Partitions and Tagged Partitions of an Interval

The vocabulary of partitions, mesh, and refinement is not merely historical scaffolding — it is the literal backbone of the formal definition of the Riemann Integral, where one requires Riemann sums over tagged partitions to converge to a single number as $\|P\|\to 0$, and of Upper and Lower Darboux Sums, where refinement (via the common-refinement construction proved here) is exactly what shows that refining a partition can only increase lower sums and decrease upper sums, driving the Squeeze-type argument that identifies the integral. In numerical analysis, the same combinatorics underlies the trapezoidal rule, Simpson's rule, and adaptive quadrature algorithms, where meshes are refined locally (adding points precisely where a function varies quickly) using exactly the single-point-insertion refinements proved to exist here. In probability and measure theory, partitions of an interval generalize to partitions of a measurable space into disjoint measurable pieces, the direct ancestor being the discretization step used to build simple functions approximating a general measurable function. In numerical PDE methods (finite element and finite difference schemes), mesh refinement — literally the same word, with literally the same monotonicity property (refining shrinks the maximum cell size) — is the central computational primitive, and common refinements of two independently generated meshes are needed whenever two different discretizations must be compared or combined.

:::quiz[If $P = \{0,2,5,8\}$ and $Q=\{0,3,5,7,8\}$ are partitions of $[0,8]$, what is $\|P \vee Q\|$?]

• 3
• 2
• 8
• 1 ✓

:::quiz[Two partitions $P$ and $Q$ of $[a,b]$ are given. What can be said about comparability under $\preceq$?]

• They are always comparable: one must refine the other.
• They are never comparable.
• They may be incomparable, but always possess a common refinement $P \vee Q$. ✓
• They are comparable exactly when $\|P\| = \|Q\|$.

:::faq[Why do tagged partitions require tags in the closed subinterval $[x_{k-1},x_k]$ rather than the open interval?] Allowing tags at the endpoints $x_{k-1}$ or $x_k$ (not just strictly inside) is essential for Riemann's original formulation, since it recovers Cauchy's left-endpoint sums ($t_k = x_{k-1}$ for every $k$) and right-endpoint sums ($t_k = x_k$ for every $k$) as special cases of tagged partitions. Riemann's key requirement is that the Riemann sum converges to the same limit regardless of which admissible tags are chosen — including these boundary choices — as the mesh shrinks to zero. Restricting to the open interval would needlessly exclude these important special cases without changing which functions end up being integrable.

:::faq[Can a partition of $[a,b]$ have $n=0$ subintervals?] No. The definition requires $n \ge 1$, i.e. at least the two points $a$ and $b$ (which are automatically distinct since $a<b$), giving at least one subinterval, namely $[a,b]$ itself. There is no meaningful "partition with zero subintervals," since $[a,b]$ is nonempty ($a<b$) and must be covered by at least one closed subinterval. The partition $P=\{a,b\}$, with $n=1$ and mesh $\|P\|=b-a$, is the coarsest possible partition and is a perfectly legitimate degenerate case.

Ask the tutor: How does the mesh of a partition control the difference between upper and lower Darboux sums as $\|P\| \to 0$?

Think of $[a,b]$ as a stick you are about to saw into pieces. A partition is simply the list of sawing points, together with the two endpoints, listed in order from left to right. It says nothing yet about how you will use the pieces — that comes later, when you form sums. Right now a partition is pure combinatorics: a finite increasing sequence of numbers. The mesh $\|P\|$ answers a single question: what is the length of the worst (longest) piece? A partition with small mesh is "fine" everywhere — every piece is short. Crucially, mesh is a maximum, not an average: a partition can have a million tiny subintervals and one huge one, and its mesh is dictated entirely by that one huge piece. This matters enormously later, because Riemann's theorem requires every subinterval to shrink, not just most of them. Refinement is the natural order on partitions: $Q$ refines $P$ if $Q$ was obtained from $P$ by adding more sawing points (never removing any). Adding a point to $P$ inside some subinterval $I_k$ splits $I_k$ into two smaller pieces whose lengths sum to $\Delta x_k$; every other subinterval of $P$ is untouched. This is the source of the key monotonicity fact: refining a partition can never make any subinterval longer, so $\|Q\| \le \|P\|$ whenever $Q \succeq P$. Two different partitions $P$ and $Q$ need not be comparable — neither need refine the other — but they always have a common refinement, formed just by pooling all their sawing points together. This is the combinatorial fact that later lets one compare, e.g., the lower sum for $P$ with the lower sum for $Q$: both are dominated by quantities computed on $P \vee Q$. A tagged partition adds one more piece of data per subinterval: a chosen point $t_k \in I_k$ used later to sample a function's value on that piece. Different tags on the same partition $P$ give different tagged partitions; this extra freedom is precisely what makes Riemann sums (built from tagged partitions) more flexible — but harder to control — than Darboux sums (built from suprema/infima over untagged partitions alone).

:::svg[A partition of $[2,9]$ with $n=5$ subintervals, mesh $=2.5$, and one refinement point added]

Example 1 (mesh, refinement, and common refinement with an asymmetric partition). Let $[a,b] = [1,10]$ and consider two partitions:

Step 1: Compute subinterval lengths and mesh of $P$. The subintervals of $P$ are $[1,2],[2,4],[4,9],[9,10]$ with lengths

So $\|P\| = \max(1,2,5,1) = 5$, attained on $[4,9]$. Step 2: Compute subinterval lengths and mesh of $Q$. The subintervals of $Q$ are $[1,3],[3,5],[5,6],[6,10]$ with lengths

So $\|Q\| = \max(2,2,1,4) = 4$, attained on $[6,10]$. Step 3: Is either a refinement of the other? $P \not\subseteq Q$ (e.g. $2 \in P$, $2\notin Q$) and $Q \not\subseteq P$ (e.g. $3 \in Q$, $3 \notin P$). So $P$ and $Q$ are incomparable in $\preceq$ — neither refines the other. This is exactly why we need common refinements rather than assuming one partition always dominates another. Step 4: Form the common refinement.

Sorted, this gives subintervals

with lengths

Check: these sum to $1+1+1+1+1+3+1 = 9 = 10-1 = b-a$. Good, consistent with $\sum \Delta x_k = b-a$ for any partition. Step 5: Verify the corollary.

Indeed $\|R\| = 3 \le \min(\|P\|,\|Q\|) = \min(5,4) = 4$, confirming the corollary; note the inequality is strict here ($3 < 4$), which is typical — merging points can break up both long subintervals simultaneously. Observe also that the longest subinterval of $R$, namely $[6,9]$ with length $3$, arose because $P$'s point $x_3=9$ landed strictly inside $Q$'s longest subinterval $[6,10]$, cutting it down from length $4$ to length $3$ (splitting $[6,10]$ into $[6,9]$ and $[9,10]$), while $Q$'s point $6$ similarly cut into $P$'s worst subinterval $[4,9]$, splitting it into $[4,5],[5,6],[6,9]$ — the last of which, length $3$, becomes the new overall worst piece.

Example 2 (building a tagged partition and checking the tag constraint carefully). Let $[a,b] = [0,6]$ and take the partition

giving subintervals $I_1 = [0,1]$, $I_2 = [1,3]$, $I_3 = [3,6]$ with $\Delta x_1 = 1$, $\Delta x_2 = 2$, $\Delta x_3 = 3$, so $\|P\| = 3$. Step 1: propose tags. Consider the tag choice

Check the constraint $t_k \in I_k$:

• $t_1 = 1 \in [0,1]$ ✓ (right endpoint is allowed — the definition requires $t_k \in [x_{k-1},x_k]$, a closed interval).
• $t_2 = 1 \in [1,3]$ ✓ (left endpoint, also allowed).
• $t_3 = 5 \in [3,6]$ ✓.

So $(P,T)$ is a valid tagged partition. Note $t_1 = t_2 = 1$ is perfectly legal: the same real number can tag two different (adjacent) subintervals, since tags are chosen independently per subinterval — there is no injectivity requirement. Step 2: an invalid tag choice, to see why the constraint matters. Suppose instead we tried $t_2 = 3.5$. Since $3.5 \notin [1,3]$, this is not a valid tag for $I_2$; $(P,(1,3.5,5))$ is not a tagged partition of $[0,6]$ relative to $P$. This illustrates that the tag constraint is per-subinterval and must be checked against the correct $I_k$, not against $[a,b]$ as a whole (it would be a common student error to think any $t_k \in [a,b]=[0,6]$ suffices). Step 3: refine $(P,T)$. Refine $P$ by inserting the point $2$ into $I_2 = [1,3]$, giving

with subintervals $[0,1],[1,2],[2,3],[3,6]$. This $Q$ refines $P$, and by Theorem part (2),

in this instance — refinement did not strictly decrease the mesh, because the worst subinterval $[3,6]$ (length $3$) was untouched by the new point. This shows the theorem's inequality $\|Q\|\le \|P\|$ can be an equality: refining is guaranteed to be non-increasing on mesh, not strictly decreasing, unless the refining points specifically break up the longest subinterval(s). To tag $Q$, we must supply tags for all four subintervals of $Q$, including the two new pieces $[1,2]$ and $[2,3]$ that replaced $I_2$; the old tag $t_2 = 1$ for $I_2$ does not automatically transfer, since $1 \notin (2,3]$'s natural role. A valid choice is

checking $1\in[0,1]$, $1 \in [1,2]$, $3\in[2,3]$, $5\in[3,6]$ — all satisfied. Then $(Q,S)$ is a tagged refinement of $(P,T)$.