Oscillation and the Darboux Integrability Criterion

Oscillation Measures the Darboux Gap

For a bounded function, the size of its values on each partition interval determines how far apart its upper and lower Darboux sums can be. Oscillation records that variation interval by interval. The Darboux integrability criterion says that Riemann integrability is exactly the possibility of making the total, length-weighted oscillation as small as desired.

Throughout, let $a<b$, and work with a bounded, real-valued function $f:[a,b]\to\mathbb{R}$. The interval may be viewed as a measure space with Lebesgue measure restricted to $[a,b]$, but the criterion below concerns the Riemann integral and does not use a Lebesgue integral. A partition is a finite sequence $P=\{a=x_0<x_1<\cdots<x_n=b\}$. On the $i$-th closed subinterval, set $M_i=\sup_{x\in[x_{i-1},x_i]}f(x)$ and $m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$. These are finite because $f$ is bounded.

Definition: The oscillation of $f$ on a closed subinterval $J\subseteq[a,b]$ is $\omega(f;J)=\sup_J f-\inf_J f$. For a partition $P$, its upper and lower Darboux sums are $U(f,P)=\sum_{i=1}^n M_i(x_i-x_{i-1})$ and $L(f,P)=\sum_{i=1}^n m_i(x_i-x_{i-1})$. Their difference, called the Darboux gap, is $U(f,P)-L(f,P)=\sum_{i=1}^n\omega(f;[x_{i-1},x_i])(x_i-x_{i-1})$.

The gap identity follows by subtracting the two finite sums term by term. Each interval contributes its oscillation multiplied by its length; thus the oscillation alone is not the contribution to the gap. Values at partition endpoints belong to the adjacent closed intervals, as required by the suprema and infima. This convention does not affect the formulas.

Refinement and the Darboux Integrals

A partition $R$ refines $P$ if it contains every point of $P$. Refining cannot increase an upper sum and cannot decrease a lower sum. Indeed, if one interval $[u,v]$ is split into intervals $J_1,\ldots,J_r$, then on each $J_j$, $\sup_{J_j}f\le\sup_{[u,v]}f$ and $\inf_{J_j}f\ge\inf_{[u,v]}f$. Multiplying by the lengths and summing, whose total is $v-u$, gives the claimed inequalities on that interval. Applying this to every split interval proves the claim for a refinement.

Any two partitions have a common refinement: take the ordered union of their finite sets of partition points. Consequently, if $P,Q$ are any partitions and $R$ is their common refinement, then $L(f,P)\le L(f,R)\le U(f,R)\le U(f,Q)$. In particular, every lower sum is at most every upper sum.

Definition: The lower Darboux integral is $\underline{\int_a^b} f=\sup_P L(f,P)$, and the upper Darboux integral is $\overline{\int_a^b} f=\inf_P U(f,P)$, where the supremum and infimum range over all partitions of $[a,b]$. The bounded function $f$ is Riemann integrable if these two finite numbers are equal; their common value is its Riemann integral.

Both numbers are finite: if $|f(x)|\le C$ for all $x$, then every upper and lower sum lies between $-C(b-a)$ and $C(b-a)$. The common-refinement inequality proves $\underline{\int_a^b}f\le\overline{\int_a^b}f$. These definitions allow the criterion to be proved directly from the meaning of the upper and lower integrals.

The Darboux Integrability Criterion

Theorem (Darboux Integrability Criterion): A bounded real-valued function $f:[a,b]\to\mathbb{R}$, where $a<b$, is Riemann integrable if and only if for every $\varepsilon>0$ there is a partition $P$ such that $U(f,P)-L(f,P)<\varepsilon$.

Proof. Suppose first that $f$ is Riemann integrable, and let $I=\underline{\int_a^b}f=\overline{\int_a^b}f$. Given $\varepsilon>0$, the definition of infimum gives a partition $P$ with $U(f,P)<I+\varepsilon/2$. The definition of supremum gives a partition $Q$ with $L(f,Q)>I-\varepsilon/2$. Let $R$ be a common refinement. Refinement gives $U(f,R)\le U(f,P)$ and $L(f,R)\ge L(f,Q)$, so $$ U(f,R)-L(f,R)\le U(f,P)-L(f,Q)<(I+\varepsilon/2)-(I-\varepsilon/2)=\varepsilon. $$ Thus the required partition exists.

Conversely, suppose that for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$. By the definitions of the upper and lower Darboux integrals, $\underline{\int_a^b}f\ge L(f,P)$ and $\overline{\int_a^b}f\le U(f,P)$. Together with $\underline{\int_a^b}f\le\overline{\int_a^b}f$, this gives $$ 0\le\overline{\int_a^b}f-\underline{\int_a^b}f \le U(f,P)-L(f,P)<\varepsilon. $$ This holds for every positive $\varepsilon$. A nonnegative real number smaller than every positive $\varepsilon$ must be zero: if it were $d>0$, choosing $\varepsilon=d/2$ would contradict the displayed inequality. Hence the upper and lower Darboux integrals are equal, so $f$ is Riemann integrable. $\square$

Worked Example: The Linear Function on [0,1]

Let $f(x)=x$ on $[0,1]$, and use the uniform partition with $n$ subintervals. Here $n$ is any positive integer, and the $k$-th interval is $[k/n,(k+1)/n]$ for $k=0,\ldots,n-1$. Since $x$ increases, the infimum on that interval is $k/n$, the supremum is $(k+1)/n$, and the oscillation is $1/n$. Thus

L(f,P)=\sum_{k=0}^{n-1}\frac{k}{n}\frac1n =\frac{n(n-1)}{2n^2} =\frac{n-1}{2n}, \qquad U(f,P)=\sum_{k=0}^{n-1}\frac{k+1}{n}\frac1n =\frac{n(n+1)}{2n^2} =\frac{n+1}{2n}.

The gap is $1/n$. Given any $\varepsilon>0$, choose an integer $n>1/\varepsilon$; then $U(f,P)-L(f,P)=1/n<\varepsilon$, so the criterion proves Riemann integrability. The two sums both tend to $1/2$, giving the integral. Check: for $n=4$, the lower sum is $3/8$, the upper sum is $5/8$, and their difference is $5/8-3/8=1/4$, which agrees with $1/n=1/4$.

Worked Example: The Quadratic Function on [0,1]

Let $f(x)=x^2$ on $[0,1]$, again with the uniform partition into $n$ intervals. The function is increasing on this interval, so on $[k/n,(k+1)/n]$ its infimum is $k^2/n^2$ and its supremum is $(k+1)^2/n^2$. The oscillation there is $((k+1)^2-k^2)/n^2=(2k+1)/n^2$. Multiplying by the interval length $1/n$ and summing gives

U(f,P)-L(f,P) =\sum_{k=0}^{n-1}\frac{2k+1}{n^3} =\frac{2\sum_{k=0}^{n-1}k+n}{n^3} =\frac{2\cdot n(n-1)/2+n}{n^3} =\frac{n^2}{n^3} =\frac1n.

For every $\varepsilon>0$, an integer $n>1/\varepsilon$ therefore gives a gap less than $\varepsilon$. For a direct check at $n=2$, the oscillations are $1/4$ and $3/4$, and each interval has length $1/2$; the gap is $(1/4)(1/2)+(3/4)(1/2)=1/2$, matching $1/n=1/2$. The criterion proves that $x^2$ is Riemann integrable. Moreover, the lower sum is $\sum_{k=0}^{n-1}k^2/n^3=(n-1)n(2n-1)/(6n^3)$, which tends to $1/3$; since the upper-minus-lower gap tends to zero, the upper sums have the same limit and the integral is $1/3$.

Worked Example: A Lipschitz Bound for Sine

Consider $f(x)=\sin x$ on $[0,\pi]$. For any $x,y\in[0,\pi]$, the mean value theorem gives $|\sin x-\sin y|=|\cos c|\,|x-y|\le|x-y|$ for some $c$ between $x$ and $y$. Therefore, on an interval of length $h$, the supremum minus the infimum is at most $h$. Divide $[0,\pi]$ into $N$ equal intervals, each of length $h=\pi/N$. The oscillation contribution from each interval is at most $h^2$, so

U(f,P)-L(f,P)\le N h^2 =N\left(\frac{\pi}{N}\right)^2 =\frac{\pi^2}{N}.

For a given $\varepsilon>0$, choose an integer $N>\pi^2/\varepsilon$. Then the gap is at most $\pi^2/N<\varepsilon$, and the criterion proves Riemann integrability. Check: with $N=10$, the bound is $\pi^2/10$, obtained by summing ten contributions each at most $(\pi/10)^2$. The argument uses only the same Lipschitz estimate on each interval, not an assumption that sine is monotone on the whole partition.

What the Criterion Does—and Does Not—Say

The theorem is a statement about the total weighted oscillation for a suitable partition. It does not require the oscillation to be small on every interval for every partition. A partition can succeed by making intervals with larger oscillation sufficiently short. The order of quantifiers matters: for each tolerance $\varepsilon$, one may choose a partition depending on $\varepsilon$.

A common mistake is to add the oscillations but forget to multiply each by its interval length. For $f(x)=x^2$ on $[0,1]$ with $n=2$, the oscillations are $1/4$ and $3/4$. Adding them gives $1$, but this is not the Darboux gap. Each interval has length $1/2$, so the correct calculation is $(1/4)(1/2)+(3/4)(1/2)=1/2$. The gap identity weights oscillation by interval length precisely because Darboux sums approximate area using rectangles whose widths are those lengths.

Another error is to infer integrability from a single small-looking gap without checking the tolerance. The criterion requires partitions with gap below every positive tolerance, not merely one finite gap. Conversely, once the theorem proves integrability, its first direction guarantees a suitable partition for each tolerance. The equivalence is useful because it turns the definition of the Riemann integral into a practical estimate: bound the oscillation on each subinterval, multiply by its length, and sum.

Check Your Understanding

Use the definition of oscillation and the Darboux criterion to justify each answer.

For $f(x)=x$ on $[0,1]$ with $n$ equal subintervals, compute the oscillation on one interval and the full Darboux gap.
Explain why a common refinement of two partitions has an upper sum no larger than either original upper sum and a lower sum no smaller than either original lower sum.
For $f(x)=x^2$ on $[0,1]$, verify the gap when $n=4$ by summing the four length-weighted oscillations.
State where the mean value theorem enters the sine example and why its estimate applies even to an interval containing $\pi/2$.
In the quadratic example with $n=2$, explain why the sum of the two oscillations is not the Darboux gap.

Oscillation and the Darboux Integrability Criterion

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