Finite Unions of Disjoint Intervals

Why Disjointness Lets Lengths Add

A single bounded interval has Lebesgue outer measure equal to its length, regardless of whether its endpoints are included. For a finite collection of disjoint intervals, the upper bound for the outer measure follows by covering each interval separately. The lower bound requires more care: a covering interval might meet several components, so it is not enough to apply the interval-cover estimate to each component and simply assume those estimates add.

We work in $\mathbb{R}$ with Lebesgue outer measure $m^*$, which is an outer measure, not a measure on a sigma-algebra in this discussion. For a bounded interval $J$ with endpoints $a\le b$, write $\ell(J)=b-a$; this applies to open, closed, and half-open intervals. A singleton has length zero, and the empty interval, if included, also has length zero. We use two earlier results: each bounded interval $J$ satisfies $m^*(J)=\ell(J)$, and whenever a countable family of intervals covers a bounded interval $J$, the sum of their lengths is at least $\ell(J)$. These results apply to all endpoint conventions.

Theorem: If $J_1,\ldots,J_N$ are pairwise-disjoint bounded intervals in $\mathbb{R}$, with $N$ finite and possibly $N=0$, then $$ m^*\left(\bigcup_{i=1}^{N}J_i\right)=\sum_{i=1}^{N}\ell(J_i). $$ The union is a subset of $\mathbb{R}$, and the equality is for Lebesgue outer measure. Empty intervals and singleton intervals are allowed.

Why a Cover Interval Cannot Be Counted Twice

The key point in the lower bound is a finite length comparison. Fix a bounded interval $I$ and finitely many pairwise-disjoint intervals $J_1,\ldots,J_N$. Each nonempty intersection $I\cap J_i$ is an interval. These intersections are disjoint subintervals of $I$. When placed in order from left to right, their endpoint spans cannot overlap: if their endpoints are $u_r\le v_r$ and $u_{r+1}\le v_{r+1}$, then $v_r\le u_{r+1}$. Consequently, their total length is no greater than the span of $I$.

\sum_{i=1}^{N}\ell(I\cap J_i)\le \ell(I).

Here the length of an empty intersection is defined to be zero. This inequality also holds if some intersections are singletons, since they contribute zero. To see the endpoint calculation explicitly, list the nonempty intersections in spatial order. If there are $q\ge1$ of them, then $$ \sum_{r=1}^{q}(v_r-u_r) \le (v_q-u_1) \le \ell(I). $$ The first inequality follows because $u_{r+1}-v_r\ge0$: the span from $u_1$ to $v_q$ includes all the listed interval lengths and also any gaps between them. If there are no nonempty intersections, the sum is zero and the inequality still holds. If $I$ is unbounded, its length is infinite and the comparison is automatic; only bounded cover intervals are needed for a finite lower-bound estimate.

Proof of the Finite-Union Formula

Proof. If $N=0$, the union is empty and its outer measure is zero, equal to the empty sum. Now suppose $N\ge1$, and set $E=\bigcup_{i=1}^{N}J_i$. For the upper bound, the finite family $J_1,\ldots,J_N$ is an interval cover of $E$. The definition of Lebesgue outer measure as an infimum over countable interval covers, together with $m^*(J_i)=\ell(J_i)$, gives $$ m^*(E)\le \sum_{i=1}^{N}\ell(J_i). $$ This is the only direction that follows immediately from using the components as a cover.

For the lower bound, consider any countable interval cover $(I_j)_{j\ge1}$ of $E$. If its total length is infinite, then $\sum_i\ell(J_i)\le\sum_j\ell(I_j)$ because the left side is finite, so this cover already has length at least the desired lower bound. Suppose instead that $\sum_j\ell(I_j)<\infty$. For each fixed $i$, the intervals $I_j\cap J_i$, with empty intersections omitted, cover $J_i$: if $x\in J_i$, then $x\in E$, so $x\in I_j$ for at least one $j$, and therefore $x\in I_j\cap J_i$. The interval-cover lower bound stated above now gives $$ \ell(J_i)\le\sum_{j=1}^{\infty}\ell(I_j\cap J_i). $$ This inequality includes a degenerate component, for which its left side is zero.

Sum these inequalities over the finitely many indices $i$. For each fixed $j$, the sets $I_j\cap J_i$ are disjoint subintervals of $I_j$; the finite length comparison gives $\sum_i\ell(I_j\cap J_i)\le\ell(I_j)$. Since the terms are nonnegative and there are only finitely many $i$, the finite sum can be interchanged with the series by the definition of the nonnegative series (equivalently, by taking limits of its finite partial sums). Thus $$ \sum_{i=1}^{N}\ell(J_i) \le \sum_{i=1}^{N}\sum_{j=1}^{\infty}\ell(I_j\cap J_i) = \sum_{j=1}^{\infty}\sum_{i=1}^{N}\ell(I_j\cap J_i) \le \sum_{j=1}^{\infty}\ell(I_j). $$ Every countable interval cover of $E$ therefore has total length at least $\sum_i\ell(J_i)$. Taking the infimum over all such covers yields $m^*(E)\ge\sum_i\ell(J_i)$. Combined with the upper bound, this proves the asserted equality. $\square$

Worked Applications

Worked Example: Two Half-Open Components

Find the outer measure of $E=[1,3)\cup(5,8]$. The intervals are disjoint: every point of the first is less than $3$, and every point of the second is greater than $5$. Their lengths are $$ \ell([1,3))=3-1=2,\qquad \ell((5,8])=8-5=3. $$ The theorem applies, so $$ m^*(E)=2+3=5. $$ The endpoint $3$ is excluded and $8$ is included, but neither endpoint convention changes the length. Check: covering the two components separately gives total length $2+3=5$, while the theorem's lower bound says every interval cover has total length at least $5$. Thus the upper and lower bounds agree.

Worked Example: A Singleton Component

Let $F=[-2,0]\cup\{1\}\cup(2,9/2)$. These three sets are pairwise disjoint: the first ends at $0$, the singleton is $1$, and the last interval contains only points strictly greater than $2$. Their lengths are $$ \ell([-2,0])=0-(-2)=2,\qquad \ell(\{1\})=1-1=0,\qquad \ell((2,9/2))=9/2-2=5/2. $$ It follows that $$ m^*(F)=2+0+5/2=9/2. $$ The point $1$ adds no length. Check: the sum is $4/2+0/2+5/2=9/2$, and a cover by the three components has this same total length. The theorem also ensures that no interval cover of $F$ can have smaller total length.

Worked Example: Adjacent Intervals with No Shared Point

Consider $G=[0,1)\cup[1,2]$. The intervals are disjoint because $1\notin[0,1)$, whereas $1\in[1,2]$. Their lengths are $1-0=1$ and $2-1=1$, respectively. Therefore $$ m^*(G)=1+1=2. $$ In fact, $G=[0,2]$: every point from $0$ up to $1$ is in the first interval, and every point from $1$ through $2$ is in the second. The earlier interval formula gives $m^*([0,2])=2-0=2$, independently confirming the answer. The membership distinction at $1$ matters for disjointness, even though changing whether an endpoint is included does not change length.

Common Mistakes and the Role of the Hypotheses

Mistake: using the length of the smallest interval containing the union. For $H=[0,1]\cup[3,5]$, the intervals are disjoint and have lengths $1-0=1$ and $5-3=2$. Hence $m^*(H)=1+2=3$. A common incorrect calculation uses the containing interval $[0,5]$ and reports $5$. That is the length of a cover, so it gives an upper bound, not the exact value; it includes the gap $(1,3)$, whose length is $3-1=2$. The correction is to use the finite-union theorem for the disjoint components, giving $3$.

Disjointness is essential. The intervals $[0,2]$ and $[1,3]$ overlap on $[1,2]$, so the theorem does not apply. Adding their lengths gives $(2-0)+(3-1)=2+2=4$, but their union is $[0,3]$, whose outer measure is $3-0=3$. The overlap has been counted twice in the sum. In particular, it is not enough that a collection consists of finitely many bounded intervals; they must be pairwise disjoint for the formula proved here.

Endpoint conventions do not repair overlap or change length. A singleton component has length zero, and including or excluding a finite number of endpoints leaves each interval length unchanged. But whether two intervals share an endpoint can determine whether they are disjoint as sets. For example, $[0,1]$ and $[1,2]$ are not disjoint because both contain $1$; $[0,1)$ and $[1,2]$ are disjoint. These two collections have the same component lengths, but only the latter meets the theorem's pairwise-disjointness hypothesis.

The result is a finite additivity statement for outer measure on this particular class of sets. It does not say that outer measure is a measure on all subsets of $\mathbb{R}$, nor does it establish any countable-union formula. Its lower-bound argument also identifies what changes when the number of components is infinite: the finite sum over components and the finite comparison inside each cover interval would need additional justification. For finite unions, however, the interval-cover lower bound and the disjointness comparison provide exactly what is needed.

Check Your Understanding

In each question, check both the interval membership conditions and the hypotheses of the finite-union formula.

Compute $m^*([{-1},2)\cup(4,6])$. State the length of each component before adding.
Are $(0,1]$ and $(1,3)$ disjoint? If so, compute the outer measure of their union.
Why does the sum of the component lengths give an upper bound even when the intervals are not disjoint, but not necessarily the exact outer measure?
Find the outer measure of $[-3,-1]\cup\{0\}\cup[2,5)$, and explain the contribution of the singleton.
Give a specific cover-based reason that an interval meeting several disjoint components cannot cause the lower-bound proof to count its length more than once.