Why Sample-Point Choice Does not Change the Limit

Why Sample-Point Choice Matters

A Riemann sum estimates accumulated change by adding the heights of selected rectangles multiplied by their widths. For a fixed partition, choosing a different point in each subinterval generally changes those heights, so the finite sums need not be equal.

The key fact is that, for a continuous function on a closed interval, this difference becomes small as the subintervals become narrow. Left endpoints, right endpoints, midpoints, and other valid sample points therefore lead to the same definite-integral limit.

Partitions, Tags, and Riemann Sums

A partition \(P\) of \([a,b]\) is a list of points \(a=x_0<x_1<\cdots<x_m=b\). Its \(i\)th subinterval is \([x_{i-1},x_i]\), with width \(\Delta x_i=x_i-x_{i-1}\). The mesh of the partition is its largest subinterval width.

A tag \(\xi_i\) for the \(i\)th subinterval is any point in that subinterval. Given a function \(f\) defined on \([a,b]\), the tagged Riemann sum is the sum of each function value at a tag multiplied by its subinterval width.

For an equal-width partition, left-endpoint sums use each subinterval's left endpoint as its tag, right-endpoint sums use the right endpoint, and midpoint sums use the midpoint. A different valid tag may be chosen independently in each subinterval.

In the Riemann-sum definition of the definite integral, the mesh tends to zero. The result we need to understand is that, for a continuous function, the limiting value does not depend on how the tags are chosen.

Why Continuity Controls the Difference

A function is uniformly continuous on \([a,b]\) if, for every \(\varepsilon>0\), there exists \(\delta>0\) such that whenever \(x,y\in[a,b]\) and \(|x-y|<\delta\), then \(|f(x)-f(y)|<\varepsilon\). The same \(\delta\) works throughout the interval, not just near one particular point.

Every function continuous on a closed, bounded interval is uniformly continuous there. One way to see why is by contradiction. If uniform continuity failed, there would be some fixed \(\varepsilon_0>0\) and pairs \(x_n,y_n\in[a,b]\) such that \(|x_n-y_n|<1/n\) but \(|f(x_n)-f(y_n)|\ge\varepsilon_0\) for every \(n\). Since \([a,b]\) is bounded and closed, the sequence \(x_n\) has a convergent subsequence with limit \(c\in[a,b]\). Along that subsequence, \(|x_n-y_n|\) tends to zero, so \(y_n\) tends to \(c\) as well. Continuity at \(c\) then makes both function values tend to \(f(c)\), contradicting their separation by at least \(\varepsilon_0\).

For a uniform-continuity bound, consider any two tags \(\xi_i,\eta_i\) in the same subinterval. Their distance cannot exceed that subinterval's width. As the mesh gets smaller, uniform continuity guarantees that the corresponding function values get closer.

The Same-Partition Bound

Take two tagged sums using the same partition \(P\), with tags \(\xi_i\) in one sum and \(\eta_i\) in the other. Subtract the sums, apply the triangle inequality, and use uniform continuity. If the mesh is small enough that function values at points within a subinterval differ by less than \(\varepsilon\), then:

The last equality follows because the subinterval widths add to the full interval length. Thus all tag choices on the same sufficiently fine partition produce sums that are close to one another. Since \(\varepsilon\) can be made arbitrarily small, their difference tends to zero as the mesh tends to zero.

Why All Fine Tagged Sums Have One Limit

The same-partition bound explains why tag choices become unimportant when the partition is fixed. We can also show that sums formed from different fine partitions approach one common value. This establishes that the limiting value exists and is independent of the partitions and tags.

Because \(f\) is uniformly continuous, for each \(\varepsilon>0\) there is a mesh size \(\delta>0\) such that any two function values at points less than \(\delta\) apart differ by less than \(\varepsilon\). Take any two tagged sums whose partitions have mesh less than \(\delta\). Combine the partition points to form a common refinement. Choose one tag in each subinterval of this refinement, and use these same refinement tags to form a refined sum.

Compare the first original sum with the refined sum. In each original subinterval, its original tag and every refinement tag lie in that subinterval, so their function values differ by less than \(\varepsilon\). Weighting by widths and adding shows that the original sum differs from the refined sum by at most \(\varepsilon(b-a)\). The same reasoning applies to the second original sum. By the triangle inequality, the two original sums differ by at most \(2\varepsilon(b-a)\).

To complete the argument, the bound \(2\varepsilon(b-a)\) can be made arbitrarily small by choosing the uniform-continuity tolerance \(\varepsilon\) small and the partitions fine enough. Thus the family of fine tagged sums is Cauchy: for any desired closeness, all sufficiently fine sums differ from one another by less than that amount. Completeness of the real numbers guarantees a common limit. Since the bound did not depend on the tag-selection rule, no valid rule can produce a different limit.

What the Result Does—and Does Not—Say

The result matters because it tells us the definite integral is not tied to a special rectangle-placement rule. A convenient choice of tags may be used to calculate or estimate a sum; for a continuous function, refining the partition makes the choice immaterial to the limit.

A common mistake is to conclude that different sample-point rules give equal sums for every fixed partition. The first example disproves that: \(L_4=14/64\), while \(R_4=30/64\), so \(R_4-L_4=16/64=1/4\). The correction is to distinguish a finite approximation from its limit. The theorem says that the difference tends to zero as the mesh tends to zero, not that it is already zero.

The continuity condition is important. The argument depends on function values becoming uniformly close when their input points are close. If a function has a discontinuity, that guarantee may fail, so one cannot apply this theorem without checking its condition.