The Riemann integral

Intuition Beginner

The area under a curve is found by filling the region with thin rectangles and adding up their areas. As the rectangles get thinner and thinner, the total area of the rectangles approaches the true area under the curve.

A Riemann sum builds each rectangle by picking one height inside each strip — any sample point will do — and multiplying that height by the strip's width. The defining demand of Riemann's integral is strong: no matter how the sample points are chosen, the total must settle to the same number as the strips thin.

When that single number exists, the function is Riemann integrable, and the number is its integral. Any curve you can draw without lifting your pen passes this test, and so do many curves with carefully placed jumps.

This concept exists because a definition of "area" must not depend on a lucky choice of sample points. Riemann's definition pins down the value by requiring every reasonable sampling to agree, which is what separates a genuine area from an artefact of measurement.

Visual Beginner

The picture shows a curve rising from left to right over the interval from $0$ to $1$. The interval is sliced into equal strips. Inside each strip a marker dot indicates the chosen sample point, and a rectangle is drawn whose height equals the curve's value at that dot. The rectangles sometimes overshoot the curve and sometimes undershoot it, depending on where the dot lands.

As the strips become thinner, the position of each dot matters less and less, and the total area of the rectangles converges to a single value regardless of where the dots were placed.

Worked example Beginner

Find the area under $f(x)=x^2$ from $0$ to $1$ using right-endpoint rectangles. With $4$ equal strips of width $1/4$, the sample points are $1/4,1/2,3/4,1$.

Step 1. The heights are $f(1/4)=1/16$, $f(1/2)=1/4$, $f(3/4)=9/16$, $f(1)=1$.

Step 2. The total is the sum of heights times the strip width: $(1/16+1/4+9/16+1)\times 1/4=(30/16)\times 1/4=30/64=15/32\approx 0.469$.

Step 3. Repeat with $8$ strips of width $1/8$. The squared heights, written over $64$, are $1/64,4/64,9/64,16/64,25/64,36/64,49/64,64/64$, adding to $204/64$. Multiply by $1/8$: the total is $204/512=51/128\approx 0.398$.

Step 4. Doubling again to $16$ strips gives approximately $0.365$, and the values keep sinking toward $1/3\approx 0.333$. The exact area is $1/3$.

What this tells us: as the strips thin, the right-endpoint totals converge to a single number, and any other sampling (left endpoints, midpoints) converges to the same number. That shared limit is the Riemann integral.

Check your understanding Beginner

Formal definition Intermediate+

A partition of $[a,b]$ is a finite ordered set $P=\{x_0,x_1,\dots ,x_n\}$ with $a=x_0<x_1<\cdots <x_n=b$. The subinterval widths are $\Delta x_i=x_i-x_{i-1}$ and the mesh (or norm) of $P$ is $\Vert P\Vert ={\max }_i\Delta x_i$. A partition $P^{\prime }$ refines $P$ when $P\subseteq P^{\prime }$.

A tagged partition is a pair $(P,\xi )$ where $P$ is a partition and $\xi =({\xi }_1,\dots ,{\xi }_n)$ is a choice of tags with ${\xi }_i\in [x_{i-1},x_i]$ for each $i$. The associated Riemann sum is

$$S(f,P,\xi )=\sum _{i=1}^{n}f({\xi }_i)\Delta x_i.$$

A bounded function $f:[a,b]\to \mathbb{R}$ is Riemann integrable on $[a,b]$ if there exists a number $L\in \mathbb{R}$ such that, for every $ϵ>0$, there is $\delta >0$ with

$$|S(f,P,\xi )-L|<ϵ$$

for every tagged partition $(P,\xi )$ with $\Vert P\Vert <\delta$. The number $L$ is unique when it exists and is called the Riemann integral of $f$ over $[a,b]$, written ${\int }_a^{b}f(x)dx$.

The link to the Darboux sums of 02.04.01 is immediate. For any tagged partition $(P,\xi )$,

$$L(f,P)\le S(f,P,\xi )\le U(f,P),$$

because $f({\xi }_i)$ always lies between $m_i={\inf }_{[x_{i-1},x_i]}f$ and $M_i={\sup }_{[x_{i-1},x_i]}f$. Moreover $L(f,P)={\inf }_{\xi }S(f,P,\xi )$ and $U(f,P)={\sup }_{\xi }S(f,P,\xi )$, since on each subinterval the values of $f$ can be made arbitrarily close to $m_i$ and $M_i$ by a suitable tag. The Riemann and Darboux theories are therefore two views of the same object.

Counterexamples to common slips Intermediate+

• Mesh versus refinement. The Riemann definition controls sums by mesh ($\Vert P\Vert \to 0$), the Darboux theory controls them by refinement ($P\subseteq P^{\prime }$). The two are not interchangeable: a sequence of partitions can refine without the mesh tending to zero (keep one wide interval), and the mesh can tend to zero without successive refinement. The equivalence theorem below reconciles them.

• Forgetting boundedness. The definition presupposes $f$ is bounded on $[a,b]$. An unbounded function has, on at least one subinterval, Riemann sums of arbitrarily large magnitude, so no finite $L$ can satisfy the defining inequality. For example $f(x)=1/x$ is not Riemann integrable on $(0,1]$.

• Requiring a single partition. Integrability is a statement about all tagged partitions of sufficiently small mesh, not about one lucky partition. A carefully chosen partition can make one Riemann sum land exactly on a target value while other samplings of the same partition diverge wildly (as for the Dirichlet function below).

Key theorem with proof Intermediate+

Theorem (Equivalence of Riemann and Darboux integrability). For a bounded function $f:[a,b]\to \mathbb{R}$ the following are equivalent.

1. $f$ is Darboux integrable on $[a,b]$.
2. $f$ is Riemann integrable on $[a,b]$.

When either condition holds, the two integrals coincide: ${\int }_a^{b}fdx$ has the same value under both definitions.

Proof. We use throughout that $L(f,P)={\inf }_{\xi }S(f,P,\xi )$ and $U(f,P)={\sup }_{\xi }S(f,P,\xi )$, so every Riemann sum lies in $[L(f,P),U(f,P)]$.

(2) $\Rightarrow$ (1): Suppose $f$ is Riemann integrable with value $L$. Let $ϵ>0$ and pick $\delta >0$ so that $|S(f,P,\xi )-L|<ϵ/3$ for every tagged partition with mesh $<\delta$. Fix any partition $P$ with $\Vert P\Vert <\delta$. On each subinterval choose a tag ${\xi }_i^{+}$ with $f({\xi }_i^{+})>M_i-ϵ/(3(b-a))$ and a tag ${\xi }_i^{-}$ with $f({\xi }_i^{-})<m_i+ϵ/(3(b-a))$, which is possible by definition of supremum and infimum. Then

$$S(f,P,{\xi }^{+})>U(f,P)-ϵ/3,S(f,P,{\xi }^{-})<L(f,P)+ϵ/3.$$

Both sums lie within $ϵ/3$ of $L$, so $U(f,P)<L+2ϵ/3$ and $L(f,P)>L-2ϵ/3$, giving $U(f,P)-L(f,P)<4ϵ/3$. By the Darboux integrability criterion of 02.04.01, $f$ is Darboux integrable. Since the Darboux integral $I$ lies in $[L(f,P),U(f,P)]$, we have $|I-L|<2ϵ/3$ for every $ϵ$, whence $I=L$.

(1) $\Rightarrow$ (2): Suppose $f$ is Darboux integrable with value $I$. Let $ϵ>0$. By the Darboux criterion choose a partition $P_0=\{t_0,\dots ,t_N\}$ with $U(f,P_0)-L(f,P_0)<ϵ/3$. Let $M=1+{\sup }_{[a,b]}|f|$ and set $\delta =ϵ/(6MN)$. Let $(P,\xi )$ be any tagged partition with $\Vert P\Vert <\delta$, and form the common refinement $P^{\ast }=P\cup P_0$. Refining $P$ by inserting the $N$ points of $P_0$ changes each of $U(f,P)$ and $L(f,P)$ by at most $2MN\Vert P\Vert$ in total (each inserted point splits a subinterval of width $<\delta$, changing the supremum sum by at most $2M\Delta x_i<2M\delta$). Hence

$$U(f,P)\le U(f,P^{\ast })+2MN\Vert P\Vert <U(f,P^{\ast })+ϵ/3\le U(f,P_0)+ϵ/3,$$ $$L(f,P)\ge L(f,P^{\ast })-2MN\Vert P\Vert >L(f,P^{\ast })-ϵ/3\ge L(f,P_0)-ϵ/3.$$

Because $L(f,P)\le S(f,P,\xi )\le U(f,P)$ and $I\in [L(f,P_0),U(f,P_0)]$,

$$S(f,P,\xi )-I<U(f,P)-L(f,P_0)<(U(f,P_0)-L(f,P_0))+ϵ/3<2ϵ/3,$$

and symmetrically $I-S(f,P,\xi )<2ϵ/3$. Thus $|S(f,P,\xi )-I|<ϵ$ for every tagged partition with mesh $<\delta$, which is precisely Riemann integrability with value $I$. $\square$

Corollary (Riemann's integrability criterion, Cauchy form). A bounded $f:[a,b]\to \mathbb{R}$ is Riemann integrable if and only if for every $ϵ>0$ there exists $\delta >0$ such that $|S(f,P,\xi )-S(f,Q,\eta )|<ϵ$ for any two tagged partitions $(P,\xi ),(Q,\eta )$ with mesh smaller than $\delta$.

This reformulation avoids presupposing the value $L$: integrability is recast as the statement that all sufficiently fine Riemann sums cluster within $ϵ$ of one another. The forward direction is the triangle inequality applied to the common limit $L$; the reverse direction selects a sequence of tagged partitions with mesh tending to zero, observes that the corresponding sums form a Cauchy sequence in $\mathbb{R}$, and uses the completeness of the real numbers 02.02.01 to extract a limit $L$. The full argument appears in the Full proof set.

Bridge. The equivalence between the Riemann and Darboux definitions is the foundational reason the two constructions of area never disagree, and this is exactly the content of the main theorem: the limit-of-sums picture (Riemann) and the squeeze-from-above-and-below picture (Darboux) describe the same number whenever either exists. The central insight is that a single Riemann sum always sits between the lower and upper Darboux sums of its partition, so collapsing the Darboux gap forces every tagged sum to the same limit. This builds toward 02.04.03, where the Darboux criterion is applied to continuous functions via uniform continuity, and appears again in 02.04.04, where the fundamental theorems of calculus are stated for the Riemann integral precisely because the equivalence guarantees a unique value to differentiate and to evaluate at endpoints.

Exercises Intermediate+

Advanced results Master

Theorem 1 (Riemann's Cauchy criterion). A bounded $f:[a,b]\to \mathbb{R}$ is Riemann integrable if and only if for every $ϵ>0$ there is $\delta >0$ such that $|S(f,P,\xi )-S(f,Q,\eta )|<ϵ$ whenever $(P,\xi )$ and $(Q,\eta )$ are tagged partitions with mesh below $\delta$.

This is Riemann's intrinsic integrability test: it dispenses with the integral value $L$ and asks only that all sufficiently fine sums agree with each other. The reverse direction constructs a candidate value by completeness of $\mathbb{R}$ 02.02.01: a sequence of tagged partitions with mesh tending to zero produces sums that form a Cauchy sequence, hence a convergent one, and that limit is the integral. The proof is given in the Full proof set.

Theorem 2 (Compatibility with Darboux-class integrability results). Every Darboux-integrable function is Riemann integrable and conversely, with equal integrals. In particular: every continuous function on $[a,b]$ is Riemann integrable 02.04.03; every monotone function on $[a,b]$ is Riemann integrable 02.04.01; and Lebesgue's criterion holds — a bounded $f$ is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero 02.04.03.

The equivalence theorem of the Key section upgrades every Darboux result to a Riemann result at no cost. The Riemann viewpoint adds nothing about which functions integrate, only a different (and often more intuitive) definition of what the integral is.

Theorem 3 (Thomae's function is Riemann integrable). The function $T:[0,1]\to \mathbb{R}$ with $T(p/q)=1/q$ in lowest terms and $T(\text{irrational})=0$ is Riemann integrable with ${\int }_0^{1}Tdx=0$.

Thomae's function is discontinuous at every rational and continuous at every irrational, so its discontinuity set is $\mathbb{Q}\cap [0,1]$, which is countable and hence of Lebesgue measure zero. Lebesgue's criterion therefore predicts integrability. The direct proof (Full proof set) exhibits, for each $ϵ>0$, a partition with upper sum below $ϵ$ by isolating the finitely many points where $T$ exceeds $ϵ/2$. The lesson is sharp: a function can be discontinuous on a dense set and still be Riemann integrable, provided the discontinuities are sparse in the sense of measure.

Theorem 4 (The Dirichlet function is not Riemann integrable). The function $D:[0,1]\to \mathbb{R}$ with $D(x)=1_{\mathbb{Q}}(x)$ is bounded but not Riemann integrable.

Proof. Suppose $D$ were Riemann integrable with value $L$. Take $ϵ=1/4$ and let $\delta >0$ be the corresponding mesh bound. Pick any partition $P$ with $\Vert P\Vert <\delta$. Choosing all tags rational gives the sum $1$; choosing all tags irrational gives the sum $0$. Both must lie within $1/4$ of $L$, so $1=|1-0|\le |1-L|+|L|<1/2$, a contradiction. $\square$

This is the canonical obstruction to Riemann integrability: the discontinuity set is all of $[0,1]$, of full measure, and the density of both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ lets opposing tag choices produce sums that differ by the entire interval length, for any partition however fine.

Theorem 5 (Linearity and additivity). If $f$ and $g$ are Riemann integrable on $[a,b]$ and $\alpha \in \mathbb{R}$, then $f+g$ and $\alpha f$ are Riemann integrable, with ${\int }_a^b(f+g)={\int }_a^{b}f+{\int }_a^{b}g$ and ${\int }_a^b(\alpha f)=\alpha {\int }_a^{b}f$. If $c\in (a,b)$ and $f$ is integrable on $[a,b]$, then $f$ is integrable on $[a,c]$ and $[c,b]$ with ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$.

These follow immediately from the Darboux versions 02.04.01 together with the equivalence theorem. They can also be proved directly from Riemann sums: Riemann sums split linearly, $S(f+g,P,\xi )=S(f,P,\xi )+S(g,P,\xi )$, so the limits add; and any partition of $[a,b]$ containing $c$ decomposes into a partition of $[a,c]$ and one of $[c,b]$.

Theorem 6 (Composition with a continuous function). If $f:[a,b]\to [c,d]$ is Riemann integrable and $g:[c,d]\to \mathbb{R}$ is continuous, then $g\circ f$ is Riemann integrable on $[a,b]$.

The composition $g\circ f$ has no more discontinuities than $f$ itself, because a continuous $g$ preserves limits. By Lebesgue's criterion the discontinuity set of $g\circ f$ is contained in that of $f$ and hence has measure zero. In particular $|f|$, $f^2$, and $e^f$ are integrable whenever $f$ is, and the product $fg=\frac{1}{4}((f+g{)}^2-(f-g{)}^2)$ is integrable whenever $f$ and $g$ are.

Synthesis. The foundational reason the Riemann integral is the right elementary notion of area is that tagged-partition convergence forces every sampling choice to agree on a single number, and this is exactly what the equivalence with the Darboux integral makes precise. The central insight is that integrability is a question about the collapse of oscillation — the gap between supremum and infimum sums — rather than about any particular sample. Putting these together, Riemann's sum definition and Darboux's supremum definition are two faces of one completeness-driven construction on the real line. The pattern generalises in two directions: the bridge is from sample-point limits to set-of-discontinuities criteria via the Lebesgue theorem 02.04.03, and onward to the Lebesgue integral 02.07.04, which extends the same area concept to every bounded measurable function. The same tagged-partition framework, with the mesh replaced by a variable gauge, underlies the Henstock-Kurzweil integral, which recovers every Riemann and Lebesgue integral while handling conditionally integrable functions such as $\sin (x)/x$ that the Riemann theory leaves out.

Full proof set Master

Proposition 1 (Riemann's Cauchy criterion). A bounded $f:[a,b]\to \mathbb{R}$ is Riemann integrable if and only if for every $ϵ>0$ there exists $\delta >0$ such that $|S(f,P,\xi )-S(f,Q,\eta )|<ϵ$ for any two tagged partitions $(P,\xi ),(Q,\eta )$ with mesh below $\delta$.

Proof. ($\Rightarrow$) Suppose $f$ is Riemann integrable with value $L$. Given $ϵ>0$, choose $\delta >0$ so that $|S(f,R,\zeta )-L|<ϵ/2$ for every tagged partition $(R,\zeta )$ with mesh below $\delta$. Then for any two such tagged partitions $(P,\xi )$ and $(Q,\eta )$,

$$|S(f,P,\xi )-S(f,Q,\eta )|\le |S(f,P,\xi )-L|+|L-S(f,Q,\eta )|<ϵ/2+ϵ/2=ϵ.$$

($\Leftarrow$) Assume the Cauchy condition. For each $n\in \mathbb{N}$ let $(P_n,{\xi }^n)$ be the tagged partition of $[a,b]$ into $n$ subintervals of equal width $(b-a)/n$, with tags at the left endpoints, and set $s_n=S(f,P_n,{\xi }^n)$. Since $\Vert P_n\Vert =(b-a)/n\to 0$, the Cauchy condition applied with $ϵ>0$ yields $\delta >0$; choosing $N$ with $(b-a)/N<\delta$, we have $|s_n-s_m|<ϵ$ for all $n,m\ge N$. Thus $(s_n)$ is a Cauchy sequence in $\mathbb{R}$, which converges by the completeness of the real numbers 02.02.01; call its limit $L$.

It remains to show that every tagged partition $(P,\xi )$ with sufficiently small mesh has sum within any prescribed tolerance of $L$. Given $ϵ>0$, let $\delta >0$ be the mesh bound supplied by the Cauchy condition with tolerance $ϵ/2$. Choose $n$ large enough that $\Vert P_n\Vert <\delta$ and $|s_n-L|<ϵ/2$. Then for every tagged partition $(P,\xi )$ with $\Vert P\Vert <\delta$,

$$|S(f,P,\xi )-L|\le |S(f,P,\xi )-s_n|+|s_n-L|<ϵ/2+ϵ/2=ϵ,$$

since both $(P,\xi )$ and $(P_n,{\xi }^n)$ have mesh below $\delta$ and the Cauchy condition applies. Therefore $f$ is Riemann integrable with value $L$. $\square$

Proposition 2 (Thomae's function is Riemann integrable with integral $0$). Define $T:[0,1]\to \mathbb{R}$ by $T(p/q)=1/q$ for $p/q\in [0,1]$ in lowest terms ($q\ge 1$), and $T(x)=0$ for $x$ irrational, with $T(0)=0$. Then $T$ is Riemann integrable and ${\int }_0^{1}Tdx=0$.

Proof. By the equivalence theorem of the Key section it suffices to verify the Darboux integrability criterion. Since $T(x)\ge 0$ for all $x$, every lower Darboux sum satisfies $L(T,P)\ge 0$, so it is enough to exhibit, for each $ϵ>0$, a partition $P$ with $U(T,P)<ϵ$.

Let $ϵ>0$. Choose $Q\in \mathbb{N}$ with $1/Q<ϵ/2$. The set $A_Q=\{x\in [0,1]:T(x)\ge 1/Q\}$ consists of the rationals $p/q\in [0,1]$ in lowest terms with $q\le Q$, which is finite; call its cardinality $N$. Cover $A_Q$ by finitely many open intervals of total length less than $ϵ/2$ (possible since $A_Q$ is finite). Form a partition $P$ of $[0,1]$ that includes the endpoints $0,1$ and all endpoints of the covering intervals, so that every subinterval of $P$ lies either inside one of the covering intervals or entirely within the complement $[0,1]\setminus A_Q$.

On the complement, every point satisfies $T(x)<1/Q<ϵ/2$, so the supremum of $T$ on any complement subinterval is at most $ϵ/2$. On the covering intervals, $T\le 1$ throughout $[0,1]$, so the supremum is at most $1$. Summing the contributions,

$$U(T,P)\le 1\cdot \sum _{\text{cover}}\Delta x_i+(ϵ/2)\cdot \sum _{\text{complement}}\Delta x_i\le 1\cdot (ϵ/2)+(ϵ/2)\cdot 1=ϵ.$$

Thus $U(T,P)-L(T,P)\le U(T,P)<ϵ$, and the Darboux criterion holds. The integral $I$ satisfies $0\le L(T,P)\le I\le U(T,P)<ϵ$ for every $ϵ>0$, whence $I=0$. $\square$

Connections Master

• Step-function integral and the Darboux integral 02.04.01. The Darboux integral is the prerequisite construction: its upper and lower sums bracket every Riemann sum, and the equivalence theorem of this unit shows that collapsing the Darboux gap is exactly the condition that forces all tagged sums to the same limit. Every integrability result proved with the Darboux machinery — monotone functions, the constant and step-function base cases, linearity and additivity — transfers to the Riemann setting through this equivalence.

• Integrability of continuous functions on $[a,b]$ 02.04.03. The theorem that continuous functions are Riemann integrable is proved there via the Heine-Cantor theorem and uniform continuity, working directly with the Darboux criterion. The equivalence of the present unit licenses that proof as a theorem about the Riemann integral, and Lebesgue's measure-zero criterion stated there identifies the precise class of Riemann integrable functions, explaining both the integrability of Thomae's function (discontinuities on a countable, measure-zero set) and the failure of the Dirichlet function (discontinuities everywhere, full measure).

• Fundamental theorems of calculus 02.04.04. Both FTC1 and FTC2 are statements about the Riemann integral: FTC1 differentiates the area function $F(x)={\int }_a^{x}f$, and FTC2 evaluates the integral through an antiderivative. The equivalence theorem guarantees the uniqueness of the value being differentiated and evaluated, which is what makes the fundamental theorems well-posed. The mean-value-theorem proof of FTC2 in that unit runs through Riemann sums, closing the loop between the limit-of-sums definition and the antiderivative formula.

• The Lebesgue integral and monotone convergence 02.07.04. The Riemann integral is the finitary precursor: its sums partition the domain into intervals, while the Lebesgue integral partitions the range and integrates over level sets. The Lebesgue integral strictly extends the Riemann integral, agreeing wherever the Riemann value exists and assigning values to functions such as $1_{\mathbb{Q}}$ (Lebesgue integral $0$) where the Riemann procedure fails. Improper Riemann integrals like $\sin (x)/x$ mark the boundary where the conditionally-convergent Riemann framework diverges from the absolutely-convergent Lebesgue framework.

Historical & philosophical context Master

Riemann introduced the integral in his 1854 Göttingen Habilitationsschrift, Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe ^{[Riemann1854]}, delivered as a Habilitationsvortrag and published posthumously in 1868. Where Cauchy had integrated continuous functions as limits of left-endpoint sums, Riemann deliberately allowed the sample points to vary freely within each subinterval and asked for convergence independent of that choice. This shift made integrability a property of the function rather than of a particular summation recipe, and it broadened the class of integrable functions to include some with infinitely many discontinuities, provided they are arranged sparsely. Riemann himself gave the first example of an integrable function with a dense discontinuity set, a precursor of Thomae's function, by a counting argument on rational approximations.

Darboux in 1875 reformulated the theory in Annales scientifiques de l'École Normale Supérieure ^{[Darboux1875]}, replacing tagged sums with upper and lower sums built from suprema and infima. The reformulation avoids the quantification over tags and exposes the role of completeness directly: the integral exists when the infimum of upper sums meets the supremum of lower sums. The equivalence theorem of this unit, that the Riemann and Darboux definitions pick out exactly the same functions with the same values, is the precise reconciliation of the two viewpoints; it appears in its modern form in Rudin ^[Rudin1976] and Dieudonné ^{[Dieudonne1960]}. The eventual refinement by Lebesgue in 1901 — that Riemann integrability is equivalent to boundedness plus a measure-zero discontinuity set — explains why Riemann's broad class is exactly what it is, and points forward to the Lebesgue integral, which abandons the domain-partition sum in favour of a range-partition sum and integrates every bounded measurable function.

Bibliography Master

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