Why Sample-Point Choices Need Not Matter
A tagged Riemann sum uses one sample point from each interval of a partition. Different choices of sample points can give different sums for the same partition. The important question is whether those differences disappear as the intervals become small.
For a Riemann-integrable function, they do. As long as the mesh tends to zero, every choice of tags produces sums approaching the same definite integral. The key is that integrability controls the total variation of the function across the partition intervals, even when the function is not continuous everywhere.
Partitions, Tags, and Oscillation
A partition \(P\) of \([a,b]\) is a finite sequence \(a=x_0<x_1<\cdots<x_n=b\). Its subintervals are \([x_{i-1},x_i]\), and their widths are \(\Delta x_i=x_i-x_{i-1}\). The mesh of \(P\) is the largest subinterval width:
A tagged partition consists of a partition together with a tag \(\xi_i\in[x_{i-1},x_i]\) for each subinterval. Its tagged Riemann sum for \(f\) is
For a bounded function \(f\), the oscillation on a subinterval \(J\) is the difference between the largest and smallest values the function takes there. More precisely, it is \(\sup_{x\in J}f(x)-\inf_{x\in J}f(x)\). This quantity measures how much changing the tag within \(J\) could change the function value.
We will also use the basic order property of the definite integral: if \(f\) is integrable on \([u,v]\), then
In particular, if \(\xi\in[u,v]\), then the difference between \(f(\xi)(v-u)\) and the integral over that interval has magnitude at most \(\operatorname{osc}(f;[u,v])(v-u)\). This is the estimate that connects oscillation to error.
Worked Example: Arbitrary Tags for a Linear Function
Worked Example: Arbitrary Tags for a Linear Function
Let \(f(x)=x\) on \([0,1]\), and use the uniform partition into \(n\) intervals. Each interval has width \(1/n\), and its tag may be any point \(\xi_i\in[(i-1)/n,i/n]\). The tagged sum is
Because each tag lies in its interval, the sum is between the sums formed by using the left and right endpoints:
Using \(\sum_{i=1}^{n}(i-1)=n(n-1)/2\) and \(\sum_{i=1}^{n}i=n(n+1)/2\), this becomes
The integral is \(\int_0^1x\,dx=1/2\). Thus, for every permitted choice of tags, \(|S_n-1/2|\le 1/(2n)\), which tends to zero as \(n\) increases. For example, when \(n=4\), every tagged sum lies between \(3/8\) and \(5/8\), so its distance from \(1/2\) is at most \(1/8\).
Check: The left and right endpoint sums are \(3/8\) and \(5/8\), respectively, and their midpoint is \(1/2\), the stated integral. The bound holds for every tag, not just for these endpoint choices.
Why the Integrability Condition Controls Tags
Suppose two tags are chosen in the same subinterval \(J\). The two corresponding function values differ by at most \(\operatorname{osc}(f;J)\). Multiplying by the subinterval width bounds the change in that term of the sum. Adding these bounds over a partition explains why the weighted oscillation condition matters.
For the full convergence proof, there is an additional issue: an arbitrary fine partition need not line up with the partition that has small weighted oscillation. The proof handles this by splitting intervals where they cross the endpoints of that chosen partition. Only a small total length is affected by those crossings.
Proof: Every Fine Tagged Sum Is Close to the Integral
Write \(I=\int_a^b f(x)\,dx\). Since \(f\) is bounded, choose \(M\ge0\) such that \(|f(x)|\le M\) for every \(x\in[a,b]\). Fix an arbitrary \(\varepsilon>0\).
By the integrability fact, there is a fixed partition \(P\) such that the sum of oscillation times width over its subintervals is less than \(\varepsilon/2\). Denote the number of internal partition points by \(r\).
Let \(Q\) be any tagged partition whose mesh is smaller than the smallest width of a subinterval of \(P\). Then no interval of \(Q\) can cross two internal points of \(P\). At most \(r\) intervals of \(Q\) cross an internal point, and their total length is at most \(r\|Q\|\).
Split each crossing interval of \(Q\) at the internal point of \(P\) that it crosses. Keep every other interval unchanged. The resulting intervals form a refinement of both partitions. Choose a tag in each refined interval, using the original tag on intervals that were not split.
On an interval of \(Q\) that is split, the original sum contribution has magnitude at most \(M\) times its width. The sum of contributions on the split pieces also has magnitude at most \(M\) times the original width. Therefore the difference between these contributions is at most \(2M\) times that width. Summing over crossing intervals gives a total change of at most \(2Mr\|Q\|\).
Every refined interval lies within one subinterval of \(P\). On each such interval, the error between its tagged contribution and its integral is at most its width times the oscillation on the containing subinterval of \(P\). Summing over all refined intervals gives an error less than \(\varepsilon/2\), because the refined intervals within each subinterval of \(P\) have total width equal to that subinterval’s width.
Combining the last two estimates yields
If \(Mr=0\), the second term is already zero. Otherwise, choose the mesh small enough that \(2Mr\|Q\|<\varepsilon/2\), while also keeping it smaller than the least width in \(P\). Then \(|S(Q,\xi)-I|<\varepsilon\). The choice of \(Q\) and its tags was arbitrary, so every tagged sum with sufficiently small mesh is within \(\varepsilon\) of \(I\). This proves the theorem.
The proof separates two sources of error. On intervals lying inside a subinterval where the function has little overall oscillation, any tag gives a contribution close to the integral. The intervals that cross the fixed partition’s boundaries can be controlled by their small total length and the bound \(M\).
Worked Example: A Jump Does Not Defeat Convergence
Worked Example: A Jump Does Not Defeat Convergence
Define \(f:[0,1]\to\mathbb R\) by \(f(x)=1\) when \(x<1/3\), and \(f(x)=0\) when \(x\ge1/3\). Its integral is \(1/3\). Consider any partition and any tags, with mesh \(h\).
Every interval wholly to the left of \(1/3\) contributes its full width, as does its part of the integral. Every interval wholly to the right contributes zero, as does its part of the integral. Only an interval containing or ending at \(1/3\) can contribute a discrepancy. Its width is at most \(h\), and the difference between its sum contribution and integral contribution is at most its width, since \(f\) only takes the values zero and one.
For instance, use partition points \(0,0.2,0.4,0.7,1\) and tags \(0.1,0.35,0.6,0.9\). The widths are \(0.2,0.2,0.3,0.3\), and the function values at the tags are \(1,0,0,0\). The tagged sum is therefore
The error is \(|0.2-1/3|=2/15\), which is less than the mesh \(h=0.3\). For any sequence of partitions with mesh tending to zero, the same error bound tends to zero, regardless of the tags.
Check: The sum includes the \(0.2\)-wide interval to the left of the jump and no other nonzero contribution. Its value and the stated error agree with the chosen tags and integral.
Worked Example: Nonuniform Partitions for \(f(x)=x\)
Worked Example: Nonuniform Partitions for \(f(x)=x\)
For \(f(x)=x\) on \([0,2]\), consider any partition with widths \(\Delta x_i\) and arbitrary tags. On an interval \([u,v]\), the oscillation is \(v-u=\Delta x_i\). The integral over that interval is \((v^2-u^2)/2\), while any tagged contribution is \(\xi_i\Delta x_i\). Since \(u\le\xi_i\le v\), their difference in absolute value is at most \((\Delta x_i)^2/2\). Consequently,
Here \(\int_0^2x\,dx=2\) and the interval widths sum to \(2\). To check a particular choice, take partition points \(0,0.5,1.5,2\) and tags \(0.1,0.8,1.8\). The widths are \(0.5,1,0.5\), so
The error is \(|1.75-2|=0.25\), no greater than the mesh \(1\). As the mesh tends to zero, the general bound shows that all tag choices converge to \(2\).
Check: The three terms sum to \(1.75\), and the integral over \([0,2]\) is \(2\), giving the stated error.
Common Pitfall: A Coarse Sum Need Not Be Accurate
The tag can matter substantially for a coarse partition. Integrability guarantees that the possible error becomes small when the mesh does, not that every individual tagged sum is already accurate.
What the Result Lets Us Do
Tag-independence is what makes Riemann sums a robust definition of accumulation. A calculation does not need to use left endpoints, right endpoints, midpoints, or any other special rule. If the function is integrable and the partition mesh tends to zero, each of these choices—and even choices that vary unpredictably from interval to interval—has the same limit.
The result does not say that all tags give the same sum for a fixed partition. It says that tag-dependent differences vanish in the limit. The small-oscillation condition controls the ordinary intervals, while boundedness controls the short pieces affected by partition-boundary crossings.
Check Your Understanding
Answer these questions using the definitions and estimates above.
- What does the mesh of a partition measure?
- How does oscillation on a subinterval bound the difference between two choices of tag there?
- Why does the proof split intervals that cross internal points of the partition with small weighted oscillation?
- For the function equal to one on \([0,1/3)\) and zero elsewhere, why is a single coarse tagged sum of zero not a counterexample to the theorem?
- Does the theorem require tags to follow one fixed rule as the partitions become finer?
