Tagged Partitions

From Riemann Sums to Tagged Partitions

A Riemann sum uses a partition and one chosen sample point from each subinterval. The partition determines the intervals and their lengths; the sample points determine which function values are used. A tagged partition records these two choices together. This notation is useful because it lets us discuss sums while keeping track of exactly where each value is sampled.

Recall that a partition of $[a,b]$ is a finite set $P=\{x_0,x_1,\ldots,x_n\}$ with $a=x_0<x_1<\cdots<x_n=b$, and that $\Delta x_i=x_i-x_{i-1}$. For each subinterval $[x_{i-1},x_i]$, choose a tag $\xi_i$ belonging to that closed subinterval. In particular, tags may be endpoints, and a shared endpoint may be chosen as the tag for either adjacent subinterval.

Definition: A tagged partition of $[a,b]$ is a partition $P=\{x_0,\ldots,x_n\}$, together with points $\xi_i\in[x_{i-1},x_i]$ for $i=1,\ldots,n$. Its tagged Riemann sum is $$ S(f;P,\xi)=\sum_{i=1}^{n}f(\xi_i)\Delta x_i. $$ The mesh is $\|P\|=\max_{1\leq i\leq n}\Delta x_i$; it depends on the partition, not on the tags.

Thus two tagged partitions can use the same partition but different tags, or different partitions and tags. The notation $S(f;P,\xi)$ makes both choices explicit. This is the Riemann-sum notation from the previous tutorial, now viewed as the sum associated with a tagged partition.

Worked Examples: Making the Tags Explicit

Worked Example: Two Tag Choices on the Same Partition

Let $f(x)=2x+1$ on $[0,1]$, and take $P=\{0,\frac{3}{10},1\}$. The subinterval lengths are $\frac{3}{10}$ and $\frac{7}{10}$. Choose tags $\xi_1=\frac{1}{10}$ and $\xi_2=\frac{7}{10}$. Since $f(\frac{1}{10})=\frac{6}{5}$ and $f(\frac{7}{10})=\frac{12}{5}$, the tagged sum is

S(f;P,\xi) =\frac65\cdot\frac3{10}+\frac{12}{5}\cdot\frac7{10} =\frac{18}{50}+\frac{84}{50} =\frac{51}{25}.

Now keep the partition but choose tags $\eta_1=\frac{3}{10}$ and $\eta_2=1$. We have $f(\frac{3}{10})=\frac85$ and $f(1)=3$, so

S(f;P,\eta) =\frac85\cdot\frac3{10}+3\cdot\frac7{10} =\frac{24}{50}+\frac{105}{50} =\frac{129}{50}.

The sums differ by $\frac{129}{50}-\frac{51}{25}=\frac{27}{50}$. The partition and interval lengths have not changed; the difference comes entirely from the tags.

Worked Example: Tags on Unequal Subintervals

Let $f(x)=x^2$ on $[0,1]$, and use $P=\{0,\frac14,1\}$. The lengths are $\frac14$ and $\frac34$. With tags $\xi_1=0$ and $\xi_2=\frac12$, the sum is

S(f;P,\xi) =0^2\cdot\frac14+\left(\frac12\right)^2\cdot\frac34 =0+\frac{3}{16} =\frac{3}{16}.

With tags $\eta_1=\frac14$ and $\eta_2=1$, instead, the sum is

S(f;P,\eta) =\left(\frac14\right)^2\cdot\frac14+1^2\cdot\frac34 =\frac{1}{64}+\frac{48}{64} =\frac{49}{64}.

The interval lengths weight the chosen function values: a tag on the longer subinterval contributes with weight $\frac34$, not $\frac14$. This is why a tagged sum is not just an unweighted average of the sampled values.

Worked Example: A Tag at a Jump

Define $g:[0,1]\to\mathbb{R}$ by $g(x)=0$ for $0\leq x<\frac12$ and $g(x)=1$ for $\frac12\leq x\leq1$. For the one-interval partition $P=\{0,1\}$, the tag $\xi_1=\frac14$ gives

S(g;P,\xi)=g\left(\frac14\right)\cdot 1=0,

whereas the tag $\eta_1=\frac34$ gives $S(g;P,\eta)=g(\frac34)\cdot1=1$. The same interval can therefore give very different sums when the mesh is large. If we use $P'=\{0,\frac12,1\}$ with tags $\frac14$ and $\frac34$, then

S(g;P',\zeta) =g\left(\frac14\right)\cdot\frac12+ g\left(\frac34\right)\cdot\frac12 =0+\frac12=\frac12.

The point $\frac12$ is permitted as the tag on either adjacent subinterval. But $g(\frac12)=1$, so choosing it on the first subinterval changes that subinterval’s contribution. Tags at endpoints matter when a function has a jump; they are not required to be interior points.

What Happens When a Tagged Partition Is Refined?

A partition $Q$ refines $P$ when it contains every point of $P$. A tagged partition based on such a $Q$ is called a tagged refinement of the original tagged partition, regardless of how the new tags are chosen. Refinement preserves the original cut points, but it does not in general preserve the tagged sum: the tags on the smaller intervals may sample different values.

The change can be controlled by the oscillation of the function on each original subinterval. For a bounded function $f$ and an interval $J\subseteq[a,b]$, write $\omega(f;J)=\sup\{|f(x)-f(y)|:x,y\in J\}$. This is the largest possible difference between two values of $f$ on $J$.

Theorem (Tagged-Sum Change Under Refinement): Let $f:[a,b]\to\mathbb{R}$ be bounded. Let $P=\{x_0,\ldots,x_n\}$, with tags $\xi_i\in[x_{i-1},x_i]$, and let $Q$ be a refinement of $P$, with arbitrary tags on the subintervals of $Q$. Then $$ \left|S(f;P,\xi)-S(f;Q,\eta)\right| \leq\sum_{i=1}^{n}\omega\bigl(f;[x_{i-1},x_i]\bigr)\Delta x_i. $$

Proof. Fix one original subinterval $[x_{i-1},x_i]$. The refinement divides it into finitely many subintervals $J_{i,1},\ldots,J_{i,r_i}$, whose lengths add to $\Delta x_i$. Let $\eta_{i,j}$ be the tag in $J_{i,j}$. The original contribution from this interval is $f(\xi_i)\Delta x_i$, and its contributions in the refined sum total $\sum_{j=1}^{r_i}f(\eta_{i,j})|J_{i,j}|$. Since $\sum_j|J_{i,j}|=\Delta x_i$, their difference is

\sum_{j=1}^{r_i}\bigl(f(\eta_{i,j})-f(\xi_i)\bigr)|J_{i,j}|.

Both $\eta_{i,j}$ and $\xi_i$ belong to $[x_{i-1},x_i]$. Therefore $\left|f(\eta_{i,j})-f(\xi_i)\right|\leq\omega(f;[x_{i-1},x_i])$. Applying the triangle inequality and using the sum of the refined lengths gives

\left|\sum_{j=1}^{r_i}\bigl(f(\eta_{i,j})-f(\xi_i)\bigr)|J_{i,j}|\right| \leq\omega(f;[x_{i-1},x_i])\sum_{j=1}^{r_i}|J_{i,j}| =\omega(f;[x_{i-1},x_i])\Delta x_i.

Now sum these bounds over $i=1,\ldots,n$. The differences of the interval contributions add to the difference of the two full sums, so another application of the triangle inequality proves the stated estimate. $\square$

This estimate does not claim that a refinement leaves the sum unchanged. Instead, it bounds the possible change using the oscillation on each parent interval. If $f$ varies little on those intervals, the sum is stable under refinement; if it varies substantially, refinement can change the sum substantially.

How Much Can Different Tags Change a Sum?

The same idea applies even when the partition stays fixed. Each tag selected for an interval samples a value in that interval, so any two such values differ by no more than the interval’s oscillation. The following estimate makes the dependence on mesh especially clear when $f$ is uniformly continuous.

Theorem (Tag-Choice Bound for a Uniformly Continuous Function): Suppose $f:[a,b]\to\mathbb{R}$ is uniformly continuous. Define $$ \omega_f(\delta)=\sup\{|f(x)-f(y)|:x,y\in[a,b],\ |x-y|\leq\delta\}. $$ For any partition $P$ and any two choices of tags $\xi$ and $\eta$ on $P$, $$ \left|S(f;P,\xi)-S(f;P,\eta)\right| \leq(b-a)\omega_f(\|P\|). $$ Moreover, $\omega_f(\delta)\to0$ as $\delta\to0^+$.

Proof. Fix an interval $[x_{i-1},x_i]$. Its tags $\xi_i$ and $\eta_i$ satisfy $|\xi_i-\eta_i|\leq\Delta x_i\leq\|P\|$. By the definition of $\omega_f$, $\left|f(\xi_i)-f(\eta_i)\right|\leq\omega_f(\|P\|)$. Thus

\begin{aligned} \left|S(f;P,\xi)-S(f;P,\eta)\right| &\leq\sum_{i=1}^{n}|f(\xi_i)-f(\eta_i)|\Delta x_i\\ &\leq\omega_f(\|P\|)\sum_{i=1}^{n}\Delta x_i\\ &=(b-a)\omega_f(\|P\|). \end{aligned}

The bound says that tag choices become less influential as the mesh decreases. In particular, for any sequence of partitions whose meshes tend to zero, the difference between any two tagged sums on each partition tends to zero. This is a statement about comparing choices on the same partition; the Convergence of Riemann Sums from the previous tutorial gives the stronger conclusion that, for an integrable function, all tagged sums on sufficiently fine partitions are close to the integral.

Why Tagged Partitions Matter

The word “tag” keeps a crucial choice visible. A partition alone specifies where the interval is cut, but it does not specify which value of the function represents each interval. For continuous functions, uniform continuity ensures that nearby tag choices lead to nearby values when the mesh is small. For discontinuous functions, a tag placed at a jump can change a finite sum, as the step-function example shows.

A common pitfall is to assume that a finer partition automatically produces the same sum as before. It does not: refining replaces a single sampled value times a full interval length by several sampled values times smaller lengths. The refinement estimate explains when the change is controlled. Another pitfall is to exclude endpoints from the allowed tag locations. The standard definition uses closed subintervals, so endpoints are allowed, including shared endpoints of adjacent intervals.

Tagged partitions provide a precise way to discuss arbitrary Riemann sums: the partition controls the mesh, while the tags record the sampling choices. The results here quantify how those choices interact with refinement and with the variation of the function across each subinterval.

Check Your Understanding

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

Can a tag be an endpoint of its subinterval? Can a shared endpoint serve as the tag for either adjacent subinterval?
In the refinement estimate, why do the lengths of the refined subintervals inside one original interval add to the original interval’s length?
What quantity controls the possible change in a tagged sum when one tagged partition is refined?
For a uniformly continuous function, what happens to the bound on the difference between two tagged sums as the mesh tends to zero?
Why can choosing a tag at the jump in the step-function example change a finite sum?

What You'll Learn