The Riemann integrability criterion

Statement

Let \( a, b \in \mathbb{R} \) with \( a \lt b \), and let \( f : [a,b] \to \mathbb{R} \) be bounded. For a partition \( P : a = x_0 \lt x_1 \lt \cdots \lt x_n = b \) write \( \Delta x_i = x_i - x_{i-1} \), \( m_i = \inf\{ f(x) : x \in [x_{i-1}, x_i] \} \), \( M_i = \sup\{ f(x) : x \in [x_{i-1}, x_i] \} \), and form the lower and upper Darboux sums \( L(f,P) = \sum_{i=1}^{n} m_i \, \Delta x_i \) and \( U(f,P) = \sum_{i=1}^{n} M_i \, \Delta x_i \). Then \( f \) is Riemann (Darboux) integrable on \( [a,b] \) — that is, \( \sup_P L(f,P) = \inf_P U(f,P) \), the suprema and infima taken over all partitions of \( [a,b] \) — if and only if \[ \forall \varepsilon \gt 0 \;\; \exists \text{ a partition } P_\varepsilon \text{ of } [a,b] \text{ such that } U(f,P_\varepsilon) - L(f,P_\varepsilon) \lt \varepsilon . \]

Why it matters

The definition of the Darboux integral asks that two infinite optimisation problems — a supremum over all lower sums and an infimum over all upper sums — happen to agree. That is almost never checkable directly. The criterion replaces it with a single finitary test: exhibit, for each tolerance \( \varepsilon \), one partition on which the upper and lower staircases squeeze to within \( \varepsilon \). Nearly every integrability proof in a first analysis course — continuous functions, monotone functions, Thomae's function, products and absolute values of integrable functions — is really an application of this criterion.

Conceptually, it recasts integrability as a statement about oscillation: since \( U(f,P) - L(f,P) = \sum_i (M_i - m_i)\,\Delta x_i \), the function is integrable exactly when its total oscillation, weighted by interval length, can be made small. This is the germ of Lebesgue's later characterisation of Riemann integrability via sets of measure zero, and the reason the criterion sits at the crossroads of the space (area, measure) and change (accumulation) threads.

Hypotheses

Proof

Result

Corollaries & converses

• Continuous \( \Rightarrow \) integrable. A continuous \( f \) on \( [a,b] \) is uniformly continuous (Heine–Cantor), so a partition of mesh \( \delta \) gives \( M_i - m_i \lt \varepsilon/(b-a) \) on each cell; the criterion applies.
• Monotone \( \Rightarrow \) integrable. Worked Example 2 below — no continuity needed, and monotone functions may have countably many jumps.
• Sequential form. \( f \in \mathcal{R}[a,b] \) iff there is a sequence of partitions \( (P_n) \) with \( U(f,P_n) - L(f,P_n) \to 0 \); then \( \int_a^b f = \lim_n U(f,P_n) = \lim_n L(f,P_n) \) (Problem 3).
• Stability of \( \mathcal{R}[a,b] \). Via the oscillation estimate \( \sup_S |f| - \inf_S |f| \le \sup_S f - \inf_S f \) and its analogues, integrability of \( f \) (and \( g \)) yields integrability of \( |f| \), \( f^2 \), \( fg \), \( \max(f,g) \) — each proof is a one-line application of the criterion (Problem 4).
• Converse. The theorem is a biconditional, so the converse is part of the statement. What genuinely fails is the quantifier-strengthened variant "\( U - L \lt \varepsilon \) for every partition of small mesh implies nothing new" — in fact that stronger-looking condition (Darboux's theorem on mesh) is equivalent to the criterion, but proving it requires a separate argument, not a formal consequence.
• What the criterion does not give. It does not compute \( \int_a^b f \); it only certifies existence. Computation needs limits of specific sums or the Fundamental Theorem of Calculus.

Fails without

Common errors

• Quantifier flip. Writing "for all partitions \( P \), \( U(f,P) - L(f,P) \lt \varepsilon \)". The criterion demands one witness partition per \( \varepsilon \). The for-all version is false as stated (coarse partitions of an integrable function can have huge \( U - L \)).
• Assuming \( M_i, m_i \) are attained. \( M_i \) is a supremum, not necessarily a maximum — \( f \) need not be continuous. Arguments that "pick \( t_i \) with \( f(t_i) = M_i \)" are invalid in general; use the approximation property of \( \sup \) instead.
• Endpoint substitution. Writing \( M_i - m_i = f(x_i) - f(x_{i-1}) \). This is only legitimate for monotone increasing \( f \); in general the oscillation on a cell has nothing to do with the endpoint values.
• Comparing incomparable partitions. In the forward direction, concluding \( U(f,P_1) - L(f,P_2) \lt \varepsilon \) and stopping. The sums belong to different partitions; one must pass to the common refinement \( P_1 \cup P_2 \) before subtracting sums of the same partition.
• "Small \( U - L \) means \( f \) is nearly continuous." False: Thomae's function is discontinuous at every rational yet satisfies the criterion (Problem 5 territory); the criterion measures aggregate oscillation, not pointwise regularity.
• Uniform partitions as a hidden assumption. The witness \( P_\varepsilon \) may need unequal cells (e.g. tiny cells around finitely many bad points and coarse cells elsewhere); restricting to uniform partitions can make proofs needlessly hard or impossible to finish.

Discussion

Riemann stated the essence of this criterion in his 1854 Göttingen Habilitationsschrift on trigonometric series — the integral was for him a tool to ask which functions are representable by Fourier series, and he needed a definition wide enough to admit badly discontinuous functions. Cauchy had integrated only continuous functions; Riemann's condition (total contribution of cells with oscillation above any threshold can be made small) widened the class dramatically. Darboux's 1875 reformulation via upper and lower sums, the version proved here, made the logic transparent: integrability is the statement that a single number is squeezed between two monotone families of approximations.

The right invariant hiding in the proof is the oscillation \( \omega_f(S) = \sup_S f - \inf_S f \), since \( U(f,P) - L(f,P) = \sum_i \omega_f([x_{i-1},x_i]) \, \Delta x_i \). The criterion says: \( f \) is integrable iff its oscillation, integrated against length, is negligible. Every classical integrability proof is a strategy for controlling this sum — uniform continuity makes every \( \omega_i \) small (continuous case), telescoping makes \( \sum \omega_i \) small even when individual terms are not (monotone case), and isolation of bad points makes the total length carrying large oscillation small (Thomae, piecewise continuity).

The endpoint of this line of thought is Lebesgue's 1904 criterion: a bounded \( f : [a,b] \to \mathbb{R} \) is Riemann integrable iff its set of discontinuities has Lebesgue measure zero. The bridge is the pointwise oscillation \( \omega_f(x) = \lim_{\delta \to 0^+} \omega_f\!\left( (x-\delta, x+\delta) \cap [a,b] \right) \): the sets \( D_\eta = \{ x : \omega_f(x) \ge \eta \} \) are compact, \( f \) is continuous at \( x \) iff \( \omega_f(x) = 0 \), and the Darboux criterion holds iff each \( D_\eta \) can be covered by finitely many intervals of arbitrarily small total length. Thus the criterion proved on this page is precisely the finitary, partition-level shadow of a measure-theoretic fact — and its limitation (Dirichlet's function, discontinuous everywhere, fails it) is what the Lebesgue integral was built to transcend: \( \chi_{\mathbb{Q}} \) is Lebesgue integrable with integral \( 0 \).

Common misconceptions. The criterion is an existence theorem, not an evaluation device — students often expect it to produce the value of the integral, which it never does. Second, "integrable" here means Riemann integrable; the same word in a measure-theory course means something strictly weaker to fail and strictly wider to hold. Third, the criterion's witness partition depends on \( \varepsilon \): there is no single partition that works for all tolerances unless \( f \) is a step function.

Worked examples

Example 1. Show \( f(x) = x^2 \) is integrable on \( [0,1] \), directly from the criterion.

Example 2. Show every monotone increasing \( f : [a,b] \to \mathbb{R} \) is integrable — continuity is not assumed.

Problems

1. Let \( f(x) = x \) on \( [0,1] \). Using the uniform partition \( P_n \), compute \( U(f,P_n) - L(f,P_n) \) exactly and deduce integrability from the criterion.
   Solution

   With \( x_i = \tfrac{i}{n} \), \( f \) is increasing, so \( m_i = \tfrac{i-1}{n} \), \( M_i = \tfrac{i}{n} \). Then \[ U(f,P_n) - L(f,P_n) = \sum_{i=1}^{n} \left( \frac{i}{n} - \frac{i-1}{n} \right)\frac{1}{n} = \sum_{i=1}^{n} \frac{1}{n^2} = \frac{1}{n}. \] Given \( \varepsilon \gt 0 \), pick \( n \gt 1/\varepsilon \) (Archimedean property); then \( U - L \lt \varepsilon \), so by the criterion \( f \in \mathcal{R}[0,1] \). (For the value: \( U(f,P_n) = \sum_{i=1}^n \frac{i}{n^2} = \frac{n+1}{2n} \to \frac{1}{2} \), so \( \int_0^1 x \, dx = \frac{1}{2} \).)
2. Prove from the criterion that Dirichlet's function \( \chi_{\mathbb{Q}} \) on \( [0,1] \) is not Riemann integrable.
   Solution

   Let \( P : 0 = x_0 \lt \cdots \lt x_n = 1 \) be any partition. Each cell \( [x_{i-1}, x_i] \) has positive length, so by density of \( \mathbb{Q} \) in \( \mathbb{R} \) it contains a rational, giving \( M_i = 1 \); by density of the irrationals it contains an irrational, giving \( m_i = 0 \). Hence \[ U(\chi_{\mathbb{Q}},P) - L(\chi_{\mathbb{Q}},P) = \sum_{i=1}^{n} (1 - 0)\,\Delta x_i = 1 \] for every partition \( P \). Taking \( \varepsilon = \tfrac{1}{2} \), no partition satisfies \( U - L \lt \varepsilon \), so the criterion fails; by the theorem (contrapositive of \( \Rightarrow \) read through the biconditional), \( \chi_{\mathbb{Q}} \notin \mathcal{R}[0,1] \).
3. (Sequential criterion.) Let \( f : [a,b] \to \mathbb{R} \) be bounded. Prove: \( f \in \mathcal{R}[a,b] \) iff there exists a sequence of partitions \( (P_n) \) with \( U(f,P_n) - L(f,P_n) \to 0 \); and in that case \( \int_a^b f = \lim_n U(f,P_n) = \lim_n L(f,P_n) \).
   Solution

   (\( \Rightarrow \)) If \( f \in \mathcal{R}[a,b] \), apply the criterion with \( \varepsilon = \tfrac{1}{n} \) to obtain \( P_n \) with \( 0 \le U(f,P_n) - L(f,P_n) \lt \tfrac{1}{n} \); by the squeeze theorem the difference tends to \( 0 \). (\( \Leftarrow \)) If such \( (P_n) \) exists, then given \( \varepsilon \gt 0 \) choose \( N \) with \( U(f,P_N) - L(f,P_N) \lt \varepsilon \); this \( P_N \) witnesses the criterion, so \( f \in \mathcal{R}[a,b] \). For the value: write \( I = \int_a^b f \). By Step 4 of the main proof, for every \( n \): \[ L(f,P_n) \le I \le U(f,P_n), \qquad L(f,P_n) \le U(f,P_n). \] Hence \( 0 \le U(f,P_n) - I \le U(f,P_n) - L(f,P_n) \to 0 \) and \( 0 \le I - L(f,P_n) \le U(f,P_n) - L(f,P_n) \to 0 \), so both sequences converge to \( I \) by the squeeze theorem.
4. Prove: if \( f \in \mathcal{R}[a,b] \) then \( |f| \in \mathcal{R}[a,b] \). Show also that the converse fails.
   Solution

   First an oscillation identity: for nonempty \( S \subseteq [a,b] \), \[ \sup_S f - \inf_S f = \sup\{ f(x) - f(y) : x, y \in S \} \ge \sup\{ \left| |f(x)| - |f(y)| \right| : x,y \in S \} = \sup_S |f| - \inf_S |f|, \] where the middle inequality uses the reverse triangle inequality \( \left| |u| - |v| \right| \le |u - v| \) and \( |f(x)-f(y)| \le \sup_S f - \inf_S f \). (For the first equality: \( f(x) - f(y) \le \sup f - \inf f \) always, and choosing \( x \) near the sup and \( y \) near the inf via the approximation properties shows the bound is sharp.) Now let \( \varepsilon \gt 0 \). Since \( f \in \mathcal{R}[a,b] \), the criterion gives \( P \) with \( U(f,P) - L(f,P) \lt \varepsilon \). Applying the oscillation inequality on each cell, \[ U(|f|,P) - L(|f|,P) = \sum_i \left( \sup_i |f| - \inf_i |f| \right) \Delta x_i \le \sum_i \left( M_i - m_i \right) \Delta x_i \lt \varepsilon , \] so \( |f| \in \mathcal{R}[a,b] \) by the criterion. \( |f| \) is bounded since \( f \) is. Converse fails: let \( f(x) = 1 \) for \( x \in \mathbb{Q} \), \( f(x) = -1 \) otherwise, on \( [0,1] \). Then \( |f| \equiv 1 \) is integrable, but \( f = 2\chi_{\mathbb{Q}} - 1 \) is not (by Problem 2, since integrability is preserved under the affine map \( g \mapsto \tfrac{g+1}{2} \), or directly: every cell has \( M_i = 1 \), \( m_i = -1 \), so \( U - L = 2 \) for all \( P \)).
5. Let \( f : [a,b] \to \mathbb{R} \) be bounded, with \( |f| \le K \), and continuous except at finitely many points \( c_1, \dots, c_k \). Prove \( f \in \mathcal{R}[a,b] \). You may use the Heine–Cantor theorem (a continuous function on a compact set is uniformly continuous).
   Solution

   Let \( \varepsilon \gt 0 \). Set \( \eta = \dfrac{\varepsilon}{8Kk} \) (if \( K = 0 \) then \( f \equiv 0 \) and we are done; assume \( K \gt 0 \)). Around each \( c_j \) place the open interval \( I_j = (c_j - \eta, c_j + \eta) \cap [a,b] \); the total length of \( \bigcup_j I_j \) is at most \( 2\eta k = \dfrac{\varepsilon}{4K} \). The set \( C = [a,b] \setminus \bigcup_j I_j \) is a finite union of closed bounded intervals \( J_1, \dots, J_r \) (removing finitely many open intervals from a closed interval leaves such a union), and \( f \) is continuous on each \( J_s \), hence uniformly continuous there by Heine–Cantor. So there is \( \delta \gt 0 \) such that for all \( s \) and all \( x, y \in J_s \) with \( |x - y| \lt \delta \), \( |f(x) - f(y)| \lt \dfrac{\varepsilon}{4(b-a)} \). Build the partition \( P \): take all endpoints of all the \( J_s \) and of the closures of the \( I_j \), then subdivide each \( J_s \) into cells of length less than \( \delta \). Every cell of \( P \) is now of one of two types. Good cells (inside some \( J_s \), length \( \lt \delta \)): here \( M_i - m_i \le \dfrac{\varepsilon}{4(b-a)} \), since for any \( x, y \) in the cell \( |f(x)-f(y)| \lt \frac{\varepsilon}{4(b-a)} \), and \( M_i - m_i = \sup_{x,y} (f(x)-f(y)) \) (approximation property of sup/inf, as in Problem 4) — note the sup of quantities bounded by \( \frac{\varepsilon}{4(b-a)} \) is at most \( \frac{\varepsilon}{4(b-a)} \). Their total contribution is \[ \sum_{\text{good}} (M_i - m_i)\,\Delta x_i \le \frac{\varepsilon}{4(b-a)} \sum_{\text{good}} \Delta x_i \le \frac{\varepsilon}{4(b-a)} (b-a) = \frac{\varepsilon}{4}. \] Bad cells (contained in the closure of some \( I_j \)): here we only know \( M_i - m_i \le 2K \), but their total length is at most \( \dfrac{\varepsilon}{4K} \), so \[ \sum_{\text{bad}} (M_i - m_i)\,\Delta x_i \le 2K \cdot \frac{\varepsilon}{4K} = \frac{\varepsilon}{2}. \] Altogether \( U(f,P) - L(f,P) \le \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{2} \lt \varepsilon \), so the criterion is satisfied and \( f \in \mathcal{R}[a,b] \). (Remark: the same two-bucket strategy — spend a length budget on bad points, an oscillation budget on the rest — proves Thomae's function integrable, with the finitely many points replaced by the finitely many rationals of small denominator.)