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The Riemann Integral: Definition and Integrability: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

01x2=1/3\int_0^1 x^{2} = 1/3, computed from the Darboux definition with uniform partitions and the closed form k<nk2=n(n1)(2n1)/6\sum_{k<n} k^{2} = n(n-1)(2n-1)/6

Example

Let f:[0,1]Rf : [0,1] \to \mathbb{R} be f(x)=x2f(x) = x^{2} (Integer powers ama^m). Then ff is Riemann integrable on [0,1][0,1] and

01x2dx  =  13.\int_0^1 x^{2}\,\mathrm{d}x \;=\; \frac{1}{3} .

Everything is computed from the definition. For the uniform partition UNU_N of [0,1][0,1] into N1N \ge 1 parts (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = t_0 < t_1 < \dots < t_n = b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions) the two Darboux sums (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i) are

L(f,UN)  =  (N1)(2N1)6N2,U(f,UN)  =  (N+1)(2N+1)6N2,U(f,UN)L(f,UN)  =  1N,L(f,U_N) \;=\; \frac{(N-1)(2N-1)}{6N^{2}}, \qquad U(f,U_N) \;=\; \frac{(N+1)(2N+1)}{6N^{2}}, \qquad U(f,U_N) - L(f,U_N) \;=\; \frac{1}{N},

where a natural NN multiplying or dividing a real stands for its canonical natural ι(N)\iota(N) (The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field), as in clause 2 of Laws of finite sums and finite products. Both expressions converge to 1/31/3 as NN grows, the lower sums from below and the upper sums from above, and the gap 1/N1/N is what Archimedes' property drives to 00.

The arithmetic rests on one closed form, valid for every nNn \in \mathbb{N}:

6k<nι(k)2  =  ι(n)(ι(n)1)(2ι(n)1),6\sum_{k<n} \iota(k)^{2} \;=\; \iota(n)\bigl(\iota(n)-1\bigr)\bigl(2\iota(n)-1\bigr) ,

which at n=0n = 0 reads 0=00 = 0 and is the familiar n(n1)(2n1)/6n(n-1)(2n-1)/6 written so that the empty sum is the case n=0n = 0 rather than an exception. Note the indexing: the sum runs over k<nk < n, so its last term is ι(n1)2\iota(n-1)^{2} and not ι(n)2\iota(n)^2.

Facts & Assumptions

Given: f:[0,1]Rf : [0,1] \to \mathbb{R} with f(x)=x2f(x) = x^{2}, and for N1N \ge 1 the uniform partition UN=(N,t)U_N = (N,t) of [0,1][0,1] with ti=ι(i)/ι(N)t_i = \iota(i)/\iota(N) for iNi \le N and lengths Δi=1/ι(N)\Delta_i = 1/\iota(N).

[L1]

For UNU_N: t0=0t_0 = 0, tN=1t_N = 1, ti<ti+1t_i < t_{i+1}, Δi=1/ι(N)\Delta_i = 1/\iota(N) for i<Ni < N, i<NΔi=1\sum_{i<N}\Delta_i = 1, and Ii=[ti,ti+1]I_i = [t_i,t_{i+1}] (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L2]

mi=inff[Ii]m_i = \inf f[I_i], Mi=supf[Ii]M_i = \sup f[I_i], L(f,P)=i<NmiΔiL(f,P) = \sum_{i<N}m_i\Delta_i, U(f,P)=i<NMiΔiU(f,P) = \sum_{i<N}M_i\Delta_i, and L(f,P)01f01fU(f,P)L(f,P) \le \underline{\int_0^1} f \le \overline{\int_0^1} f \le U(f,P) for every partition PP; when ff is integrable, 01f\int_0^1 f is the common value of the two integrals (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f).

[L3]

For 0a<b0 \le a < b one has a2<b2a^{2} < b^{2}; a set with a least element has it as its infimum and one with a greatest element has it as its supremum (Monotonicity of xxnx \mapsto x^n and of nann \mapsto a^n, Integer powers ama^m, Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L4]

The closed form. For every nNn \in \mathbb{N}, 6k<nι(k)2=ι(n)(ι(n)1)(2ι(n)1)6\sum_{k<n}\iota(k)^{2} = \iota(n)(\iota(n)-1)(2\iota(n)-1). This is an induction on nn (The principle of mathematical induction) from the recursion clause k<n+1=k<n+ι(n)2\sum_{k<n+1} = \sum_{k<n} + \iota(n)^2 of Finite sums and finite products, by recursion and the identity x(x1)(2x1)+6x2=x(x+1)(2x+1)x(x-1)(2x-1) + 6x^{2} = x(x+1)(2x+1), valid in any commutative ring and in particular for x=ι(n)x = \iota(n) (Ordered field, Canonical naturals are positive and strictly increasing).

[L5]

Finite sums: scaling, additivity, telescoping i<N(ci+1ci)=cNc0\sum_{i<N}(c_{i+1}-c_i) = c_N - c_0, and ι(i+1)=ι(i)+1\iota(i+1) = \iota(i)+1 (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing).

[L8]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive; uvw|u - v| \le w whenever cudc \le u \le d, cvdc \le v \le d and dcwd - c \le w (Basic properties of the absolute value, 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.

Verification

technique · direct
1.1

ff is continuous on [0,1][0,1] and 0<10 < 1, so ff is Riemann integrable on [0,1][0,1] by [L6]; write J:=01fJ := \int_0^1 f.

givenL6
1.2

On Ii=[ti,ti+1]I_i = [t_i,t_{i+1}] with 0ti<ti+10 \le t_i < t_{i+1} the function ff takes its least value at tit_i and its greatest at ti+1t_{i+1}, by [L3]; both values are attained, so mi=ti2=ι(i)2/ι(N)2m_i = t_i^{2} = \iota(i)^{2}/\iota(N)^{2} and Mi=ti+12=ι(i+1)2/ι(N)2M_i = t_{i+1}^{2} = \iota(i+1)^{2}/\iota(N)^{2} by [L3] and [L1].

givenL1L3
2.1

By [L2], [L5] and [L1], L(f,UN)=i<Nι(i)2ι(N)21ι(N)=1ι(N)3i<Nι(i)2L(f,U_N) = \sum_{i<N}\frac{\iota(i)^{2}}{\iota(N)^{2}}\cdot\frac{1}{\iota(N)} = \frac{1}{\iota(N)^{3}}\sum_{i<N}\iota(i)^{2}, and by [L4] this equals ι(N)(ι(N)1)(2ι(N)1)6ι(N)3=(ι(N)1)(2ι(N)1)6ι(N)2\frac{\iota(N)(\iota(N)-1)(2\iota(N)-1)}{6\,\iota(N)^{3}} = \frac{(\iota(N)-1)(2\iota(N)-1)}{6\,\iota(N)^{2}}.

step 1.2L1L2L4L5
2.2

Likewise U(f,UN)=1ι(N)3i<Nι(i+1)2U(f,U_N) = \frac{1}{\iota(N)^{3}}\sum_{i<N}\iota(i+1)^{2}. Since i<Nι(i+1)2=k<N+1ι(k)2ι(0)2=k<N+1ι(k)2\sum_{i<N}\iota(i+1)^{2} = \sum_{k<N+1}\iota(k)^{2} - \iota(0)^{2} = \sum_{k<N+1}\iota(k)^{2} by [L5] and ι(0)=0\iota(0) = 0, [L4] applied at N+1N+1 gives U(f,UN)=(ι(N)+1)(2ι(N)+1)6ι(N)2U(f,U_N) = \frac{(\iota(N)+1)(2\iota(N)+1)}{6\,\iota(N)^{2}}.

step 1.2L1L2L4L5
2.3

Directly, U(f,UN)L(f,UN)=1ι(N)3i<N(ι(i+1)2ι(i)2)=ι(N)2ι(0)2ι(N)3=1ι(N)U(f,U_N) - L(f,U_N) = \frac{1}{\iota(N)^{3}}\sum_{i<N}\bigl(\iota(i+1)^{2} - \iota(i)^{2}\bigr) = \frac{\iota(N)^{2} - \iota(0)^{2}}{\iota(N)^{3}} = \frac{1}{\iota(N)}, by telescoping in [L5].

step 1.2L1L2L5
3.1

Expanding by [L8], 13L(f,UN)=3ι(N)16ι(N)2>0\frac{1}{3} - L(f,U_N) = \frac{3\iota(N)-1}{6\,\iota(N)^{2}} > 0 and U(f,UN)13=3ι(N)+16ι(N)2>0U(f,U_N) - \frac{1}{3} = \frac{3\iota(N)+1}{6\,\iota(N)^{2}} > 0 for N1N \ge 1, so L(f,UN)<13<U(f,UN)L(f,U_N) < \frac{1}{3} < U(f,U_N).

step 2.1step 2.2L7L8
3.2

By [L2], L(f,UN)JU(f,UN)L(f,U_N) \le J \le U(f,U_N) as well.

step 1.1step 2.1step 2.2L2
4.1

Both JJ and 1/31/3 lie in the interval with endpoints L(f,UN)L(f,U_N) and U(f,UN)U(f,U_N), whose length is 1/ι(N)1/\iota(N) by step 2.3, so J1/31/ι(N)|J - 1/3| \le 1/\iota(N) for every natural N1N \ge 1, by [L8].

step 2.3step 3.1step 3.2L8
5.1

If J1/3J \ne 1/3 then J1/3>0|J - 1/3| > 0, and [L7] supplies N1N \ge 1 with 1/ι(N)<J1/31/\iota(N) < |J - 1/3|, contradicting step 4.1. Hence J=1/3J = 1/3, that is 01x2dx=1/3\int_0^1 x^{2}\,\mathrm{d}x = 1/3.

step 4.1L7L8

Remarks

  • The first index is 00 and the last term of the lower sum is not 11. The lower sum uses mi=ti2m_i = t_i^{2} with ii running from 00, so its first term is 00 and its last is ((N1)/N)2(1/N)\bigl((N-1)/N\bigr)^{2}\cdot(1/N); the upper sum uses Mi=ti+12M_i = t_{i+1}^{2} and so ends at 1(1/N)1 \cdot (1/N). Reading the closed form with the other convention, k=1nk2=n(n+1)(2n+1)/6\sum_{k=1}^{n}k^{2} = n(n+1)(2n+1)/6, and attaching it to the wrong sum is the standard way to lose the factor and land on 1/21/2 or 1/61/6 instead of 1/31/3.

  • The gap is exactly 1/N1/N, not merely O(1/N)O(1/N). Step 2.3 computes it by telescoping without evaluating either sum, which is also the cheapest route to integrability through Riemann's criterion: a bounded ff on [a,b][a,b] is Darboux integrable if and only if for every real ε>0\varepsilon > 0 there is a partition PP with U(f,P)L(f,P)<εU(f,P) - L(f,P) < \varepsilon: the continuity of x2x^2 is not needed for that, only for the shortcut taken in step 1.1.

  • What this example does not do. No antiderivative and no fundamental theorem of calculus is used or available at this point in the reading order; the value 1/31/3 is extracted from the two sums and the Archimedean property alone. The same computation with x3x^{3} needs k<nk3\sum_{k<n}k^{3} and is no harder, but it is not carried out here.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

One refinement worked out for f(x)=x2f(x) = x^{2} on [0,1][0,1]: adding the point 1/21/2 to the trivial partition raises the lower sum from 00 to 1/81/8 and lowers the upper sum from 11 to 5/85/8

Example

Let f:[0,1]Rf : [0,1] \to \mathbb{R} be f(x)=x2f(x) = x^{2} (Integer powers ama^m). Let P0P_0 be the trivial partition of [0,1][0,1], with point set {0,1}\{0,1\}, and let P1:=P0+12P_1 := P_0 + \tfrac12 be the partition obtained by inserting the point 1/21/2, with point set {0,12,1}\{0, \tfrac12, 1\} (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = t_0 < t_1 < \dots < t_n = b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). Then

L(f,P0)=0,L(f,P1)=18,U(f,P1)=58,U(f,P0)=1,L(f,P_0) = 0, \qquad L(f,P_1) = \tfrac18, \qquad U(f,P_1) = \tfrac58, \qquad U(f,P_0) = 1 ,

so that

0  =  L(f,P0)    L(f,P1)    U(f,P1)    U(f,P0)  =  1,0 \;=\; L(f,P_0) \;\le\; L(f,P_1) \;\le\; U(f,P_1) \;\le\; U(f,P_0) \;=\; 1 ,

which is claim 1 of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: L(f,P)L(f,P)U(f,P)U(f,P)L(f,P) \le L(f,P') \le U(f,P') \le U(f,P) when PP' refines PP, and L(f,P)U(f,Q)L(f,P) \le U(f,Q) for arbitrary partitions PP and QQ; moreover the two changes are at most 2M(nn)P2M(n' - n)\|P\| with both inequalities strict. The gap drops from U(f,P0)L(f,P0)=1U(f,P_0) - L(f,P_0) = 1 to U(f,P1)L(f,P1)=12U(f,P_1) - L(f,P_1) = \tfrac12: exactly one refinement halves it, and 01x2=1/3\int_0^1 x^{2} = 1/3, computed from the Darboux definition with uniform partitions and the closed form k<nk2=n(n1)(2n1)/6\sum_{k<n} k^{2} = n(n-1)(2n-1)/6 shows the uniform partitions drive it to 00.

Facts & Assumptions

Given: f:[0,1]Rf : [0,1] \to \mathbb{R} with f(x)=x2f(x) = x^{2}; the partition P0=(1,t)P_0 = (1,t) with t0=0t_0 = 0 and tk=1t_k = 1 for k1k \ge 1; and P1=(2,s)P_1 = (2,s) with s0=0s_0 = 0, s1=1/2s_1 = 1/2 and sk=1s_k = 1 for k2k \ge 2.

[L1]

Both P0P_0 and P1P_1 are partitions of [0,1][0,1], and P1P_1 refines P0P_0, since pts(P0)={0,1}{0,12,1}=pts(P1)\operatorname{pts}(P_0) = \{0,1\} \subseteq \{0,\tfrac12,1\} = \operatorname{pts}(P_1); the subintervals of P0P_0 are [0,1][0,1] with length 11, and those of P1P_1 are [0,12][0,\tfrac12] and [12,1][\tfrac12,1], each of length 12\tfrac12 (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L2]

mi=inff[Ii]m_i = \inf f[I_i], Mi=supf[Ii]M_i = \sup f[I_i], L(f,P)=i<nmiΔiL(f,P) = \sum_{i<n}m_i\Delta_i and U(f,P)=i<nMiΔiU(f,P) = \sum_{i<n}M_i\Delta_i (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i).

[L3]

For 0a<b0 \le a < b one has a2<b2a^{2} < b^{2}, so on an interval [c,d][0,1][c,d] \subseteq [0,1] with c<dc < d the function ff has least value c2c^{2} and greatest value d2d^{2}, both attained; a set with a least element has it as its infimum and one with a greatest element has it as its supremum (Monotonicity of xxnx \mapsto x^n and of nann \mapsto a^n, Integer powers ama^m, Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L5]

Finite sums of one and of two terms: i<1ai=a0\sum_{i<1}a_i = a_0 and i<2ai=a0+a1\sum_{i<2}a_i = a_0 + a_1 (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: (12)2=14(\tfrac12)^{2} = \tfrac14, 1412=18\tfrac14 \cdot \tfrac12 = \tfrac18, 18+12=58\tfrac18 + \tfrac12 = \tfrac58, and 0<18<58<10 < \tfrac18 < \tfrac58 < 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)). 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

For P0P_0 the single subinterval is [0,1][0,1], so by [L3] m0=02=0m_0 = 0^{2} = 0 and M0=12=1M_0 = 1^{2} = 1, and by [L2] and [L5], L(f,P0)=01=0L(f,P_0) = 0 \cdot 1 = 0 and U(f,P0)=11=1U(f,P_0) = 1 \cdot 1 = 1.

givenL1L2L3L5
1.2

For P1P_1 the two subintervals are [0,12][0,\tfrac12] and [12,1][\tfrac12,1], each of length 12\tfrac12, so by [L3] the extreme values are m0=0m_0 = 0, M0=(12)2=14M_0 = (\tfrac12)^{2} = \tfrac14 on the first and m1=14m_1 = \tfrac14, M1=1M_1 = 1 on the second.

givenL1L3L6
2.1

Hence by [L2], [L5] and [L6], L(f,P1)=012+1412=18L(f,P_1) = 0\cdot\tfrac12 + \tfrac14\cdot\tfrac12 = \tfrac18 and U(f,P1)=1412+112=18+12=58U(f,P_1) = \tfrac14\cdot\tfrac12 + 1\cdot\tfrac12 = \tfrac18 + \tfrac12 = \tfrac58.

step 1.2L2L5L6
3.1

Comparing with step 1.1 and using [L6]: L(f,P0)=0<18=L(f,P1)L(f,P_0) = 0 < \tfrac18 = L(f,P_1) and U(f,P1)=58<1=U(f,P0)U(f,P_1) = \tfrac58 < 1 = U(f,P_0), while L(f,P1)=18<58=U(f,P1)L(f,P_1) = \tfrac18 < \tfrac58 = U(f,P_1). This is the chain of [L4] for the refinement P1P_1 of P0P_0, here with every inequality strict.

step 1.1step 2.1L1L4L6

Remarks

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

03x=3\int_0^3 \lfloor x \rfloor = 3: the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

Example

Let g:[0,3]Rg : [0,3] \to \mathbb{R} be g(x)=xg(x) = \lfloor x \rfloor, the integer part (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1). Then gg is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences), hence Riemann integrable on [0,3][0,3] (A monotone function on [a,b][a,b] is Riemann integrable: for the uniform partition into NN parts the upper minus lower sum telescopes to f(b)f(a)(ba)/ι(N)|f(b) - f(a)|\,(b-a)/\iota(N)), and

03xdx  =  3.\int_0^3 \lfloor x \rfloor \,\mathrm{d}x \;=\; 3 .

gg is discontinuous at x=1x = 1, x=2x = 2 and x=3x = 3 and continuous elsewhere on [0,3][0,3] (Discontinuity of ff 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), so this is an integrable function with genuine jumps, not a continuous one in disguise; the value 33 is 01+11+210\cdot 1 + 1\cdot 1 + 2\cdot 1, the three constant pieces weighted by their lengths.

The computation below uses the uniform partition into N=3MN = 3M parts, M1M \ge 1, for which the lower sum is exactly 33 at every MM and the upper sum is 3+3/M3 + 3/M. So the lower sums do not merely approach the integral, they attain it.

Facts & Assumptions

Given: g:[0,3]Rg : [0,3] \to \mathbb{R} with g(x)=xg(x) = \lfloor x \rfloor; a natural M1M \ge 1; N:=3MN := 3M; and the uniform partition UN=(N,t)U_N = (N,t) of [0,3][0,3] with ti=3ι(i)/ι(N)=ι(i)/ι(M)t_i = 3\iota(i)/\iota(N) = \iota(i)/\iota(M) for iNi \le N and lengths Δi=3/ι(N)=1/ι(M)\Delta_i = 3/\iota(N) = 1/\iota(M).

[L1]

For every real xx there is exactly one integer x\lfloor x \rfloor with xx<x+1\lfloor x \rfloor \le x < \lfloor x \rfloor + 1 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1).

[L2]

gg is nondecreasing: for xyx \le y, xxy<y+1\lfloor x \rfloor \le x \le y < \lfloor y \rfloor + 1, and x\lfloor x \rfloor, y\lfloor y \rfloor are integers, so xy\lfloor x \rfloor \le \lfloor y \rfloor, no integer lying strictly between y\lfloor y \rfloor and y+1\lfloor y \rfloor + 1 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences).

[L5]

For a nondecreasing gg and a subinterval Ii=[ti,ti+1]I_i = [t_i,t_{i+1}]: mi=g(ti)m_i = g(t_i) and Mi=g(ti+1)M_i = g(t_{i+1}), both attained; L(g,P)=i<NmiΔiL(g,P) = \sum_{i<N}m_i\Delta_i, U(g,P)=i<NMiΔiU(g,P) = \sum_{i<N}M_i\Delta_i, and L(g,P)03g03gU(g,P)L(g,P) \le \underline{\int_0^3}g \le \overline{\int_0^3}g \le U(g,P) (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f, Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L6]

Finite sums: splitting a sum over i<3Mi < 3M into the three blocks i<Mi < M, Mi<2MM \le i < 2M and 2Mi<3M2M \le i < 3M; scaling; and i<Mc=ι(M)c\sum_{i<M}c = \iota(M)c (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

For iNi \in \mathbb{N}: ι(i)/ι(M)=0\lfloor \iota(i)/\iota(M) \rfloor = 0 when i<Mi < M, =1= 1 when Mi<2MM \le i < 2M, and =2= 2 when 2Mi<3M2M \le i < 3M; and ι(j)/ι(M)=3\lfloor \iota(j)/\iota(M)\rfloor = 3 for j=3Mj = 3M. Each case is [L1] applied to the displayed inequalities 0ι(i)/ι(M)<10 \le \iota(i)/\iota(M) < 1, 1ι(i)/ι(M)<21 \le \iota(i)/\iota(M) < 2, 2ι(i)/ι(M)<32 \le \iota(i)/\iota(M) < 3 and ι(3M)/ι(M)=3\iota(3M)/\iota(M) = 3, which follow from ι\iota being strictly increasing and additive (Canonical naturals are positive and strictly increasing, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Order is preserved by adding a constant and by adding inequalities). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used here follow by adjoining the equality case, in which the two sides coincide.

[L9]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive (Basic properties of the absolute value, 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.

Verification

technique · direct
1.1

gg is nondecreasing by [L2], and 0<30 < 3, so gg is Riemann integrable on [0,3][0,3] by [L3]; write J:=03gJ := \int_0^3 g.

givenL2L3
1.2

By [L5] and [L7], for i<N=3Mi < N = 3M the lower value is mi=g(ti)=ι(i)/ι(M)m_i = g(t_i) = \lfloor \iota(i)/\iota(M)\rfloor, which is 00 for i<Mi < M, 11 for Mi<2MM \le i < 2M and 22 for 2Mi<3M2M \le i < 3M; and the upper value is Mi=g(ti+1)=ι(i+1)/ι(M)M_i = g(t_{i+1}) = \lfloor \iota(i+1)/\iota(M)\rfloor, which is 00 for i+1<Mi+1 < M, 11 for Mi+1<2MM \le i+1 < 2M, 22 for 2Mi+1<3M2M \le i+1 < 3M and 33 for i+1=3Mi+1 = 3M.

givenL1L4L5L7
2.1

By [L6] and step 1.2, L(g,UN)=1ι(M)(i<M0+Mi<2M1+2Mi<3M2)=1ι(M)(0+ι(M)+2ι(M))=3L(g,U_N) = \frac{1}{\iota(M)}\Bigl(\sum_{i<M}0 + \sum_{M \le i < 2M}1 + \sum_{2M \le i < 3M}2\Bigr) = \frac{1}{\iota(M)}\bigl(0 + \iota(M) + 2\iota(M)\bigr) = 3.

step 1.2L4L5L6L9
2.2

By [L6] and step 1.2, the upper values run over j=i+1j = i+1 from 11 to 3M3M, giving ι(M)1\iota(M)-1 indices with value 00, then ι(M)\iota(M) with value 11, then ι(M)\iota(M) with value 22, and the single index j=3Mj = 3M with value 33; hence U(g,UN)=1ι(M)(0+ι(M)+2ι(M)+3)=3+3ι(M)U(g,U_N) = \frac{1}{\iota(M)}\bigl(0 + \iota(M) + 2\iota(M) + 3\bigr) = 3 + \frac{3}{\iota(M)}.

step 1.2L4L5L6L9
3.1

By [L5], 3=L(g,UN)03gJ03gU(g,UN)=3+3/ι(M)3 = L(g,U_N) \le \underline{\int_0^3} g \le J \le \overline{\int_0^3} g \le U(g,U_N) = 3 + 3/\iota(M) for every natural M1M \ge 1.

step 1.1step 2.1step 2.2L5
4.1

Hence 0J33/ι(M)0 \le J - 3 \le 3/\iota(M) for every M1M \ge 1. If J>3J > 3 then (J3)31>0(J-3)\cdot 3^{-1} > 0 and [L8] supplies M1M \ge 1 with 1/ι(M)<(J3)311/\iota(M) < (J-3)\cdot 3^{-1}, that is 3/ι(M)<J33/\iota(M) < J - 3, contradicting step 3.1. So J=3J = 3, that is 03xdx=3\int_0^3 \lfloor x \rfloor\,\mathrm{d}x = 3.

step 3.1L8L9

Remarks

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral 00 even though it is discontinuous at uncountably many points

Example

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

  1. 1C\mathbf{1}_C is discontinuous at every point of CC and continuous at every point of [0,1]C[0,1] \setminus C, so its set of discontinuities is exactly CC (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point, Discontinuity of ff 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);
  2. 1C\mathbf{1}_C is Riemann integrable on [0,1][0,1], because CC 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 ff on [a,b][a,b] is Riemann integrable if and only if its set of discontinuities has measure zero);
  3. 011C=0\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][a,b] whose set of discontinuities is at most countable is Riemann integrable applies; what makes the function integrable is that CC 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 ff on [a,b][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[0,1]C \subseteq [0,1] and its indicator 1C:[0,1]R\mathbf{1}_C : [0,1] \to \mathbb{R}.

[L3]

A bounded ff on [a,b][a,b] with a<ba < 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 ff on [a,b][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)P = (n,t) of [0,1][0,1]: n1n \ge 1, ti<ti+1t_i < t_{i+1}, Δi>0\Delta_i > 0, Ii=[ti,ti+1][0,1]I_i = [t_i,t_{i+1}] \subseteq [0,1], and (ti,ti+1)(t_i,t_{i+1}) is a nonempty open subset of [0,1][0,1] (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length, Open subset of R\mathbb{R} (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L6]

A set with a least element has it as its infimum; the supremum of {0}\{0\} is 00 (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L7]
[L8]

Ordered-field arithmetic: for 0x10 \le x \le 1 and a real ρ>0\rho > 0 the reals u:=max{0,xρ}u := \max\{0,x-\rho\} and v:=min{1,x+ρ}v := \min\{1,x+\rho\} satisfy u<vu < v and (u,v)Nρ(x)[0,1](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 ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}, Intervals of R\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

1C\mathbf{1}_C takes only the values 00 and 11, so it is bounded on [0,1][0,1] and its Darboux sums and integrals are defined by [L5].

givenL5
1.2

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

givenL1L8
1.3

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

givenL1L2
2.1

So the set of discontinuities of 1C\mathbf{1}_C in [0,1][0,1] is exactly CC, which has measure zero by [L1]; by [L3] and 0<10 < 1, 1C\mathbf{1}_C is Riemann integrable on [0,1][0,1].

step 1.1step 1.2step 1.3L1L3
2.2

Every lower sum is 00. Let P=(n,t)P = (n,t) be a partition of [0,1][0,1] and i<ni < n. By [L4] the interval (ti,ti+1)(t_i,t_{i+1}) is a nonempty open subset of [0,1][0,1], so by [L1] it contains a point outside CC, at which 1C\mathbf{1}_C takes the value 00; since 1C0\mathbf{1}_C \ge 0, the value 00 is the least element of 1C[Ii]\mathbf{1}_C[I_i] and mi=0m_i = 0 by [L6]. Hence L(1C,P)=0L(\mathbf{1}_C,P) = 0 by [L5] and [L7].

step 1.1L1L4L5L6L7
3.1

The set of lower sums is {0}\{0\}, so 011C=0\underline{\int_0^1}\mathbf{1}_C = 0 by [L6]; and 1C\mathbf{1}_C is integrable by step 2.1, so 011C=0\int_0^1 \mathbf{1}_C = 0 by [L5].

step 2.1step 2.2L5L6

Remarks

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Thomae's function is Riemann integrable on [0,1][0,1] with integral 00: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is 00

Example

Let t:[0,1]Rt : [0,1] \to \mathbb{R} be Thomae's function restricted to [0,1][0,1]: at a rational xx with least denominator q(x)1q(x) \ge 1 it takes the value 1/ι(q(x))1/\iota(q(x)), and at an irrational xx the value 00 (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field). Then tt is Riemann integrable on [0,1][0,1] and

01t  =  0.\int_0^1 t \;=\; 0 .

Two ingredients, and they pull in opposite directions. tt is continuous at every irrational and discontinuous at every rational (The Dirichlet function is continuous at no point of R\mathbb{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 cc equals t(c)t(c)), so its discontinuity set is Q[0,1]\mathbb{Q}\cap[0,1], which is infinite and dense — and countable, which is what A bounded function on [a,b][a,b] whose set of discontinuities is at most countable is Riemann integrable needs. The value is then read off the lower sums, every one of which is 00 because every subinterval contains an irrational (Both Q\mathbb{Q} and RQ\mathbb{R} \setminus \mathbb{Q} are dense in R\mathbb{R}, and every nonempty open subset of R\mathbb{R} is uncountable).

Every upper sum, by contrast, is strictly positive, since every subinterval contains a rational; the upper integral is nevertheless 00, an infimum of positive numbers.

Facts & Assumptions

Given: Thomae's function t:[0,1]Rt : [0,1] \to \mathbb{R} as above.

[L1]

t(x)=1/ι(q(x))t(x) = 1/\iota(q(x)) with ι(q(x))1>0\iota(q(x)) \ge 1 > 0 at a rational xx, and t(x)=0t(x) = 0 at an irrational xx; hence 0t(x)10 \le t(x) \le 1 everywhere and t(x)>0t(x) > 0 at every rational xx (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing).

[L3]

Q\mathbb{Q} is countably infinite and every subset of an at most countable set is at most countable (Q\mathbb{Q} is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L4]

A bounded function on [a,b][a,b] with a<ba < b whose set of discontinuities is at most countable is Riemann integrable (A bounded function on [a,b][a,b] whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set).

[L6]

For a partition P=(n,s)P = (n,s) of [0,1][0,1]: n1n \ge 1, si<si+1s_i < s_{i+1}, Δi>0\Delta_i > 0, Ii=[si,si+1][0,1]I_i = [s_i,s_{i+1}] \subseteq [0,1], and (si,si+1)(s_i,s_{i+1}) is a nonempty open interval (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L8]

A set with a least element has it as its infimum; the supremum of {0}\{0\} is 00 (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L9]
[L10]

Ordered-field arithmetic: the order is total and transitive, and a reciprocal of a positive quantity is positive (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.

Verification

technique · direct
1.1

tt is bounded on [0,1][0,1], with 0t(x)10 \le t(x) \le 1 for every xx, by [L1].

givenL1L10
1.2

By [L2], tt is continuous at every irrational point of [0,1][0,1], so its set of discontinuities in [0,1][0,1] is contained in Q[0,1]\mathbb{Q}\cap[0,1], which is at most countable by [L3]; a subset of it is then at most countable as well.

givenL2L3
2.1

By [L4] applied on [0,1][0,1], with 0<10 < 1, tt is Riemann integrable on [0,1][0,1].

step 1.1step 1.2L4
2.2

Every lower sum is 00. Let P=(n,s)P = (n,s) be a partition of [0,1][0,1] and i<ni < n. By [L6] the interval (si,si+1)(s_i,s_{i+1}) is nonempty and open, so by [L5] it contains an irrational yy, and yIi[0,1]y \in I_i \subseteq [0,1], so t(y)=0t(y) = 0 by [L1]. Since t0t \ge 0 by [L1], the value 00 is the least element of t[Ii]t[I_i] and mi=0m_i = 0 by [L8]. Hence L(t,P)=0L(t,P) = 0 by [L7] and [L9].

step 1.1L1L5L6L7L8L9
3.1

The set of lower sums is therefore {0}\{0\} and 01t=0\underline{\int_0^1}t = 0 by [L8]; since tt is integrable by step 2.1, 01t=0\int_0^1 t = 0 by [L7].

step 2.1step 2.2L7L8

Remarks

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

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

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][a,b] is Riemann integrable, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f).

The witness is the Dirichlet function 1Q\mathbf{1}_{\mathbb{Q}} restricted to [0,1][0,1] (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx). It takes only the values 00 and 11, so it is bounded (Lower bound, bounded below, bounded set); every lower Darboux sum is 00 and every upper Darboux sum is 11; hence

011Q  =  0    1  =  011Q,\underline{\int_0^1}\mathbf{1}_{\mathbb{Q}} \;=\; 0 \;\ne\; 1 \;=\; \overline{\int_0^1}\mathbf{1}_{\mathbb{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

[A1]

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

[L2]

For a partition P=(n,t)P = (n,t) of [0,1][0,1]: n1n \ge 1, ti<ti+1t_i < t_{i+1}, Δi>0\Delta_i > 0, i<nΔi=1\sum_{i<n}\Delta_i = 1, Ii=[ti,ti+1]I_i = [t_i,t_{i+1}], and (ti,ti+1)(t_i,t_{i+1}) is a nonempty open interval contained in IiI_i (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L3]

mi=infg[Ii]m_i = \inf g[I_i], Mi=supg[Ii]M_i = \sup g[I_i], L(g,P)=i<nmiΔiL(g,P) = \sum_{i<n}m_i\Delta_i, U(g,P)=i<nMiΔiU(g,P) = \sum_{i<n}M_i\Delta_i; 01g\underline{\int_0^1}g is the supremum of the lower sums and 01g\overline{\int_0^1}g the infimum of the upper sums; gg is integrable exactly when they agree (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\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}\{c\} are both cc (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L5]

Finite sums: scaling and i<n0=0\sum_{i<n}0 = 0 (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: 010 \ne 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

gg is bounded, with 0g(x)10 \le g(x) \le 1 for every x[0,1]x \in [0,1], so its Darboux sums and integrals are defined by [L3].

givenL3
1.2

Let P=(n,t)P = (n,t) be any partition of [0,1][0,1] and i<ni < n. By [L2] the interval (ti,ti+1)(t_i,t_{i+1}) is nonempty and open, so by [L1] it contains a rational and an irrational, both lying in IiI_i. Hence g[Ii]={0,1}g[I_i] = \{0,1\} and, by [L4], mi=0m_i = 0 and Mi=1M_i = 1.

givenL1L2L4
2.1

Therefore L(g,P)=i<n0Δi=0L(g,P) = \sum_{i<n}0\cdot\Delta_i = 0 and U(g,P)=i<n1Δi=i<nΔi=1U(g,P) = \sum_{i<n}1\cdot\Delta_i = \sum_{i<n}\Delta_i = 1, for every partition PP of [0,1][0,1], by [L3], [L5] and [L2].

step 1.2L2L3L5
3.1

The set of lower sums is {0}\{0\} and the set of upper sums is {1}\{1\}, so 01g=0\underline{\int_0^1}g = 0 and 01g=1\overline{\int_0^1}g = 1 by [L4] and [L3]. Since 010 \ne 1 by [L6], gg is not Riemann integrable on [0,1][0,1].

step 2.1L3L4L6
4.1

So gg is bounded on [0,1][0,1], an interval with 0<10 < 1, and is not Riemann integrable: [A1] is refuted.

step 1.1step 3.1A1

Remarks

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

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][a,b] is Riemann integrable whenever its set of discontinuities is nowhere dense (FALSE: a bounded function on [a,b][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 R\mathbb{R}).

The witness is the indicator gg of the Smith-Volterra-Cantor set S[0,1]S \subseteq [0,1] (The Smith-Volterra-Cantor set: the same construction removing, at stage n1n \ge 1, an open middle interval of length 4n4^{-n} from each of the 2n12^{n-1} remaining intervals). Its set of discontinuities is exactly SS, 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 gg is not Riemann integrable, with

01g  =  0,01g    12.\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 00 (The indicator of the Cantor set is discontinuous exactly on the Cantor set, which is null, so it is Riemann integrable with integral 00 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 ff on [a,b][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 ff on [a,b][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[0,1]S \subseteq [0,1] and its indicator g:[0,1]Rg : [0,1] \to \mathbb{R}, with g(x)=1g(x) = 1 for xSx \in S and g(x)=0g(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]

SS is closed, bounded and nowhere dense, so SS contains no nonempty open set; and if (ak)(a_k), (bk)(b_k) are sequences of reals with akbka_k \le b_k, Sk[ak,bk]S \subseteq \bigcup_k[a_k,b_k] and k<i(bkak)M\sum_{k<i}(b_k-a_k) \le M for every iNi \in \mathbb{N}, then M21M \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 n1n \ge 1, an open middle interval of length 4n4^{-n} from each of the 2n12^{n-1} remaining intervals, Nowhere dense, meager (first category), residual, and second category subsets of R\mathbb{R}, Interior, closure, boundary and exterior of a subset of R\mathbb{R}, Measure zero (a countable cover by intervals of total length below every ε\varepsilon) and content zero (a finite such cover)).

[L3]

For a partition P=(n,t)P = (n,t) of [0,1][0,1]: n1n \ge 1, ti<ti+1t_i < t_{i+1}, Δi>0\Delta_i > 0, Ii=[ti,ti+1]I_i = [t_i,t_{i+1}], [0,1]=i<nIi[0,1] = \bigcup_{i<n}I_i, and (ti,ti+1)(t_i,t_{i+1}) is a nonempty open subset of [0,1][0,1] (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length, Open subset of R\mathbb{R} (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L4]

mi=infg[Ii]m_i = \inf g[I_i], Mi=supg[Ii]M_i = \sup g[I_i], L(g,P)=i<nmiΔiL(g,P) = \sum_{i<n}m_i\Delta_i, U(g,P)=i<nMiΔiU(g,P) = \sum_{i<n}M_i\Delta_i; 01g\underline{\int_0^1}g is the supremum of the lower sums and 01g\overline{\int_0^1}g the infimum of the upper sums; gg is integrable exactly when they agree (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\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 c\ge c is c\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, i<n0=0\sum_{i<n}0 = 0; and a finite list of closed intervals extends to a sequence by degenerate intervals of length 00 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\varepsilon) and content zero (a finite such cover)).

[L7]

Ordered-field arithmetic: for 0x10 \le x \le 1 and a real ρ>0\rho > 0 the reals u:=max{0,xρ}u := \max\{0,x-\rho\} and v:=min{1,x+ρ}v := \min\{1,x+\rho\} satisfy u<vu < v and (u,v)Nρ(x)[0,1](u,v) \subseteq N_\rho(x)\cap[0,1]; and 0<21<10 < 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 ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}, Intervals of R\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

gg takes only the values 00 and 11, so it is bounded on [0,1][0,1] and its Darboux sums and integrals are defined by [L4].

givenL4
1.2

Discontinuity on SS. Let xSx \in S, so g(x)=1g(x) = 1, and let a real ρ>0\rho > 0 be given. By [L7] the interval (u,v)Nρ(x)[0,1](u,v) \subseteq N_\rho(x)\cap[0,1] is nonempty and open, so by [L1] it contains a point ySy \notin S; then yx<ρ|y-x| < \rho, y[0,1]y \in [0,1] and g(x)g(y)=1|g(x)-g(y)| = 1, so continuity fails at xx for ε:=1\varepsilon := 1 (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point, Discontinuity of ff 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 SS. Let x[0,1]x \in [0,1] with xSx \notin S. Since SS is closed, [L2] gives a real ρ>0\rho > 0 with Nρ(x)S=N_\rho(x)\cap S = \varnothing, so gg vanishes on Nρ(x)[0,1]N_\rho(x)\cap[0,1] and continuity holds at xx.

givenL1L2
2.1

So the set of discontinuities of gg in [0,1][0,1] is exactly SS, which is nowhere dense by [L1].

step 1.2step 1.3L1
2.2

Every lower sum is 00. Let P=(n,t)P = (n,t) be a partition of [0,1][0,1] and i<ni < n. By [L3] the interval (ti,ti+1)(t_i,t_{i+1}) is a nonempty open subset of [0,1][0,1], so by [L1] it contains a point outside SS, at which gg takes the value 00; as g0g \ge 0, that value is the least element of g[Ii]g[I_i] and mi=0m_i = 0 by [L5]. Hence L(g,P)=0L(g,P) = 0 by [L4] and [L6].

step 1.1L1L3L4L5L6
2.3

Every upper sum is at least 212^{-1}. With PP as above put B:={i<n:IiS}B := \{\, i < n : I_i \cap S \ne \varnothing \,\}. For iBi \in B the set g[Ii]g[I_i] contains 11, so Mi=1M_i = 1 by [L5]; for iBi \notin B one has g[Ii]={0}g[I_i] = \{0\} and Mi=0M_i = 0. Hence U(g,P)U(g,P) is the sum of the Δi\Delta_i with iBi \in B, by [L4] and [L6].

step 1.1L4L5L6
3.1

The intervals IiI_i with iBi \in B cover SS, since S[0,1]=i<nIiS \subseteq [0,1] = \bigcup_{i<n}I_i by [L3]. Extending that finite list to a sequence by degenerate intervals [0,0][0,0] ([L6]) gives a cover of SS all of whose partial total lengths are at most U(g,P)U(g,P), so [L1] gives U(g,P)21U(g,P) \ge 2^{-1}.

step 2.3L1L3L6
4.1

By [L5] and step 2.2, 01g=0\underline{\int_0^1}g = 0; by [L5] and step 3.1, 01g21>0\overline{\int_0^1}g \ge 2^{-1} > 0. The two differ, so gg is not Riemann integrable by [L4].

step 2.2step 3.1L4L5L7
5.1

So gg is bounded on [0,1][0,1], an interval with 0<10 < 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

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

For every FσF_\sigma subset EE of [0,1][0,1] of measure zero there is a bounded Riemann integrable function on [0,1][0,1] whose set of discontinuities is exactly EE

Example

Let E[0,1]E \subseteq [0,1] be an FσF_\sigma subset of R\mathbb{R} (FσF_\sigma and GδG_\delta subsets of R\mathbb{R}) of measure zero (Measure zero (a countable cover by intervals of total length below every ε\varepsilon) and content zero (a finite such cover)). Then there is a bounded function h:[0,1]Rh : [0,1] \to \mathbb{R}, with values in [0,1][0,1], that is Riemann integrable on [0,1][0,1] and whose set of discontinuities is exactly EE (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point, Discontinuity of ff 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).

The construction. Fix closed sets F0,F1,F2,F_0, F_1, F_2, \dots with E=nNFnE = \bigcup_{n \in \mathbb{N}} F_n and put

h(x)  :=  2n(x)  for xE,h(x)  :=  0  for x[0,1]E,h(x) \;:=\; 2^{-n(x)} \ \text{ for } x \in E, \qquad h(x) \;:=\; 0 \ \text{ for } x \in [0,1]\setminus E,

where n(x):=min{nN:xFn}n(x) := \min\{\, n \in \mathbb{N} : x \in F_n \,\} is the least index of a closed set containing xx (The well-ordering principle). Nothing is selected: n(x)n(x) is the least element of a set determined by xx and the fixed sequence (Fn)(F_n).

Why this is worth stating. Together with For f:ARf : A \to \mathbb{R} the set of points of AA at which ff is discontinuous is the intersection with AA of an FσF_\sigma subset of R\mathbb{R}, and the set of points at which ff is continuous is the intersection with AA of a GδG_\delta subset; for A=RA = \mathbb{R} the two sets are FσF_\sigma and GδG_\delta outright, which shows that a discontinuity set is always the trace of an FσF_\sigma set, and with Lebesgue's criterion for Riemann integrability: a bounded ff on [a,b][a,b] is Riemann integrable if and only if its set of discontinuities has measure zero, which shows that an integrable function has a null discontinuity set, the example says that the two necessary conditions are also jointly sufficient: null and FσF_\sigma is exactly what a discontinuity set of a Riemann integrable function on [0,1][0,1] can be. The Cantor set and any at most countable subset of [0,1][0,1] are instances.

Choice. The construction uses none; the only choice principle in the statement comes from the direction of Lebesgue's criterion for Riemann integrability: a bounded ff on [a,b][a,b] is Riemann integrable if and only if its set of discontinuities has measure zero used at the end, and that direction, "null \Rightarrow integrable", is a theorem of ZF.

Facts & Assumptions

Given: An FσF_\sigma set E[0,1]E \subseteq [0,1] of measure zero, and a sequence (Fn)nN(F_n)_{n \in \mathbb{N}} of closed subsets of R\mathbb{R} with E=nFnE = \bigcup_n F_n.

[L2]

Every nonempty subset of N\mathbb{N} has a least element (The well-ordering principle).

[L5]

Powers: 2n>02^{-n} > 0 for every nNn \in \mathbb{N}, 20=12^{0} = 1, 2n12^{-n} \le 1, and m<nm < n implies 2n<2m2^{-n} < 2^{-m} (Integer powers ama^m, Monotonicity of xxnx \mapsto x^n and of nann \mapsto a^n).

[L7]

For every real η>0\eta > 0 there is a natural N1N \ge 1 with 1/ι(N)<η1/\iota(N) < \eta, and 2N1/ι(N)2^{-N} \le 1/\iota(N) for N1N \ge 1, since ι(N)2N\iota(N) \le 2^{N} by induction (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Every complete ordered field is Archimedean, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing, Monotonicity of xxnx \mapsto x^n and of nann \mapsto a^n).

[L9]

Ordered-field arithmetic and the absolute value: the order is total and transitive; for 0x10 \le x \le 1 and a real ρ>0\rho > 0 the reals u:=max{0,xρ}u := \max\{0,x-\rho\} and v:=min{1,x+ρ}v := \min\{1,x+\rho\} satisfy u<vu < v and (u,v)Nρ(x)[0,1](u,v) \subseteq N_\rho(x)\cap[0,1]; a nonempty open interval is a nondegenerate interval (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Maximum and minimum of a set, Ordered field, Complete ordered field (least-upper-bound property), Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length, The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}, Interior, closure, boundary and exterior of a subset of R\mathbb{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). 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 · constructive
1.1

Fix the sequence (Fn)(F_n) of [L1] and define n(x):=min{nN:xFn}n(x) := \min\{\, n \in \mathbb{N} : x \in F_n \,\} for xEx \in E, which exists by [L2] since the set is a nonempty subset of N\mathbb{N}; define h:[0,1]Rh : [0,1] \to \mathbb{R} by h(x):=2n(x)h(x) := 2^{-n(x)} for xEx \in E and h(x):=0h(x) := 0 otherwise.

givenL1L2L5construct
2.1

hh is bounded with values in [0,1][0,1]: 0<2n(x)20=10 < 2^{-n(x)} \le 2^{0} = 1 for xEx \in E by [L5], and h=0h = 0 off EE.

step 1.1L5
2.2

hh is discontinuous at every point of EE. Let xEx \in E, so h(x)=2n(x)>0h(x) = 2^{-n(x)} > 0 by [L5], and let a real ρ>0\rho > 0 be given. By [L9] the interval (u,v)Nρ(x)[0,1](u,v) \subseteq N_\rho(x)\cap[0,1] is nonempty with u<vu < v, hence is a nondegenerate interval, so by [L4] it is not contained in EE: there is y(u,v)y \in (u,v) with yEy \notin E, and then h(y)=0h(y) = 0 and h(x)h(y)=2n(x)|h(x)-h(y)| = 2^{-n(x)}. So the continuity condition fails at xx for ε:=2n(x)\varepsilon := 2^{-n(x)}.

step 1.1L4L5L9
2.3

hh is continuous at every point of [0,1]E[0,1]\setminus E. Let x[0,1]x \in [0,1] with xEx \notin E, so h(x)=0h(x) = 0, and let a real ε>0\varepsilon > 0 be given. By [L7] fix a natural N1N \ge 1 with 2N1/ι(N)<ε2^{-N} \le 1/\iota(N) < \varepsilon. For each nNn \le N one has xFnx \notin F_n, since FnEF_n \subseteq E, so [L3] supplies a real ρn>0\rho_n > 0 with Nρn(x)Fn=N_{\rho_n}(x) \cap F_n = \varnothing; put ρ:=min{ρ0,,ρN}>0\rho := \min\{\rho_0,\dots,\rho_N\} > 0, which exists by [L6].

step 1.1L3L6L7L9choose
3.1

For y[0,1]y \in [0,1] with yx<ρ|y - x| < \rho: if yEy \notin E then h(y)=0h(y) = 0; and if yEy \in E then yFny \notin F_n for every nNn \le N by step 2.3, so n(y)>Nn(y) > N and h(y)=2n(y)<2N<εh(y) = 2^{-n(y)} < 2^{-N} < \varepsilon by [L5] and step 2.3. In both cases h(y)h(x)=h(y)<ε|h(y) - h(x)| = h(y) < \varepsilon, so hh is continuous at xx.

step 1.1step 2.3L5L9
4.1

By steps 2.2 and 3.1 the set of discontinuities of hh in [0,1][0,1] is exactly EE, which has measure zero by hypothesis; hh is bounded by step 2.1 and 0<10 < 1, so [L8] gives that hh is Riemann integrable on [0,1][0,1]. The function hh constructed in step 1.1 therefore has all the stated properties.

step 2.1step 2.2step 3.1givenL8discharge-construct

Remarks

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

For the Dirichlet function every uniform partition with rational tags gives Riemann sum 11, 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

Statement refuted

Refuted: that a bounded ff on [a,b][a,b] is Riemann integrable with integral II as soon as there is one sequence (PN,ξN)(P_N,\xi^N) of tagged partitions with PN0\|P_N\| \to 0 and S(f,PN,ξN)IS(f,P_N,\xi^N) \to I (Tagged partitions of [a,b][a,b], with a tag ξi\xi_i in each subinterval, and the Riemann sum S(f,P,ξ)=if(ξi)ΔiS(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f).

The witness is again the Dirichlet function on [0,1][0,1] (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx). Take PN:=UNP_N := U_N, the uniform partition into N1N \ge 1 parts, and tag each subinterval by its left endpoint ξiN:=ι(i)/ι(N)\xi^N_i := \iota(i)/\iota(N), a rational. Then

UN=1ι(N)0,S(1Q,UN,ξN)=1  for every N1,\|U_N\| = \frac{1}{\iota(N)} \longrightarrow 0, \qquad S(\mathbf{1}_{\mathbb{Q}}, U_N, \xi^N) = 1 \ \text{ for every } N \ge 1,

so the Riemann sums converge, to 11; and yet 1Q\mathbf{1}_{\mathbb{Q}} is not Riemann integrable on [0,1][0,1] (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).

What this shows about The Darboux and Riemann definitions agree: a bounded ff on [a,b][a,b] is Darboux integrable with integral II if and only if for every real ε>0\varepsilon > 0 there is a real δ>0\delta > 0 such that S(f,P,ξ)I<ε|S(f,P,\xi) - I| < \varepsilon for every tagged partition of mesh below δ\delta. Condition 2 there quantifies over every tagged partition of mesh below δ\delta, tags included, and the quantifier cannot be replaced by the existence of one good sequence. Tagging the very same partition by irrationals instead gives Riemann sum 00, so for each NN the two taggings of UNU_N give the two values 11 and 00; no real number is within 1/21/2 of both.

Facts & Assumptions

Given: g:[0,1]Rg : [0,1] \to \mathbb{R} with g(x)=1g(x) = 1 for rational xx and g(x)=0g(x) = 0 for irrational xx; for N1N \ge 1 the uniform partition UN=(N,t)U_N = (N,t) of [0,1][0,1] with ti=ι(i)/ι(N)t_i = \iota(i)/\iota(N) and Δi=1/ι(N)\Delta_i = 1/\iota(N); and the tagging ξN\xi^N with ξiN:=ti\xi^N_i := t_i for i<Ni < N.

[A1]

The refuted claim: if some sequence of tagged partitions of [0,1][0,1] has meshes tending to 00 and Riemann sums tending to II, then gg is Riemann integrable with integral II.

[L1]

For UNU_N: ti=ι(i)/ι(N)[0,1]t_i = \iota(i)/\iota(N) \in [0,1], ti<ti+1t_i < t_{i+1}, Δi=1/ι(N)\Delta_i = 1/\iota(N), i<NΔi=1\sum_{i<N}\Delta_i = 1, and UN=1/ι(N)\|U_N\| = 1/\iota(N) (Partition of [a,b][a,b] as a finite strictly increasing list a=t0<t1<<tn=ba = 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\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L2]

ξiN:=tiIi\xi^N_i := t_i \in I_i defines a tagging of UNU_N, and S(g,UN,ξN)=i<Ng(ξiN)ΔiS(g,U_N,\xi^N) = \sum_{i<N}g(\xi^N_i)\Delta_i (Tagged partitions of [a,b][a,b], with a tag ξi\xi_i in each subinterval, and the Riemann sum S(f,P,ξ)=if(ξi)ΔiS(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i).

[L4]

Finite sums: scaling and i<N1Δi=i<NΔi\sum_{i<N}1 \cdot \Delta_i = \sum_{i<N}\Delta_i (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

A constant sequence converges to its value; and for every real η>0\eta > 0 there is N1N \ge 1 with 1/ι(N)<η1/\iota(N) < \eta, so the sequence N1/ι(N)N \mapsto 1/\iota(N) converges to 00, its terms being positive and decreasing below every positive bound (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, Basic properties of the absolute value).

[L9]

Ordered-field arithmetic: the order is total and transitive, ι(N)>0\iota(N) > 0 for N1N \ge 1, and 010 \ne 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)). 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

For each N1N \ge 1, (UN,ξN)(U_N,\xi^N) is a tagged partition of [0,1][0,1] of mesh 1/ι(N)1/\iota(N), by [L1] and [L2].

givenL1L2
1.2

By [L3] every tag ξiN=ti\xi^N_i = t_i is rational, so g(ξiN)=1g(\xi^N_i) = 1; hence by [L2], [L4] and [L1], S(g,UN,ξN)=i<N1Δi=i<NΔi=1S(g,U_N,\xi^N) = \sum_{i<N}1\cdot\Delta_i = \sum_{i<N}\Delta_i = 1.

givenL1L2L3L4
2.1

The sequence NUN=1/ι(N)N \mapsto \|U_N\| = 1/\iota(N) converges to 00 and the sequence NS(g,UN,ξN)N \mapsto S(g,U_N,\xi^N) is constantly 11, hence converges to 11, by [L5].

step 1.1step 1.2L5
3.1

So the hypothesis of [A1] is met with I:=1I := 1; but gg is not Riemann integrable on [0,1][0,1] by [L6]. [A1] is therefore refuted.

step 2.1A1L6
4.1

Moreover, for each N1N \ge 1 the same partition carries a tagging whose Riemann sum is 00: by [L7] each open interval (ti,ti+1)(t_i,t_{i+1}) contains an irrational, and choosing one in each of the NN subintervals is a finite selection, giving a tagging ζ\zeta of UNU_N with g(ζi)=0g(\zeta_i) = 0 for every i<Ni < N and hence S(g,UN,ζ)=0S(g,U_N,\zeta) = 0 by [L2] and [L4]. So for every real δ>0\delta > 0 there are tagged partitions of mesh below δ\delta with Riemann sum 11 and others with Riemann sum 00, and by [L9] no single real II can satisfy the condition of [L8] at ε:=21\varepsilon := 2^{-1}.

step 1.1step 1.2L2L4L5L7L8L9

Remarks

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28Open item page →

Thomae's function is nonnegative, Riemann integrable on [0,1][0,1] with integral 00, and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish

Statement refuted

Refuted: that a nonnegative Riemann integrable function on [a,b][a,b] with abf=0\int_a^b f = 0 vanishes identically (FALSE: a nonnegative Riemann integrable function on [a,b][a,b] with abf=0\int_a^b f = 0 is identically zero, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f).

The witness is Thomae's function tt on [0,1][0,1] (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx). It satisfies 0t10 \le t \le 1, it is Riemann integrable with 01t=0\int_0^1 t = 0 (Thomae's function is Riemann integrable on [0,1][0,1] with integral 00: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is 00), and it is positive at every rational point of [0,1][0,1] — a dense set, and an infinite one. So the failure is not at a single stray point: the function is nonzero on a dense subset of the interval and the integral still vanishes.

The repaired statement asks for continuity. With ff continuous the conclusion is true, and the reason is exactly what fails here: a continuous function positive at one point is positive on a whole subinterval, whereas tt is positive only on a set that contains no interval (Both Q\mathbb{Q} and RQ\mathbb{R} \setminus \mathbb{Q} are dense in R\mathbb{R}, and every nonempty open subset of R\mathbb{R} is uncountable). That repaired statement is not proved here, since the additivity of the integral over subintervals is not available at this point in the reading order.

Facts & Assumptions

Given: Thomae's function t:[0,1]Rt : [0,1] \to \mathbb{R}, with t(x)=1/ι(q(x))t(x) = 1/\iota(q(x)) at a rational xx of least denominator q(x)1q(x) \ge 1 and t(x)=0t(x) = 0 at an irrational xx (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field).

[A1]

The refuted claim: a nonnegative Riemann integrable function on a closed bounded interval with distinct endpoints whose integral is 00 vanishes identically.

[L3]

tt is discontinuous at every rational point and continuous at every irrational point, so its discontinuity set in [0,1][0,1] is Q[0,1]\mathbb{Q}\cap[0,1] (The Dirichlet function is continuous at no point of R\mathbb{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 cc equals t(c)t(c)).

[L5]

Ordered-field arithmetic: 0<21<10 < 2^{-1} < 1, so 212^{-1} lies in [0,1][0,1] and is rational; 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), Intervals of R\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

tt is nonnegative on [0,1][0,1] by [L1], and it is Riemann integrable there with 01t=0\int_0^1 t = 0 by [L2]; the interval [0,1][0,1] has 0<10 < 1 by [L5].

givenL1L2L5
1.2

tt does not vanish identically: 212^{-1} is a rational point of [0,1][0,1] by [L5], so t(21)>0t(2^{-1}) > 0 by [L1].

givenL1L5
2.1

The hypotheses of [A1] hold for tt on [0,1][0,1] and its conclusion fails, so [A1] is refuted.

step 1.1step 1.2A1
3.1

The failure is dense, not isolated: by [L4] every nonempty open subinterval of [0,1][0,1] contains a rational, at which tt is positive by [L1]; so {x[0,1]:t(x)>0}\{\, x \in [0,1] : t(x) > 0 \,\} meets every subinterval of [0,1][0,1] with distinct endpoints. It is also exactly the set of discontinuities of tt, by [L3] and [L1].

step 1.2L1L3L4

Remarks

Sources