Outer Measure of Finite Interval Configurations

From a List of Intervals to Its Outer Measure

A finite union of intervals may be presented with overlapping pieces, redundant pieces, and endpoints that are included in one interval but omitted from another. Adding the lengths as written can count some regions more than once. The reliable procedure is to order the intervals by their left endpoints, merge intervals whose spans overlap or touch, and add the lengths of the resulting separated spans.

All sets in this tutorial are subsets of $\mathbb{R}$. For $E\subseteq\mathbb{R}$, its Lebesgue outer measure $m^*(E)$ is the infimum of $\sum_{j=1}^{\infty}(b_j-a_j)$ over countable covers of $E$ by open intervals $(a_j,b_j)$. We use the established facts that $m^*$ is monotone and countably subadditive, that a bounded interval with endpoints $a\le b$ has outer measure $b-a$ regardless of endpoint inclusion, and that a finite union of pairwise disjoint intervals has outer measure equal to the sum of their lengths. These are statements about outer measure; no measurability assumption is needed here.

Definition: For a nonempty bounded interval with endpoints $a\le b$, its span is $[a,b]$, and its length is $b-a$. The span records the extreme endpoints, whether or not the interval itself contains them. An empty interval has length zero. A singleton $\{a\}$ has length zero. An unbounded interval has infinite length.

For a finite list of bounded intervals, the calculation uses the spans. Sort the nonempty intervals so that their left endpoints are nondecreasing. Start with the first span $[a,b]$. If the next span is $[c,d]$ with $c\le b$, merge it into the current span, replacing the right endpoint by $\max(b,d)$. If $c>b$, finish the current span and start a new one. Continue until the list is exhausted. A singleton span $[a,a]$ is permitted and contributes length zero. Empty intervals are skipped.

Why Endpoint Spans Give the Correct Answer

Lemma: A finite set $F\subseteq\mathbb{R}$ has $m^*(F)=0$. Consequently, if $A,B\subseteq\mathbb{R}$ and their symmetric difference $A\,\triangle\,B$ is finite, then $m^*(A)=m^*(B)$.

Proof. If $F$ has $N\ge1$ points and $\varepsilon>0$, cover each point by an open interval of length $\varepsilon/N$ centered at that point. The total length of this finite cover is $N(\varepsilon/N)=\varepsilon$. Thus $m^*(F)\le\varepsilon$ for every $\varepsilon>0$, and nonnegativity gives $m^*(F)=0$. If $F$ is empty, its outer measure is zero directly. Now suppose $A\,\triangle\,B=F$ is finite. Then $A\subseteq B\cup F$, so monotonicity and subadditivity give $m^*(A)\le m^*(B)+m^*(F)=m^*(B)$. Interchanging $A$ and $B$ gives $m^*(B)\le m^*(A)$. The two outer measures are equal, including when both are infinite. $\square$

Theorem: Let $I_1,\ldots,I_N$ be a finite list of bounded intervals in $\mathbb{R}$, with any choice of open, closed, or half-open endpoints. Remove empty intervals, sort the remaining intervals by their left endpoints, and merge successive spans whenever the next left endpoint is less than or equal to the current right endpoint. If the resulting spans are $[a_1,b_1],\ldots,[a_k,b_k]$, then $$ m^*\!\left(\bigcup_{i=1}^{N} I_i\right)=\sum_{j=1}^{k}(b_j-a_j). $$ If all intervals are empty, the sum is zero.

Proof. First replace each nonempty bounded interval $I_i$ by its span $H_i=[a_i,b_i]$. The set $H_i\setminus I_i$ is contained in the two endpoints $\{a_i,b_i\}$, and $I_i\subseteq H_i$. Therefore the union of the spans $H=\bigcup_i H_i$ contains $U=\bigcup_i I_i$, and $H\setminus U$ is contained in the finite set of all listed endpoints. The lemma implies $m^*(U)=m^*(H)$.

It remains to calculate $m^*(H)$. After sorting, process the closed spans from left to right. Suppose the current span is $[a,b]$. If the next span is $[c,d]$ and $c\le b$, then their union is the single closed interval $[a,\max(b,d)]$. Replacing those spans by that interval preserves their union. If instead $c>b$, the current span and the next one are disjoint and separated by a positive gap. Finish the current span and begin a new one. Repeating these steps gives closed intervals $[a_1,b_1],\ldots,[a_k,b_k]$ whose union is $H$, with distinct intervals separated by positive gaps. They are pairwise disjoint, so the established finite-disjoint-interval formula gives $m^*(H)=\sum_{j=1}^k(b_j-a_j)$. Together with $m^*(U)=m^*(H)$, this proves the formula. If every listed interval is empty, $U$ is empty and $m^*(U)=0$. $\square$

The condition $c\le b$ includes both actual overlap and endpoint contact. Even if two original intervals omit their common endpoint, replacing them by their spans changes the union by at most that endpoint, which has outer measure zero by the lemma. The theorem therefore computes outer measure, not necessarily the exact set-theoretic union of the closed spans.

Worked Calculations

Worked Example: Overlap Followed by a Separate Component

Calculate the outer measure of $U=(0,3]\cup[2,5)\cup[7,8]$. The three intervals have spans $[0,3]$, $[2,5]$, and $[7,8]$, with left endpoints $0,2,7$ already in increasing order. The first two merge because $2\le3$, producing $[0,\max(3,5)]=[0,5]$. The last span starts at $7$, and $7>5$, so it starts a separate component. The lengths are $5-0=5$ and $8-7=1$. Hence

$$ m^*(U)=(5-0)+(8-7)=5+1=6. $$

Check: The first interval has length $3-0=3$, the second has length $5-2=3$, and their overlap has length $3-2=1$. Their union therefore has length $3+3-1=5$; the separate interval contributes $8-7=1$, giving $5+1=6$, in agreement with the merge calculation.

Worked Example: Touching Spans with a Missing Endpoint

Let $V=[-2,1)\cup(1,4]\cup[4,6)$. The first two intervals both omit $1$, while the second and third both contain $4$. Their spans are $[-2,1]$, $[1,4]$, and $[4,6]$, in sorted order. The second span starts at $1\le1$, so it merges with the first to give $[-2,4]$. The third starts at $4\le4$, so it merges again, giving $[-2,6]$. The resulting length is $6-(-2)=8$, and thus $m^*(V)=8$.

Check: The three original intervals have lengths $1-(-2)=3$, $4-1=3$, and $6-4=2$. Their total before accounting for overlaps is $3+3+2=8$. The only missing point between the first two is $1$, which contributes no positive length, and the third interval shares only the endpoint $4$ with the second. The total outer measure remains $8$.

Worked Example: Nested, Degenerate, and Empty Intervals

Consider $W=[1,9]\cup(3,5)\cup[8,11)\cup\{14\}\cup(20,20)$. The last interval is empty because its left and right endpoints agree and both endpoints are excluded, so it is skipped. The remaining spans in order are $[1,9]$, $[3,5]$, $[8,11]$, and $[14,14]$. The span $[3,5]$ is contained in $[1,9]$, so merging it leaves $[1,9]$. Since $8\le9$, the next span merges to give $[1,11]$. The singleton span $[14,14]$ is separate because $14>11$. The total is $(11-1)+(14-14)=10+0=10$, so $m^*(W)=10$.

Check: The intervals $[1,9]$, $(3,5)$, and $[8,11)$ together lie within the span from $1$ to $11$, and their union contains points arbitrarily close to both endpoints. Its outer measure is therefore $11-1=10$. The singleton $\{14\}$ has outer measure zero, and the empty interval contributes nothing. The total is $10+0+0=10$, as calculated.

Limits of the Procedure and a Common Error

Common mistake: add every listed length. For $(0,3]\cup[2,5)$, adding the interval lengths gives $(3-0)+(5-2)=3+3=6$. That is incorrect because the region from $2$ to $3$ is counted twice. The spans merge to $[0,5]$, whose length is $5-0=5$. Equivalently, subtract the overlap length $3-2=1$ from the naive total: $6-1=5$. This correction is safe for this two-interval example; for a longer list, sorting and merging avoids having to track multiple overlapping regions separately.

Unbounded intervals. If a finite configuration contains an unbounded interval, its union has infinite outer measure: the union contains that interval, whose outer measure is infinite, and monotonicity gives the same lower bound for the union. Thus the finite endpoint-merging calculation is for bounded intervals; in the unbounded case the answer is already $+\infty$. An empty interval is not unbounded, and a singleton has outer measure zero.

Outer measure is not being treated as additive on arbitrary sets. The calculation is justified by reducing the finite union to pairwise disjoint interval spans and using the established exact formula for finite disjoint interval unions. The original intervals need not be disjoint, and no general additivity assertion for outer measure is used. Endpoint changes affect at most finitely many points, which the lemma shows have outer measure zero.

Check Your Understanding

For each finite bounded configuration, sort the spans, show where merging occurs, and add the resulting lengths.

Find the outer measure of $[0,2)\cup(1,4]\cup[6,7]$.
Find the outer measure of $(-3,-1)\cup[-1,2)$, and explain why the shared endpoint does not change the answer.
What is the outer measure of $[2,5]\cup\{8\}\cup(10,10)$? Identify which piece is empty and which contributes zero length.
Give the incorrect result from adding the lengths of $[0,4]$ and $[3,6]$, then calculate the correct outer measure by merging their spans.
Explain why any finite union containing $(5,\infty)$ has infinite outer measure.

Outer Measure of Finite Interval Configurations

What You'll Learn