The Riemann Integral: Definition and Integrability: Examples and Counterexamples
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1 · Prerequisites
Abstract Algebra
Real Analysis
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Continuity, IVT, EVT, and Uniform Continuity
- Countability and Uncountability
- Foundations of the Real Numbers for Analysis
- Limits of Real Functions
- limsup, liminf, and Subsequential Limits
- Monotone Functions, Discontinuities, and Continuity Sets
- Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness
- Roots, Rational Powers, and Classical Inequalities
- Sequences and Limits
- Series: Convergence and the Nonnegative Tests
- Suprema and Infima
- The Cantor Set, Baire Category, and Measure Zero in ℝ
- The Riemann Integral: Definition and Integrability
- Topology of ℝ
Set Theory
2 · Logical flowchart
3 · Items
Example
∫₀¹ x² = 1/3, computed from the Darboux definition with uniform partitions and the closed form ∑_k<n k² = n(n-1)(2n-1)/6
Example
One refinement worked out for f(x) = x² on [0,1]: adding the point 1/2 to the trivial partition raises the lower sum from 0 to 1/8 and lowers the upper sum from 1 to 5/8
Example
∫₀³ ⌊ x ⌋ = 3: the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions
Example
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
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
Counterexample
The Dirichlet function on [0,1] has lower Darboux integral 0 and upper Darboux integral 1, so it is bounded and not Riemann integrable
Counterexample
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
Example
For every F_σ subset E of [0,1] of measure zero there is a bounded Riemann integrable function on [0,1] whose set of discontinuities is exactly E
Counterexample
For the Dirichlet function every uniform partition with rational tags gives Riemann sum 1, 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
Counterexample
Thomae's function is nonnegative, Riemann integrable on [0,1] with integral 0, and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish
Sources
- Riemann integral (Wikipedia)
- Square pyramidal number (Wikipedia)
- J. Hunter, Chapter 11: The Riemann Integral
- Darboux integral (Wikipedia)
- J. Lebl, Basic Analysis I, The Riemann Integral
- Floor and ceiling functions (Wikipedia)
- MTH 421 Homework 1 (Michigan State University)
- Cantor set (Wikipedia)
- MAT425 Lecture Notes (Princeton University)
- Thomae's function (Wikipedia)
- MAT 125B Discussion 3 (UC Davis)
- Dirichlet function (Wikipedia)
- MIT 18.013A, Nonintegrable Functions
- Smith-Volterra-Cantor set (Wikipedia)
- Fsigma set (Wikipedia)
- M. Wodzicki, The Riemann Integral
- Sets of discontinuity (University of Richmond MATH 320)
- Riemann sum (Wikipedia)