Equivalence of Riemann and Darboux Integrals

Every calculus student meets two seemingly different definitions of "the integral." One says a function is integrable if Riemann sums converge to a single number. These sums are built from tagged partitions, and convergence happens as the mesh shrinks. The other says a function is integrable if the supremum of lower sums equals the infimum of upper sums. This second definition uses no tags and no limiting mesh at all. Students are told these agree, usually without proof, and move on. But the reconciliation is not a triviality. It is the moment where two independently plausible formalizations of "area under a curve" are shown to be logically identical. It is worth doing carefully. The stakes are higher than bookkeeping. The Darboux Integral is technically easier to work with. It reduces integrability to a single inequality about suprema and infima. Its proofs (linearity, additivity, the fundamental theorem of calculus) are cleaner. There is no limit over tagged partitions to control. The Riemann Sums and the Riemann Integral definition, by contrast, is the one that generalizes. Riemann sums with tags are what you differentiate. They are also what physicists actually compute when they chop a region into pieces and pick sample points. Knowing the two definitions coincide lets you use whichever is more convenient at each stage of an argument. You need not fear a hidden gap. The proof has two directions, and they are not symmetric in difficulty. Showing that Darboux integrability implies Riemann integrability is the harder direction. Given that upper and lower Darboux sums can be squeezed together, more must be shown. One must show that every tagged Riemann sum, regardless of how the tags are chosen, is forced close to the same limit. This holds once the mesh is small enough. The converse — Riemann integrability implies Darboux integrability — is comparatively short. It exploits the fact that upper and lower sums are suprema and infima of Riemann sums over all possible tag choices. This article gives complete, self-contained proofs of both directions. It builds on the Cauchy Integrability Criterion for the Darboux side. The result is the theorem that licenses every calculus textbook's silent switching. It moves between " $\lim \sum f(t_i)\Delta x_i$ " and " $\sup L(f,P) = \inf U(f,P)$ ." It also sets up the transition to Lebesgue's criterion. There, a third, measure-theoretic characterization of Riemann integrability enters the picture.

Key takeaways

Two definitions of the integral exist: Riemann's (limit of tagged sums as mesh $\to 0$ ) and Darboux's (sup of lower sums = inf of upper sums).
The theorem states: $f$ is Riemann integrable on $[a,b]$ iff $f$ is Darboux integrable on $[a,b]$ , and the values agree.
Darboux $\Rightarrow$ Riemann is the subtle direction: it requires bounding the effect of the endpoints of a fine partition on any coarser tagged sum.
Riemann $\Rightarrow$ Darboux is the easy direction: upper and lower sums are sup/inf of Riemann sums over tag choices, so they inherit the same limit.
Both directions require $f$ bounded on $[a,b]$ ; unbounded functions break the Darboux definition entirely (sums become $\pm\infty$ ).
Key idea: controlling how many subintervals of a fine partition can straddle the endpoints of a coarser one converts "sup/inf agree" into "every tagged sum is close," and vice versa.

Where this lives today

The Darboux formulation is the workhorse behind essentially every rigorous proof in an introductory real analysis course: linearity of the integral, additivity over subintervals (Basic Properties: Linearity and Additivity), integrability of continuous and monotonic functions, and the Fundamental Theorem of Calculus all have their cleanest proofs phrased in terms of $U(f,P)$ and $L(f,P)$ . The equivalence theorem is what licenses transferring these results back to statements about Riemann sums, which is the form used whenever an integral is computed via a limit of sample sums — in numerical quadrature, in physics derivations of work, mass, or center-of-mass as limits of Riemann sums, and in the definition of the integral that most students actually see first. The theorem also serves as a template for later, harder equivalences in the theory of integration: the equivalence between Riemann integrability and Lebesgue's characterization via the set of discontinuities having measure zero (Lebesgue's Criterion for Riemann Integrability) uses a similar strategy of comparing a partition-based quantity to a set-theoretic one. In numerical analysis, Riemann sums with specific tag rules (left, right, midpoint, Simpson's rule) are the starting point for quadrature algorithms, and the Darboux perspective supplies the error bounds that guarantee convergence. In probability and stochastic calculus, Itô and Stratonovich integrals generalize the tagged-sum idea to random integrators, where — strikingly — the choice of tag does affect the value, in sharp contrast to the classical setting where this theorem guarantees tag-independence for any Riemann-integrable function.

:::quiz[Why does the function $f(x)=1/x$ on $(0,1]$ (extended by $f(0)=0$ ) fail to satisfy the theorem's hypotheses?]

It is discontinuous at countably many points.
It fails to be monotonic.
It is unbounded near $x=0$ , so upper Darboux sums are $+\infty$ for every partition. ✓
It takes only rational values.

Ask the tutor: Can you show me how the "Class A / Class B" bookkeeping argument would change if the coarse partition $P_0$ has a point where $f$ has a jump discontinuity exactly at an interior grid point?

f:[a,b]\to\mathbb{R}

Intuition

Picture the graph of a bounded function over $[a,b]$ . The Darboux approach boxes the region between two staircases: a lower staircase built from the infima $m_i$ on each subinterval, always below the graph, and an upper staircase built from suprema $M_i$ , always above it. Refining the partition can only push the lower staircase up and the upper staircase down, squeezing the true area from both sides. Integrability means this squeeze has no gap left at the end. The Riemann approach instead samples the function at a single point $t_i$ in each subinterval and adds up rectangles of height $f(t_i)$ . Because $t_i$ can be any point of $[x_{i-1},x_i]$ , a Riemann sum can realize any value between $L(f,P)$ and $U(f,P)$ for a fixed partition — actually it can get arbitrarily close to $m_i$ or $M_i$ on each piece by choosing $t_i$ appropriately, though it may not exactly hit the sup or inf if they are not attained. The theorem says that once you also let the mesh shrink to zero, "some tag choice traps the sum near $L$ " and "every tag choice traps the sum near $U(f,P)=L(f,P)$ 's common limit" become the same statement.

Figure

Lower and upper staircases sandwich every tagged Riemann sum

f:[a,b]\to\mathbb{R}

f

P=\{x_0<\dots<x_n\}

P'\supseteq P

P'

We also record, without reproving it here, the standard reformulation of Darboux integrability that we import from the Cauchy Integrability Criterion:

f:[a,b]\to\mathbb{R}

Now we prove the main theorem, in two directions.

\Rightarrow

Worked examples

Example 1: $f(x) = x^2$ on $[0,3]$ , using a non-uniform partition. We verify Darboux integrability directly and see the equivalence theorem in action, using an asymmetric partition rather than the usual uniform one. Take $P_n$ to be the partition of $[0,3]$ into $n$ subintervals defined by $x_i = 3(i/n)$ for $i=0,\dots,n$ (uniform in this case for computability, but we will illustrate the non-uniform straddling argument numerically). Since $f(x)=x^2$ is increasing on $[0,3]$ ,

M_i = x_i^2 = \left(\frac{3i}{n}\right)^2, \qquad m_i = x_{i-1}^2=\left(\frac{3(i-1)}{n}\right)^2.

Then

U(f,P_n) = \sum_{i=1}^n \left(\frac{3i}{n}\right)^2 \cdot \frac{3}{n} = \frac{27}{n^3}\sum_{i=1}^n i^2 = \frac{27}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}.

As $n\to\infty$ , this tends to $\frac{27}{6}\cdot 2 = 9$ . Similarly $L(f,P_n) \to 9$ . Since $U(f,P_n)-L(f,P_n) = \frac{27}{n^3}\left(\sum i^2 - \sum (i-1)^2\right) = \frac{27}{n^3}\cdot n^2 = \frac{27}{n} \to 0$ , the Cauchy criterion is satisfied for $n$ large, confirming $f$ is Darboux integrable with $\int_0^3 x^2\,dx = 9$ . Now invoke the equivalence theorem (Direction 1): because $f$ is Darboux integrable with value $9$ , every Riemann sum with mesh below an appropriate $\delta(\varepsilon)$ is within $\varepsilon$ of $9$ — including wildly asymmetric tag choices. For concreteness, take $n=4$ , so $P_4 = \{0, 3/4, 3/2, 9/4, 3\}$ , and choose the tags not at midpoints or endpoints but at, say, $t_1=0.1$ , $t_2=1.4$ , $t_3=1.6$ , $t_4=2.9$ (each inside its respective subinterval $[0,0.75],[0.75,1.5],[1.5,2.25],[2.25,3]$ ). Then

S(f,P_4,\mathbf t) = (0.1)^2(0.75)+(1.4)^2(0.75)+(1.6)^2(0.75)+(2.9)^2(0.75).

Computing: $0.01\cdot0.75 = 0.0075$ ; $1.96\cdot0.75=1.47$ ; $2.56\cdot0.75=1.92$ ; $8.41\cdot0.75=6.3075$ . Sum: $0.0075+1.47+1.92+6.3075 = 9.705$ . Compare to $U(f,P_4) = (0.75)^2\cdot0.75+(1.5)^2\cdot0.75+(2.25)^2\cdot0.75+3^2\cdot0.75 = 0.421875+1.6875+3.796875+6.75=12.65625$ and $L(f,P_4)=0+ (0.75)^2(0.75)+(1.5)^2(0.75)+(2.25)^2(0.75) = 0.421875+1.6875+3.796875 = 5.90625$ . Indeed $5.906 \le 9.705 \le 12.656$ , consistent with the Sandwich Lemma, and as $n$ grows this asymmetric-tag sum will be forced toward $9$ regardless of how erratically the tags are chosen — that is the content of Direction 1.

Example 2: A function integrable "by inspection" of Darboux sums where a naive Riemann-sum limit is not obviously well-defined a priori. Let $f:[0,1]\to\mathbb{R}$ be defined by

f(x) = \begin{cases} 1 & x = 1/n \text{ for some } n\in\mathbb{N} \\ 0 & \text{otherwise.} \end{cases}

This function is discontinuous at infinitely many points $\{1, 1/2, 1/3, \dots\}$ accumulating at $0$ , so it is not at all obvious from the Riemann-sum definition alone that a single limiting value $I$ exists — one must worry that tags landing exactly on some $1/n$ could perturb sums unpredictably. The Darboux approach settles this cleanly. For any partition $P=\{0=x_0<\dots<x_n=1\}$ , every subinterval $[x_{i-1},x_i]$ with $x_{i-1}>0$ contains only finitely many points of the form $1/k$ (since these accumulate only at $0$ ), but may contain one; and $f\ge 0$ everywhere with $f=0$ off a countable set, so $m_i = 0$ for every $i$ (since every subinterval, having positive length, contains irrational points, or more simply points not of the form $1/n$ , where $f=0$ ; and it also may contain a point where $f=1$ , but $\inf$ over the interval is still $0$ since $0$ values are present and $f\ge0$ ). Hence $L(f,P)=0$ for every $P$ , so $\underline\int_0^1 f = 0$ . For the upper sum: given $\varepsilon>0$ , choose $N$ so large that $1/N < \varepsilon/2$ . There are exactly $N$ points of the form $1/n$ with $1/n \ge \varepsilon/2$ lying in $[\varepsilon/2, 1]$ (namely $n=1,\dots, \lfloor 2/\varepsilon\rfloor$ , a finite set), plus infinitely many accumulating in $(0,\varepsilon/2)$ . Build a partition $P$ as follows: put one small subinterval of length $\eta$ around each of the finitely many points $1/n \ge \varepsilon/2$ where $\eta$ is chosen small enough that the total length of these "spike" intervals is $< \varepsilon/2$ , and let $[0,\varepsilon/2)$ be covered by a single subinterval $[0,\varepsilon/2]$ (on which $M_i \le 1$ trivially, contributing at most $\varepsilon/2$ to $U(f,P)$ ), with the remaining subintervals covering the rest of $[\varepsilon/2,1]$ avoiding all spike points, on which $M_i=0$ . Then

U(f,P) \le 1\cdot\frac\varepsilon2 \;(\text{from } [0,\varepsilon/2]) \;+\; 1\cdot\frac\varepsilon2\;(\text{total spike length}) \;+\;0 = \varepsilon.

Since $\varepsilon>0$ was arbitrary, $\overline\int_0^1 f = \inf_P U(f,P) = 0$ . Thus $\overline\int_0^1 f = \underline\int_0^1 f = 0$ , so $f$ is Darboux integrable with integral $0$ . By the equivalence theorem, $f$ is also Riemann integrable with Riemann integral $0$ : for any $\varepsilon>0$ , once the mesh is fine enough (per the $\delta$ constructed in Direction 1's proof, built from a Darboux partition like the one above with $U(f,P_0)-L(f,P_0)<\varepsilon/2$ ), every tagged Riemann sum — even one that deliberately tags every subinterval at a point $1/n$ where possible — lands within $\varepsilon$ of $0$ . This is a genuinely useful conclusion that would be awkward to establish by wrestling with Riemann sums directly, illustrating why one proves Darboux integrability and imports Riemann integrability via the theorem, rather than the reverse.

f

Scope, edge cases & common traps

Situation	What happens
$f$ bounded, finitely many discontinuities	Both integrals exist and agree (special case of the theorem, also provable directly)
$f$ bounded, countably many discontinuities accumulating (Example 2)	Both integrals exist and agree; Darboux route is far easier to verify
$f$ unbounded on $[a,b]$	Neither Riemann nor Darboux integral is defined; this is improper integral territory, outside this theorem
$f = \mathbf{1}_{\mathbb{Q}}$ (Dirichlet function) on $[0,1]$	$U(f,P)=1$ , $L(f,P)=0$ for every $P$ ; neither integral exists — consistent, not a counterexample
Tags chosen only at right endpoints (a common simplification)	Gives a valid Riemann sum but does not by itself establish Riemann integrability, which requires all tag choices to work

Common trapThe quantifier order is the whole theorem

The single most common error is conflating "there exists a tag choice for which $S(f,P,\mathbf t)$ converges" with Riemann integrability, which demands convergence for every tag choice as the mesh shrinks. The Darboux upper and lower sums are precisely the extreme values over all tag choices at fixed $P$ ( $\sup$ and $\inf$ , approximated arbitrarily closely by actual tag choices as in Direction 2's proof). Riemann integrability is the strictly stronger statement that these extremes are forced together not just at one partition but uniformly as $\|P\|\to0$ . Reversing "for every $\delta$ there's a tag" and "for every tag there's a $\delta$ " produces a false, much weaker notion that would make far more functions "integrable."

The mathematician behind it

Origins · 1 minBernhard Riemann and how this was discovered

Bernhard Riemann

1826–1866 · Göttingen, mid-19th century analysis

Riemann introduced his integral in his 1854 habilitation thesis on trigonometric series, seeking the most general class of functions representable by Fourier series. He needed a rigorous notion of integrability that did not presuppose continuity, since Fourier series can converge to badly discontinuous functions. His definition, in terms of sums over arbitrary tagged partitions, was the first fully rigorous treatment of integration for a general class of bounded functions, decades before measure theory existed.

Gaston Darboux, working roughly two decades later, recast Riemann's integral using upper and lower sums built from suprema and infima rather than sampled function values. His formulation, appearing in an 1875 memoir, made proofs of the basic properties of the integral—linearity, monotonicity, the fact that continuous functions are integrable—dramatically more transparent, and it is Darboux's version, not Riemann's original tagged-sum definition, that dominates most modern analysis textbooks under the name "the Riemann integral."

Origins · 1 minhow this was discovered

History

- 1854 — Riemann's habilitation thesis defines the tagged-partition sum criterion for integrability, motivated by Fourier series.

- 1868 — Riemann's thesis is published posthumously, two years after his death, making the definition widely accessible.

- 1875 — Darboux introduces upper and lower sums and proves they characterize the same class of functions, streamlining the theory.

- 1881 — Darboux's full treatise on discontinuous functions cements the upper/lower sum approach as the standard technical tool.

- 1902 — Lebesgue's thesis extends integration beyond bounded functions and gives a measure-theoretic characterization of Riemann integrability itself.

Do the Riemann and Darboux definitions ever give different answers for the same function?

No. The theorem proved here shows they are logically equivalent: a bounded function is integrable under one definition if and only if it is integrable under the other, and when both apply, they produce the same numerical value for the integral. There is no bounded function that is Riemann integrable but not Darboux integrable, or vice versa. The two formalisms are different routes to literally the same mathematical object — this is why textbooks freely call the common result "the Riemann integral" while using Darboux's upper/lower sum machinery to prove its properties.

Why bother with two definitions if they're equivalent?

Each definition is suited to different tasks. The Darboux definition is far easier to use when proving general theorems (linearity, additivity, integrability of continuous or monotonic functions), because it reduces to comparing suprema and infima rather than controlling a limit over all possible tag choices. The Riemann definition, meanwhile, is the one that matches physical and computational intuition — sampling a function at points and summing — and it is the definition that generalizes to stochastic integrals, where tag choice genuinely matters. Knowing they coincide lets you prove theorems the easy (Darboux) way and then legitimately apply them in the Riemann-sum language used in applications.

Does the equivalence theorem require

f

to be continuous?

No. The theorem applies to any bounded function on $[a,b]$ , continuous or not. Continuity is not assumed; it is a sufficient (not necessary) condition for integrability, proved as a separate, later result using the Darboux criterion (uniform continuity controls the oscillation $M_i - m_i$ on fine partitions). The equivalence theorem itself is prior to and independent of any discussion of which functions are integrable — it simply says that whichever bounded functions turn out to be integrable, both definitions select the same class and assign the same value.

What role does the Cauchy Integrability Criterion play in this proof?

It supplies the key reformulation of Darboux integrability used in Direction 1: instead of working directly with $\sup_P L(f,P) = \inf_P U(f,P)$ , we use the equivalent statement that for every $\varepsilon>0$ there is a single partition $P_0$ with $U(f,P_0)-L(f,P_0)<\varepsilon$ . This finite, concrete partition is exactly what lets us bound arbitrary fine tagged partitions by comparing them to $P_0$ via the "Class A / Class B" splitting argument. Without this criterion, one would have to work directly with suprema and infima over the (uncountable) set of all partitions, which is far more awkward.

Is one direction of the proof really harder than the other, or is that just a rhetorical framing?

It is genuinely harder, not just rhetorically so. Riemann-implies-Darboux only needs to exhibit particular tag choices (via the definition of sup/inf) that approximate $U(f,P)$ and $L(f,P)$ arbitrarily well — a constructive, one-partition argument. Darboux-implies-Riemann must show a uniform statement: for a fixed small mesh, every tag choice on every sufficiently fine partition is trapped near the limit. This requires the delicate bookkeeping of how many subintervals of a fine partition can straddle the finitely many breakpoints of a coarser, "good" Darboux partition — the Class A/Class B argument — which has no analogue in the other direction.

Does this theorem extend to unbounded intervals or unbounded functions?

Not directly. Both definitions as stated require $f$ bounded on a bounded interval $[a,b]$ ; the counterexample above shows the Darboux upper sum becomes $+\infty$ immediately if $f$ is unbounded, and the Riemann sum definition also fails since tag values can be forced arbitrarily large. Extensions to unbounded functions or infinite intervals are handled by improper Riemann integrals, defined as limits of proper integrals over $[a, b-\eta]$ or $[a,R]$ as $\eta\to 0$ or $R\to\infty$ , which is a separate construction layered on top of, not covered by, this equivalence theorem.