Refining a Partition

Refinement as a Controlled Operation

A partition is often improved by inserting additional points, but “improved” needs care: a refinement cannot worsen the upper and lower sums, yet it need not make them strictly closer. The previous tutorial used common refinements to combine estimates. Here we examine the operation itself, one inserted point at a time, and calculate exactly what happens to the sums on the interval that is split.

Let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be bounded. Write a partition as $P=\{x_0,\ldots,x_n\}$, where $a=x_0<x_1<\cdots<x_n=b$. As in “Partitions of an Interval,” a partition $Q$ refines $P$ when every point of $P$ is also a point of $Q$. We use the established result that upper sums do not increase and lower sums do not decrease under refinement. Our focus is the local calculation behind a single insertion.

Splitting One Partition Interval

Suppose $u$ and $v$ are consecutive points of $P$, and choose $c$ with $u<c<v$. The new partition $P'$ is obtained by inserting $c$. Define the supremum and infimum of $f$ on the original interval and its two pieces by

M=\sup_{x\in[u,v]}f(x),\qquad m=\inf_{x\in[u,v]}f(x),

and

M_1=\sup_{x\in[u,c]}f(x),\quad m_1=\inf_{x\in[u,c]}f(x), \qquad M_2=\sup_{x\in[c,v]}f(x),\quad m_2=\inf_{x\in[c,v]}f(x).

These quantities are finite because $f$ is bounded. Since each smaller interval lies in $[u,v]$, we have $M_1,M_2\leq M$ and $m\leq m_1,m_2$. The contribution of $[u,v]$ to the old upper sum is $M(v-u)$; after insertion, it is $M_1(c-u)+M_2(v-c)$. The corresponding lower-sum contributions are $m(v-u)$ and $m_1(c-u)+m_2(v-c)$.

Proposition (Exact Change Under a Single Insertion): If $P'$ is obtained from $P$ by inserting $c$ between consecutive points $u<v$, then $$ U(f,P)-U(f,P') =(M-M_1)(c-u)+(M-M_2)(v-c), $$ and $$ L(f,P')-L(f,P) =(m_1-m)(c-u)+(m_2-m)(v-c), $$ where the suprema and infima are taken on $[u,v]$, $[u,c]$, and $[c,v]$ as defined above. Consequently, $$ \bigl(U(f,P)-L(f,P)\bigr)-\bigl(U(f,P')-L(f,P')\bigr) =U(f,P)-U(f,P')+L(f,P')-L(f,P)\geq 0. $$

Proof. Every interval of $P$ other than $[u,v]$ remains unchanged, so its contribution to either sum cancels when the old and new sums are subtracted. The upper-sum difference is therefore

M(v-u)-\bigl(M_1(c-u)+M_2(v-c)\bigr).

Since $v-u=(c-u)+(v-c)$, this equals

$$ (M-M_1)(c-u)+(M-M_2)(v-c). $$

The lower-sum difference in the other direction is

\bigl(m_1(c-u)+m_2(v-c)\bigr)-m(v-u) =(m_1-m)(c-u)+(m_2-m)(v-c).

Both expressions are nonnegative: $M\geq M_1,M_2$, $m_1,m_2\geq m$, and both subinterval lengths are positive. Finally, subtracting the new gap from the old gap gives the sum of these two differences. This proves all three identities and the nonnegativity claim. $\square$

The formula gives more information than the general theorem on upper and lower sums under refinement. It shows that only the interval being split can change the sums, and it identifies the amount of change using the local suprema and infima. The gap shrinks by the upper-sum decrease plus the lower-sum increase. Either contribution, or both, may be zero.

Worked Examples: What One Insertion Can Do

Worked Example: Inserting a Point for a Cubic

Let $f(x)=x^3$ on $[0,2]$, and begin with $P=\{0,2\}$. This function is increasing there, so its infimum on the interval is $0$ and its supremum is $8$. The interval has length $2$, giving

L(f,P)=0\cdot2=0,\qquad U(f,P)=8\cdot2=16.

Insert $c=1$, so $P'=\{0,1,2\}$. On $[0,1]$, the infimum is $0$ and the supremum is $1$; on $[1,2]$, they are $1$ and $8$. Both pieces have length $1$, so

L(f,P')=0\cdot1+1\cdot1=1,\qquad U(f,P')=1\cdot1+8\cdot1=9.

The gap falls from $16-0=16$ to $9-1=8$. The insertion identities verify the changes directly: $M=8$, $M_1=1$, $M_2=8$, and $m=0$, $m_1=0$, $m_2=1$. Thus the upper sum decreases by $(8-1)\cdot1+(8-8)\cdot1=7$, while the lower sum increases by $(0-0)\cdot1+(1-0)\cdot1=1$. The total gap reduction is $7+1=8$, as calculated from the sums.

Worked Example: An Insertion Near a Jump

Define $f:[0,1]\to\mathbb{R}$ by $f(x)=0$ for $x<2/5$ and $f(x)=1$ for $x\geq2/5$. For $P=\{0,1\}$, the infimum is $0$ and the supremum is $1$ on the whole interval. Hence $L(f,P)=0$, $U(f,P)=1$, and the gap is $1$.

Insert $c=2/5$. On $[0,2/5]$, the function takes both values $0$ and $1$: the value at $2/5$ is $1$. On $[2/5,1]$, it is identically $1$. Therefore,

L(f,P')=0\cdot\frac25+1\cdot\frac35=\frac35, \qquad U(f,P')=1\cdot\frac25+1\cdot\frac35=1.

The new gap is $1-\frac35=\frac25$. The interval ending at the jump still contributes an oscillation, because its right endpoint belongs to that interval. To reduce this remaining contribution, choose $0<\delta<2/5$ and use $R=\{0,2/5-\delta,2/5,1\}$. The first interval has $f=0$ throughout. The middle interval has infimum $0$, supremum $1$, and length $\delta$; the last interval is constant at $1$. Thus

L(f,R)=\frac35,\qquad U(f,R)=\frac35+\delta, \qquad U(f,R)-L(f,R)=\delta.

The calculation shows why placing a partition point at a jump is not, by itself, enough to eliminate the gap. The interval just to its left still contains the jump value at its endpoint. Instead, making that interval short controls its contribution.

Worked Example: A Refinement That Does Not Improve the Gap

Let $f(x)=|x-\frac12|$ on $[0,1]$, with $P=\{0,1\}$. The minimum is $0$, the maximum is $1/2$, and

L(f,P)=0,\qquad U(f,P)=\frac12,\qquad U(f,P)-L(f,P)=\frac12.

Insert $1/2$. On each half-interval, the minimum is $0$, the supremum is $1/2$, and the length is $1/2$. Consequently $L(f,P')=0$ and $U(f,P')=\frac12\cdot\frac12+\frac12\cdot\frac12=\frac12$: the gap is unchanged. In the single-insertion formula, the old supremum $M=1/2$ is also the supremum on both new pieces, while the old infimum and both new infima are all $0$. Each change is therefore zero.

Now refine further using $R=\{0,1/4,1/2,3/4,1\}$. On each of these four intervals, the oscillation is $1/4$, and each interval has length $1/4$. Its contribution to the gap is $(1/4)(1/4)=1/16$. Adding the four contributions gives $U(f,R)-L(f,R)=1/4$. Refinement did not strictly reduce the gap at the first insertion, but later insertions did reduce it.

Building a Refinement One Point at a Time

A finite refinement can always be viewed as a sequence of single-point insertions. This observation lets us apply the local calculation repeatedly and is also useful when constructing a partition subject to several requirements.

Theorem (Sequential Construction of a Finite Refinement): Let $P$ and $Q$ be partitions of $[a,b]$, with $Q$ a refinement of $P$. Then there is a finite sequence of partitions beginning with $P$ and ending with $Q$, in which each partition after the first is obtained from the preceding one by inserting a single point. At every stage the upper sum does not increase and the lower sum does not decrease.

Proof. Since $P$ and $Q$ are finite and $P\subseteq Q$, the set $Q\setminus P$ is finite. List its points as $y_1,\ldots,y_k$, with no repetitions. Starting from $P$, insert $y_1$, then $y_2$, and continue through $y_k$. At every stage the current set of points is a subset of $Q$, so the point about to be inserted is not already present. It lies strictly between $a$ and $b$, and hence between two consecutive points of the current partition. Each insertion therefore gives a partition. After all $k$ insertions, the point set is $P\cup(Q\setminus P)=Q$. If $k=0$, then $P=Q$ and the sequence consists just of $P$. The single-insertion identities show that upper sums do not increase and lower sums do not decrease at each step; chaining these inequalities gives the same comparison between the initial and final sums. $\square$

The order of insertion is not important for the conclusion, although it can matter for the intermediate values of the sums. At each step the identity applies to the particular interval currently being split. This gives a practical way to analyze a complicated refinement without recomputing every sum from scratch.

Proposition (Refining While Preserving the Original Points): Given a partition $P$ of $[a,b]$ and $\eta>0$, there is a finite refinement $R$ of $P$ with $\|R\|<\eta$.

Proof. Write $P=\{x_0,\ldots,x_n\}$, and let $\ell_i=x_i-x_{i-1}>0$ be the length of its $i$th interval. For each $i$, choose a positive integer $N_i$ with $N_i>\ell_i/\eta$. Divide $[x_{i-1},x_i]$ into $N_i$ equal subintervals by adding the points

x_{i-1}+\frac{j\ell_i}{N_i}, \qquad 1\leq j<N_i.

Retain all points of $P$, and call the resulting set $R$. Each new interval has length $\ell_i/N_i<\eta$, because $N_i>\ell_i/\eta$. Thus $R$ refines $P$ and its mesh, the largest subinterval length, is less than $\eta$. If $N_i=1$, no new point is needed in that original interval, and its length is already less than $\eta$. This proves the proposition. $\square$

This construction differs from choosing an arbitrary fine partition: it guarantees that every point already in $P$ remains present. That matters when the original points were chosen to mark locations where the function changes behavior or where other conditions must hold.

What Refinement Guarantees—and What It Does Not

The exact insertion identities explain the established theorem on upper and lower sums under refinement, but the theorem alone does not promise strict improvement. If the supremum on each new piece equals the old supremum and the infimum on each new piece equals the old infimum, neither sum changes. The absolute-value example exhibits this possibility. Nor does a finer mesh, by itself, give a numerical bound on the Darboux gap for every bounded function: the oscillations must also be controlled.

For an integrable function, the Darboux Criterion says that suitable partitions can make the gap arbitrarily small. Once such a partition has been found, refining it preserves an already small gap: the upper sum cannot rise and the lower sum cannot fall. This is useful when additional partition points are needed, since they can be inserted without losing the existing estimate. The point is not that every insertion helps, but that none spoils the bound.

A common mistake is to treat a partition point as if it removed the function’s endpoint value from neighboring intervals. Suprema and infima are taken over the entire closed subinterval, including its endpoints. The jump example shows how that detail can leave a nonzero oscillation on an interval ending at the jump. The correct response is to estimate that interval’s length or otherwise control its contribution, not to disregard its endpoint.

Check Your Understanding

Use the insertion identities and constructions above to answer the following questions.

Which interval contributions cancel when comparing the sums before and after inserting one point?
Why are the upper-sum decrease and the lower-sum increase both nonnegative?
In the jump example, why does the interval ending at $2/5$ still have supremum $1$?
Give a condition under which a single insertion leaves both sums unchanged.
How does subdividing each original partition interval ensure that the final partition retains every point of the original one?

What You'll Learn