in B$ the set $g[I_i]$ contains $1$ , so $M_i = 1$ by [L5]; for $i \notin B$ one has $g[I_i] = \{0\}$ and $M_i = 0$ . Hence $U(g,P)$ is the sum of the $\Delta_i$ with $i \in B$ , by [L4] and [L6].

step 1.1L4L5L6

3.1

The intervals $I_i$ with $i \in B$ cover $S$ , since $S \subseteq [0,1] = \bigcup_{i<n}I_i$ by [L3]. Extending that finite list to a sequence by degenerate intervals $[0,0]$ ([L6]) gives a cover of $S$ all of whose partial total lengths are at most $U(g,P)$ , so [L1] gives $U(g,P) \ge 2^{-1}$ .

step 2.3L1L3L6

4.1

By [L5] and step 2.2, $\underline{\int_0^1}g = 0$ ; by [L5] and step 3.1, $\overline{\int_0^1}g \ge 2^{-1} > 0$ . The two differ, so $g$ is not Riemann integrable by [L4].

step 2.2step 3.1L4L5L7

5.1

So $g$ is bounded on $[0,1]$ , an interval with $0 < 1$ , its set of discontinuities is nowhere dense by step 2.1, and it is not Riemann integrable: [A1] is refuted.

step 2.1step 4.1A1∎

Remarks

Nowhere dense and null are independent, and only the second matters here. $S$ is nowhere dense and not null; $\mathbb{Q}\cap[0,1]$ is null and dense. Thomae's function has the second as its discontinuity set and is integrable (Thomae's function is Riemann integrable on $[0,1]$ with integral $0$ : it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$ ); $g$ has the first and is not. Neither notion of smallness implies the other, and 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 names the one that decides.
The constant $2^{-1}$ is what this library can state, and it is enough. No outer measure is defined here, so "the measure of $S$ is $1/2$ " is not a statement available; claim 4 of The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero gives the quantitative form actually used, that no interval cover of $S$ has total length below $2^{-1}$ . The upper integral is therefore at least $2^{-1}$ ; whether it equals $2^{-1}$ is not asserted.
What the argument uses about $S$ , and nothing more. Only three properties enter: $S$ is closed, $S$ has empty interior, and no interval cover of $S$ has total length below $2^{-1}$ . Any set with those three properties would serve as a witness in exactly the same way, and the argument is written so that the particular construction of The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$ , an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals is used only through The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero.

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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)