Equivalence of Darboux and Riemann Definitions

Two Definitions of the Same Integral

Riemann integrability can be described using either tagged sums or Darboux upper and lower sums. The tagged-sum definition asks whether every choice of tags on every sufficiently fine partition produces nearly the same sum. The Darboux definition asks whether the best possible lower and upper approximations have the same limiting value. These descriptions emphasize different features, so their equivalence is useful: it lets us move from control of sample values to control of suprema and infima, and back again.

We will compare the definitions for a bounded function $f:[a,b]\to\mathbb{R}$, with $a<b$. Write $L(f,P)$ and $U(f,P)$ for the lower and upper sums on a partition $P$, and write $\underline{\int_a^b}f$ and $\overline{\int_a^b}f$ for the Darboux lower and upper integrals. For a tagged partition $(P,\xi)$, write $S(f;P,\xi)$ for its tagged sum.

Definition (Riemann Definition): A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable with value $A$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that every tagged partition $(P,\xi)$ with $\|P\|<\delta$ satisfies $$ |S(f;P,\xi)-A|<\varepsilon. $$

Definition (Darboux Definition): A bounded function $f:[a,b]\to\mathbb{R}$ is Darboux integrable if its lower and upper integrals are equal: $$ \underline{\int_a^b} f=\overline{\int_a^b} f. $$ Their common value is the Darboux integral.

The Riemann definition requires uniform control over both partitions and tags. The Darboux definition compares the supremum of all lower sums with the infimum of all upper sums. We will prove that each definition implies the other, including agreement of the values.

A Bridge Between Tagged Sums and Darboux Sums

For any fixed partition, every tagged sum lies between its lower and upper sums. This familiar bound is one half of the connection. The other half is that tags can approximate the suprema and infima defining those sums, even when the function does not attain them.

Lemma (Tagged Approximation of Partition Sums): Let $P=\{x_0,\ldots,x_n\}$ be a partition of $[a,b]$, and let $\eta>0$. There are choices of tags $\xi_i,\zeta_i\in[x_{i-1},x_i]$ such that $$ S(f;P,\xi)>U(f,P)-\eta(b-a) \qquad\text{and}\qquad S(f;P,\zeta)<L(f,P)+\eta(b-a). $$

Proof. On the $i$-th subinterval, let $M_i$ and $m_i$ be the supremum and infimum of $f$. By the definitions of supremum and infimum, choose $\xi_i,\zeta_i$ with $f(\xi_i)>M_i-\eta$ and $f(\zeta_i)<m_i+\eta$. Multiplying by $\Delta x_i=x_i-x_{i-1}>0$ and summing gives the two inequalities, since $\sum_{i=1}^n\Delta x_i=b-a$. No assumption that either extremum is attained is needed. $\square$

The lemma is particularly useful when the hypothesis controls every tagged sum: the near-supremum and near-infimum choices are then controlled too. To prove the converse implication, we will also need to pass from one partition with a small Darboux gap to control of the gaps for all sufficiently fine partitions.

Lemma (Fine-Partition Gap Bound from a Fixed Partition): Suppose $|f(x)|\leq B$ on $[a,b]$, and let $P$ have $m$ subintervals. For any partition $Q$, $$ U(f,Q)-L(f,Q) \leq U(f,P)-L(f,P)+2B(m-1)\|Q\|. $$

Proof. Form the common refinement $R=P\cup Q$. By monotonicity of upper and lower sums under refinement, $U(f,R)-L(f,R)\leq U(f,P)-L(f,P)$. Consider a subinterval $J$ of $Q$ whose interior contains no interior point of $P$. It is not split when passing from $Q$ to $R$, so its contribution to the Darboux gap is unchanged. Any other subinterval of $Q$ contains at least one interior point of $P$ in its interior. There are at most $m-1$ such subintervals, since $P$ has $m-1$ interior points. Their total length is at most $(m-1)\|Q\|$. On each, the oscillation of $f$ is at most $2B$. Thus the sum of their contributions to the gap for $Q$ is at most $2B(m-1)\|Q\|$. The corresponding contributions to the gap for $R$ are nonnegative. Adding the unchanged contributions and these bounds proves the inequality. $\square$

This estimate addresses a quantifier issue that matters in the proof: a small gap on one selected partition does not, by itself, say that every fine partition has a small gap. The estimate shows how refinement and boundedness supply that uniform control.

Equivalence Theorem

Theorem (Equivalence of Darboux and Riemann Definitions): A bounded function $f:[a,b]\to\mathbb{R}$, where $a<b$, satisfies the Riemann definition if and only if it satisfies the Darboux definition. In either case, the tagged-sum value equals the common Darboux integral.

Proof. First suppose the tagged-sum definition holds with value $A$. We show that the lower and upper integrals both equal $A$. Let $\tau>0$. Apply the definition with tolerance $\tau/3$, and choose a partition $P$ with mesh below the resulting threshold. For any $\eta>0$, use the Tagged Approximation of Partition Sums to choose near-supremum tags and near-infimum tags. Both resulting tagged sums differ from $A$ by less than $\tau/3$. The upper approximation gives

U(f,P)<S(f;P,\xi)+\eta(b-a)<A+\frac{\tau}{3}+\eta(b-a),

and the lower approximation gives

L(f,P)>S(f;P,\zeta)-\eta(b-a)>A-\frac{\tau}{3}-\eta(b-a).

Choose $\eta>0$ so that $\eta(b-a)<\tau/3$. Since the upper integral is at most $U(f,P)$, and the lower integral is at least $L(f,P)$, we obtain

\overline{\int_a^b}f<A+\frac{2\tau}{3} \qquad\text{and}\qquad \underline{\int_a^b}f>A-\frac{2\tau}{3}.

The lower integral never exceeds the upper integral. These inequalities, for every $\tau>0$, force both to equal $A$: the lower integral is at least $A$ and the upper integral is at most $A$, while their order is the reverse. Thus the Darboux definition holds with common value $A$.

Now suppose the Darboux definition holds, with common value $A$. Fix $\varepsilon>0$, and choose a number $\alpha>0$ with $2\alpha<\varepsilon/2$. By the definitions of the upper and lower integrals, there are partitions $P_U$ and $P_L$ such that

U(f,P_U)<A+\alpha \qquad\text{and}\qquad L(f,P_L)>A-\alpha.

Take a common refinement $P$ of $P_U$ and $P_L$. Refinement does not increase upper sums or decrease lower sums, so

U(f,P)-L(f,P)<2\alpha<\frac{\varepsilon}{2}.

Let $m$ be the number of subintervals of $P$, and choose $B\geq0$ with $|f(x)|\leq B$ everywhere. By the Fine-Partition Gap Bound from a Fixed Partition, every partition $Q$ satisfies

U(f,Q)-L(f,Q) < \frac{\varepsilon}{2}+2B(m-1)\|Q\|.

If $B(m-1)>0$, choose $\delta>0$ so small that $2B(m-1)\delta<\varepsilon/2$. If $B(m-1)=0$, choose any $\delta>0$. Then every $Q$ with $\|Q\|<\delta$ has $U(f,Q)-L(f,Q)<\varepsilon$.

For every partition $Q$, the Darboux integral $A$ lies between $L(f,Q)$ and $U(f,Q)$: the lower integral is a supremum of lower sums, and the upper integral is an infimum of upper sums. Every tagged sum on $Q$ lies between those same sums. Therefore, whenever $\|Q\|<\delta$,

|S(f;Q,\xi)-A|\leq U(f,Q)-L(f,Q)<\varepsilon.

This is the Riemann definition with value $A$, completing both directions and proving that the values agree. $\square$

Worked Examples: Using the Equivalence

Worked Example: A Polynomial on a Closed Interval

Let $f(x)=x^2$ on $[0,1]$. On a subinterval $[u,v]\subseteq[0,1]$, the function is increasing, so its oscillation is $v^2-u^2=(v-u)(v+u)\leq2(v-u)$. For a partition $P$, it follows that

U(f,P)-L(f,P) \leq 2\sum_{i=1}^n(\Delta x_i)^2 \leq 2\|P\|\sum_{i=1}^n\Delta x_i =2\|P\|.

Thus partitions of sufficiently small mesh have arbitrarily small gaps. By the Darboux Criterion, $f$ is Darboux integrable. The equivalence theorem then says that every tagged sum on every sufficiently fine partition approaches the same value, namely the common upper and lower integral. This argument does not require evaluating that value.

Worked Example: A Jump with the Value at the Jump Included

Fix $c\in(0,1)$ and define $h:[0,1]\to\mathbb{R}$ by $h(x)=0$ for $x<c$ and $h(x)=1$ for $x\geq c$. Consider any partition $P$. If $c$ lies in the interior of a subinterval, that subinterval has oscillation $1$, and all other subintervals have oscillation $0$. Its contribution to the Darboux gap is at most $\|P\|$.

If $c$ is a partition point, the subinterval immediately to its left has oscillation $1$, because it contains points $x<c$ where $h(x)=0$ as well as the endpoint $c$ where $h(c)=1$. Its contribution is its length, at most $\|P\|$. The subinterval immediately to the right is constant with value $1$, and all other subintervals are constant as well. Thus in this case too,

U(h,P)-L(h,P)\leq\|P\|.

The Darboux Criterion gives integrability. To identify the common value, compare any tagged sum with $1-c$. Intervals strictly to the left of $c$ contribute zero, and intervals strictly to the right contribute their full lengths. If $c$ lies inside one interval, that interval contributes either zero or its full length, while the portion of $[c,1]$ inside it has length at most the interval’s length. If $c$ is a partition point, the interval to its left can contribute at most its length, including when the tag is $c$. In all cases,

|S(h;P,\xi)-(1-c)|\leq\|P\|.

Hence the tagged-sum definition holds with value $1-c$, which is also the common Darboux integral. The endpoint value at the jump is why the cell immediately to the left must be included when $c$ is a partition point.

Worked Example: Thomae’s Function

On $[0,1]$, define Thomae’s function by $t(x)=1/q$ when $x=p/q$ is rational in lowest terms with $q\geq1$, and $t(x)=0$ when $x$ is irrational. In particular, $t(0)=t(1)=1$. We show directly that its Darboux gaps can be made small, despite its discontinuities at every rational point.

Fix a positive integer $N$, and let $E_N$ be the finite set of rationals in $[0,1]$ whose reduced denominator is at most $N$. Write $K_N$ for the number of points in $E_N$. Use the uniform partition into $n$ subintervals of length $1/n$. Mark each closed subinterval containing a point of $E_N$. Each point belongs to at most two such subintervals, so there are at most $2K_N$ marked subintervals; their total length is at most $2K_N/n$. Since $0\leq t\leq1$, their total contribution to the Darboux gap is at most $2K_N/n$.

Every unmarked subinterval contains no rational with reduced denominator at most $N$. At every point of such a subinterval, the function is either zero or at most $1/(N+1)$. Its oscillation is therefore at most $1/(N+1)$, and the total contribution from all unmarked subintervals is at most $1/(N+1)$. Consequently,

U(t,P)-L(t,P)\leq\frac{2K_N}{n}+\frac{1}{N+1}.

Given $\varepsilon>0$, first choose $N$ so that $1/(N+1)<\varepsilon/2$, and then choose $n$ large enough that $2K_N/n<\varepsilon/2$. The Darboux Criterion proves integrability. Every interval of positive length contains irrational points, so the infimum of $t$ on each partition subinterval is zero; thus every lower sum is zero. The lower integral is zero, and equality of the upper and lower integrals shows that the common value is zero. By the equivalence theorem, every tagged sum on a sufficiently fine partition is close to zero, regardless of its tags.

Why the Equivalence Matters

The two definitions make different tasks convenient. Tagged sums are natural when studying sampling rules or numerical approximations. Darboux sums are useful when estimates on oscillation, suprema, or infima are available. The equivalence theorem justifies switching between them without changing either the integrability conclusion or the value of the integral.

A common pitfall is to prove that one sequence of tagged sums converges and conclude that the function is Riemann integrable. The Riemann definition requires every tag choice on every sufficiently fine partition to be controlled by the same mesh threshold. The Darboux formulation exposes the same uniformity through the gap $U(f,P)-L(f,P)$. In the reverse direction of the theorem, the fixed-partition gap estimate is what converts a favorable partition into control over all sufficiently fine partitions.

Check Your Understanding

Use the definitions and arguments in this tutorial to answer the following questions.

Why can tags near the supremum and infimum be chosen even when neither is attained?
When the jump point $c$ is a partition point for $h$, which adjacent subinterval has nonzero oscillation, and why?
In the Fine-Partition Gap Bound, why are at most $m-1$ subintervals of $Q$ split by the points of $P$?
Why does equality of the Darboux integrals place their common value between $L(f,Q)$ and $U(f,Q)$ for every partition $Q$?
In the Thomae function estimate, why can each rational in $E_N$ mark at most two cells of the uniform partition?