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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

Example

Let $C \subseteq [0,1]$ be the Cantor middle-thirds set (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) and let $\mathbf{1}_C : [0,1] \to \mathbb{R}$ be its indicator, $\mathbf{1}_C(x) = 1$ for $x \in C$ and $\mathbf{1}_C(x) = 0$ otherwise. Then:

$\mathbf{1}_C$ is discontinuous at every point of $C$ and continuous at every point of $[0,1] \setminus C$ , so its set of discontinuities is exactly $C$ (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);
$\mathbf{1}_C$ is Riemann integrable on $[0,1]$ , because $C$ has measure zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, 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);
$\int_0^1 \mathbf{1}_C = 0$ .

The point of the example is claim 2 against claim 1. The discontinuity set is uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, Finite, countably infinite, countable, uncountable), so no cardinality argument such as A bounded function on $[a,b]$ whose set of discontinuities is at most countable is Riemann integrable applies; what makes the function integrable is that $C$ can be covered by intervals of arbitrarily small total length, and nothing else.

Only the implication "measure zero $\Rightarrow$ integrable" 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 is used, so no choice principle is involved.

Facts & Assumptions

Given: The Cantor set $C \subseteq [0,1]$ and its indicator $\mathbf{1}_C : [0,1] \to \mathbb{R}$ .

[L1]

$C$ is closed and bounded, has measure zero, is uncountable, and contains no interval with two distinct endpoints; in particular $C$ is nowhere dense, so the interior of $C$ is empty and every nonempty open subset of $\mathbb{R}$ contains a point outside $C$ (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds, Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover), Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ , Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , Finite, countably infinite, countable, uncountable).

[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]

A bounded $f$ on $[a,b]$ with $a < b$ is Riemann integrable if and only if its set of discontinuities has measure zero; the implication from "measure zero" to "integrable" uses no choice principle (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, Lower bound, bounded below, bounded set).

[L4]

For a partition $P = (n,t)$ of $[0,1]$ : $n \ge 1$ , $t_i < t_{i+1}$ , $\Delta_i > 0$ , $I_i = [t_i,t_{i+1}] \subseteq [0,1]$ , 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).

[L5]

$m_i = \inf f[I_i]$ , $L(f,P) = \sum_{i<n}m_i\Delta_i$ , $\underline{\int_0^1}f$ is the supremum of the lower sums, $\overline{\int_0^1}f$ the infimum of the upper sums, and the integral is their common value 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 $\int_a^b f$ ).

[L6]

A set with a least element has it as its infimum; the supremum of $\{0\}$ is $0$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L7]

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

[L8]

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 $(u,v) \subseteq N_\rho(x)\cap[0,1]$ (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.

Verification

technique · direct

1.1

$\mathbf{1}_C$ takes only the values $0$ and $1$ , so it is bounded on $[0,1]$ and its Darboux sums and integrals are defined by [L5].

givenL5

1.2

Discontinuity on $C$ . Let $x \in C$ , so $\mathbf{1}_C(x) = 1$ , and let a real $\rho > 0$ be given. By [L8] the set $(u,v) \subseteq N_\rho(x)\cap[0,1]$ is a nonempty open interval, so by [L1] it contains a point $y \notin C$ ; then $y \in [0,1]$ , $|y-x| < \rho$ and $|\mathbf{1}_C(x) - \mathbf{1}_C(y)| = 1$ . So the continuity condition fails at $x$ for $\varepsilon := 1$ .

givenL1L8

1.3

Continuity off $C$ . Let $x \in [0,1]$ with $x \notin C$ . Since $C$ is closed, [L2] gives a real $\rho > 0$ with $N_\rho(x) \cap C = \varnothing$ , so $\mathbf{1}_C$ vanishes on $N_\rho(x)\cap[0,1]$ and $|\mathbf{1}_C(y) - \mathbf{1}_C(x)| = 0 < \varepsilon$ there for every $\varepsilon > 0$ .

givenL1L2

2.1

So the set of discontinuities of $\mathbf{1}_C$ in $[0,1]$ is exactly $C$ , which has measure zero by [L1]; by [L3] and $0 < 1$ , $\mathbf{1}_C$ is Riemann integrable on $[0,1]$ .

step 1.1step 1.2step 1.3L1L3

2.2

Every lower sum is $0$ . Let $P = (n,t)$ be a partition of $[0,1]$ and $i < n$ . By [L4] the interval $(t_i,t_{i+1})$ is a nonempty open subset of $[0,1]$ , so by [L1] it contains a point outside $C$ , at which $\mathbf{1}_C$ takes the value $0$ ; since $\mathbf{1}_C \ge 0$ , the value $0$ is the least element of $\mathbf{1}_C[I_i]$ and $m_i = 0$ by [L6]. Hence $L(\mathbf{1}_C,P) = 0$ by [L5] and [L7].

step 1.1L1L4L5L6L7

3.1

The set of lower sums is $\{0\}$ , so $\underline{\int_0^1}\mathbf{1}_C = 0$ by [L6]; and $\mathbf{1}_C$ is integrable by step 2.1, so $\int_0^1 \mathbf{1}_C = 0$ by [L5].

step 2.1step 2.2L5L6∎

Remarks

Uncountably many discontinuities, and the integral does not notice. $C$ is uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), so this function is outside the reach of A bounded function on $[a,b]$ whose set of discontinuities is at most countable is Riemann integrable and is the standard demonstration that the Lebesgue criterion is strictly stronger than the countable one.
Every upper sum is at least the total length of the subintervals meeting $C$ , and that total goes to $0$ . The proof above does not need this, since integrability comes from the criterion and the value from the lower sums alone; but it is the reason the upper sums also converge to $0$ , and it is precisely where the Smith-Volterra-Cantor set behaves differently (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).
The two hypotheses of claim 1 are the two properties of $C$ that matter. Closedness gives continuity off $C$ ; empty interior gives discontinuity on $C$ . A set with both is exactly a closed nowhere dense set, and any such set is the discontinuity set of its own indicator. Whether that indicator is integrable then depends only on whether the set is null, which is what FALSE: a bounded function on $[a,b]$ is Riemann integrable exactly when its set of discontinuities is nowhere dense settles in the negative for category.

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Sources

Cantor set (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
MAT425 Lecture Notes (Princeton University) (standard reference, not scraped)

The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral 000 even though it is discontinuous at uncountably many points