Why the Darboux Criterion Needs a Proof
A tagged Riemann sum uses one selected function value on each subinterval. Upper and lower Darboux sums instead use the supremum and infimum of the function there. The two constructions are related, but they are not the same: a single choice of tags does not reveal the full range of values on each subinterval.
The Darboux integrability criterion says that a bounded function is Riemann integrable precisely when partitions can make the difference between the upper and lower sums as small as desired. We will prove both directions. The key idea is that sufficiently fine tagged sums must be trapped between upper and lower sums whose gap is small.
Partitions and Their Darboux Sums
A partition \(P\) of \([a,b]\) is a finite increasing list \(a=x_0<x_1<\cdots<x_n=b\). Its subintervals are \([x_{i-1},x_i]\), with widths \(\Delta x_i=x_i-x_{i-1}\). On each subinterval define \(m_i=\inf\{f(x):x\in[x_{i-1},x_i]\}\) and \(M_i=\sup\{f(x):x\in[x_{i-1},x_i]\}\).
The lower Darboux sum uses the infimum on each subinterval, and the upper Darboux sum uses the supremum:
Since \(m_i\le M_i\) and every width is positive, \(L(f,P)\le U(f,P)\). The difference can also be written as
Each term measures the range of values on one subinterval, weighted by that subinterval’s width. A small total gap means that, across the partition, the possible variation has small weighted contribution.
A tagged partition consists of a partition \(P\) together with tags \(\xi_i\in[x_{i-1},x_i]\). Its tagged Riemann sum is \(S(f;P,\xi)=\sum_{i=1}^{n}f(\xi_i)\Delta x_i\). The mesh of \(P\), written \(\lVert P\rVert\), is the largest subinterval width. Riemann integrability means there is a real number \(I\) such that, for every \(\varepsilon>0\), some \(\delta>0\) makes \(|S(f;P,\xi)-I|<\varepsilon\) for every tagged partition with \(\lVert P\rVert<\delta\).
Two Basic Comparisons
First, every tagged sum for a given partition lies between that partition’s lower and upper sums. For each \(i\), the tag belongs to the subinterval, so \(m_i\le f(\xi_i)\le M_i\). Multiplying by \(\Delta x_i>0\) and adding gives
Second, refining a partition cannot decrease its lower sum or increase its upper sum. When a subinterval is split, the infimum on each smaller piece is at least the infimum on the original piece, and each smaller-piece supremum is at most the original supremum. Multiplying these inequalities by the smaller widths and adding proves the claim.
Any two partitions \(P\) and \(Q\) have a common refinement: take all of their partition points together, in increasing order. Calling this refinement \(R\), the refinement comparison gives \(L(f,P)\le L(f,R)\le U(f,R)\le U(f,Q)\). Thus every lower sum is at most every upper sum.
From Small Darboux Gaps to Tagged-Sum Convergence
Suppose that for every positive tolerance there is a partition with a smaller upper-lower gap. Define \(\alpha\) as the supremum of all lower sums \(L(f,P)\), and \(\beta\) as the infimum of all upper sums \(U(f,P)\), as \(P\) ranges over partitions of \([a,b]\). These are finite because \(f\) is bounded and the interval has finite length. The comparison between arbitrary lower and upper sums shows that \(\alpha\le\beta\).
For any \(\varepsilon>0\), choose a partition \(P\) with \(U(f,P)-L(f,P)<\varepsilon\). Since \(\alpha\ge L(f,P)\) and \(\beta\le U(f,P)\),
This holds for every \(\varepsilon>0\), so \(\alpha=\beta\). Write their common value as \(I\). For every partition \(T\), the definitions of supremum and infimum give \(L(f,T)\le I\le U(f,T)\).
Now fix \(\varepsilon>0\) and choose \(P\) with \(U(f,P)-L(f,P)<\varepsilon\). If \(T\) refines \(P\), every tagged sum on \(T\) lies between \(L(f,T)\) and \(U(f,T)\), and refinement gives \(L(f,T)\ge L(f,P)\) and \(U(f,T)\le U(f,P)\). Therefore every such sum lies within the interval \([L(f,P),U(f,P)]\), whose width is less than \(\varepsilon\). However, integrability requires a guarantee for all sufficiently fine partitions, not only refinements. To obtain that guarantee, we control the subintervals of a fine partition that cross points of \(P\).
Let \(P\) have \(r\) subintervals, and let \(|f(x)|\le B\) on \([a,b]\). For an arbitrary partition \(T\), call a subinterval of \(T\) a boundary subinterval if its interior contains an interior partition point of \(P\). There are at most \(r-1\) such subintervals. Their total width is at most \((r-1)\lVert T\rVert\), and the difference between the supremum and infimum of \(f\) on any one of them is at most \(2B\).
Every other subinterval of \(T\) lies within one subinterval of \(P\). Its contribution to \(U(f,T)-L(f,T)\) is bounded by the oscillation \(M_i-m_i\) on that parent subinterval, multiplied by its width. Summing these contributions over each parent subinterval gives at most its contribution to \(U(f,P)-L(f,P)\). Consequently,
Choose \(P\) with gap less than \(\varepsilon/2\). If \(B>0\) and \(r>1\), choose \(\delta>0\) small enough that \(2B(r-1)\delta<\varepsilon/2\). If \(B=0\) or \(r=1\), the boundary contribution is zero and any positive \(\delta\) suffices. For \(\lVert T\rVert<\delta\), the displayed bound gives \(U(f,T)-L(f,T)<\varepsilon\). Since \(L(f,T)\le I\le U(f,T)\), every tagged sum on \(T\) satisfies \(|S(f;T,\xi)-I|<\varepsilon\). This proves Riemann integrability.
Worked Example: Estimating a Darboux Gap from Interval Data
Suppose a partition of \([0,1]\) has two subintervals, each of width \(1/2\), and the oscillation \(M_i-m_i\) on them is at most \(1/10\) and \(1/5\), respectively. The gap is bounded by the sum of width times oscillation:
Since \(3/20<1/5\), this partition meets the criterion’s requirement for tolerance \(1/5\). As a check, the two individual contributions are \(1/20\) and \(1/10\), and their sum is \(3/20\), as computed. The calculation uses bounds on the oscillations; if these are strict upper bounds rather than exact values, the resulting inequality remains valid.
Worked Example: Using a Lipschitz Bound
Suppose \(f:[-1,2]\to\mathbb{R}\) satisfies \(|f(x)-f(y)|\le 2|x-y|\) for all \(x,y\in[-1,2]\). Take a uniform partition into \(n\) pieces, each of width \(3/n\). On a single piece, any two function values differ by at most \(2(3/n)=6/n\), so the supremum minus infimum there is at most \(6/n\). Thus
For tolerance \(3/2\), choose \(n=13\). Then the gap is at most \(18/13\), and \(18/13<3/2\) because \(36<39\). The partition therefore gives a gap strictly below the requested tolerance. More generally, for any \(\varepsilon>0\), choosing an integer \(n>18/\varepsilon\) makes \(18/n<\varepsilon\). This verifies the criterion for any function with the stated Lipschitz bound.
From Tagged-Sum Convergence to Small Darboux Gaps
For the reverse direction, suppose \(f\) is Riemann integrable with value \(I\). Given a desired gap \(\varepsilon>0\), use integrability with the smaller tolerance \(\varepsilon/4\). There is a \(\delta>0\) such that every tagged sum on a partition with mesh below \(\delta\) differs from \(I\) by less than \(\varepsilon/4\). Choose any partition \(P\) with mesh below \(\delta\).
Let \(\eta=\varepsilon/(8(b-a))\). For each subinterval, the definition of supremum permits a tag \(\xi_i^+\) such that \(f(\xi_i^+)>M_i-\eta\); the definition of infimum permits a tag \(\xi_i^-\) such that \(f(\xi_i^-)<m_i+\eta\). These choices need not attain the supremum or infimum. They only need to approximate them closely enough.
Let \(S^+\) and \(S^-\) be the tagged sums formed from the respective tags. Since each is a tagged sum on \(P\), both are within \(\varepsilon/4\) of \(I\), so \(S^+-S^-<\varepsilon/2\). The approximations on each interval also give \(U(f,P)-S^+<\eta(b-a)\) and \(S^--L(f,P)<\eta(b-a)\). Adding these bounds yields
Thus an appropriate partition has an upper-lower gap below every prescribed positive tolerance. This completes the other direction and proves the criterion.
Worked Example: Turning Tagged-Sum Control into a Gap Bound
Suppose \(f\) is integrable on \([0,2]\), and every tagged sum on a partition with mesh below some \(\delta\) differs from \(I\) by less than \(0.01\). To obtain a Darboux gap below \(0.03\), use \(\eta=0.001\), since \(2\eta(b-a)=2(0.001)(2)=0.004\). Choose a partition \(P\) with mesh below \(\delta\), and on each subinterval choose tags within \(0.001\) of its supremum and infimum in the required directions. The two resulting sums differ by less than \(0.02\), because each is within \(0.01\) of \(I\). Therefore,
The check is that the two tag-approximation errors contribute \(0.001\cdot2\) each, for a total of \(0.004\); adding this to the \(0.02\) sum separation gives \(0.024\). This is strictly below the target \(0.03\).
A Common Mistake: One Tag Does Not Give the Upper-Lower Gap
A tempting but incorrect argument is to take one tagged sum and claim that it determines the upper and lower sums. For example, let \(f(x)=x\) on \([0,1]\) and use the partition consisting of the single subinterval \([0,1]\). Choosing the midpoint as tag gives the tagged sum \(1/2\). It would be wrong to infer that the upper-lower gap is zero just because this one sum has a single value.
In fact, on this subinterval the infimum is \(0\), the supremum is \(1\), and the width is \(1\). Hence \(L(f,P)=0\), \(U(f,P)=1\), and \(U(f,P)-L(f,P)=1\). The correction is to compare a tagged sum with the lower and upper sums, not to identify it with either one. To prove the reverse implication, use two carefully chosen tagged sums that approximate the suprema and infima on every subinterval.
Why the Criterion Is Useful
The tagged-sum definition requires control of every possible choice of tags once the mesh is sufficiently small. The Darboux criterion expresses the same integrability question through a partition-specific estimate: make the total weighted oscillation small. The proof explains the connection. A small Darboux gap traps every tagged sum for that partition, while sufficiently fine arbitrary partitions have only a small extra contribution from intervals crossing the fixed partition’s boundaries.
For the opposite direction, integrability controls all tag choices on a sufficiently fine partition. Tags can be selected close to each interval’s supremum and infimum, so the upper and lower sums cannot remain far apart. Both directions depend on boundedness and on the ability to choose tags inside each subinterval; neither requires the function to attain its supremum or infimum.
Check Your Understanding
Answer these questions using the definitions and proof above.
- Why does every tagged sum for a partition lie between its lower and upper Darboux sums?
- Why does refining a partition raise or preserve the lower sum and lower or preserve the upper sum?
- In the proof from the criterion to tagged-sum convergence, why must boundary subintervals be controlled separately?
- Why can tags approximating a supremum and infimum be chosen even when those values are not attained?
- Explain why one tagged sum on a partition cannot, by itself, determine the upper-lower gap.
