From Interval Lengths to a Bound for the Union
A countable list of intervals gives a direct way to control the outer measure of the set they cover. The control is an upper bound: if the intervals overlap, adding their lengths may count some locations more than once. Even without knowing whether the intervals overlap, their total length provides a useful estimate. When that length series converges, its tails also bound the part of the union left beyond any finite collection.
We work on \(\mathbb{R}\) with Lebesgue outer measure \(m^*\). For an interval \(I\), write \(\ell(I)\) for its length: if \(I\) is bounded and has endpoints \(a\le b\), then \(\ell(I)=b-a\), regardless of whether either endpoint belongs to \(I\); a singleton and the empty interval have length \(0\); an unbounded interval has length \(+\infty\). The established interval formula is \(m^*(I)=\ell(I)\), including the unbounded case. We also use the definition of outer measure as an infimum of costs of admissible countable interval covers. In particular, when \(m^*(I)<\infty\), for every \(\eta>0\) there is such a cover of \(I\) whose total length is less than \(m^*(I)+\eta\).
Proof. Let \(S=\sum_{k=1}^{\infty}\ell(I_k)\). If \(S=+\infty\), the asserted inequality holds because every outer measure is at most \(+\infty\). Suppose instead that \(S<\infty\), and fix \(\varepsilon>0\). Each \(\ell(I_k)\le S<\infty\), so \(m^*(I_k)=\ell(I_k)<\infty\). For every \(k\), use the infimum property of interval covers to choose a cover of \(I_k\) whose total length is less than \(\ell(I_k)+\varepsilon 2^{-k}\). All these covers together form a countable interval cover of \(U\): a countable union of countable collections is countable. Its total length is at most $$ \sum_{k=1}^{\infty}\bigl(\ell(I_k)+\varepsilon 2^{-k}\bigr) =S+\varepsilon\sum_{k=1}^{\infty}2^{-k} =S+\varepsilon. $$ The last equality uses \(\sum_{k=1}^{\infty}2^{-k}=1\). By the definition of \(m^*(U)\) as an infimum over cover costs, \(m^*(U)\le S+\varepsilon\). This holds for every \(\varepsilon>0\), so \(m^*(U)\le S\): if \(m^*(U)>S\), choosing \(\varepsilon=(m^*(U)-S)/2\) would contradict \(m^*(U)\le S+\varepsilon\). \(\square\)
The proof allocates an error \(\varepsilon 2^{-k}\) to the \(k\)-th interval. The allocations add to exactly \(\varepsilon\), so infinitely many near-optimal covers still produce one cover with controlled total cost. The conclusion concerns outer measure; it does not require a separate measurability assumption on \(U\).
Finite Stages and the Remaining Tail
Proof. Apply the theorem to the family \(I_{N+1},I_{N+2},\ldots\) to obtain the displayed bound. Because the nonnegative series of lengths converges, its tail sums tend to zero. Therefore the outer measures of the tail unions are bounded above by numbers tending to zero, and so they tend to zero as well. \(\square\)
This estimate also applies to the portion of the full union not already in its first \(N\) intervals. If \(U_N=\bigcup_{k=1}^{N}I_k\), then \(U\setminus U_N\subseteq\bigcup_{k>N}I_k\). Monotonicity of outer measure therefore gives \(m^*(U\setminus U_N)\le\sum_{k>N}\ell(I_k)\). This is an outer-measure estimate for the remainder; it does not assert that the total length of all intervals equals the outer measure of the union.
Worked Example: A Telescoping Family That Attains the Bound
For \(n\ge2\), let \(I_n=[1/(n+1),1/n]\). Compute the sum of the interval lengths and the outer measure of their union.
The length of the \(n\)-th interval is \(\ell(I_n)=1/n-1/(n+1)=1/[n(n+1)]\). For \(N\ge2\), the finite sum telescopes: $$ \sum_{n=2}^{N}\ell(I_n) =\sum_{n=2}^{N}\left(\frac1n-\frac1{n+1}\right) =\frac12-\frac1{N+1}. $$ As \(N\to\infty\), this tends to \(1/2\), so the total length series is \(1/2\).
The intervals cover exactly \((0,1/2]\). Each lies in this set. Conversely, if \(0<x\le1/2\), choose an integer \(n\ge2\) with \(1/(n+1)\le x\le1/n\); then \(x\in I_n\). Thus the established interval formula gives \(m^*((0,1/2])=1/2\), and the covering bound gives \(m^*(\bigcup_{n\ge2}I_n)\le1/2\). The two values agree. Check: the union of the first \(N-1\) intervals, from \(n=2\) through \(n=N\), is \([1/(N+1),1/2]\), whose length is \(1/2-1/(N+1)\), exactly the finite sum computed above.
Worked Example: A Convergent Bound That Is Not an Equality
For \(n\ge1\), set \(I_n=[0,2^{-n}]\). Find the length-series bound and compare it with the outer measure of the union.
The lengths are \(\ell(I_n)=2^{-n}\). For each \(N\ge1\), $$ \sum_{n=1}^{N}\ell(I_n) =\frac12+\frac14+\cdots+\frac1{2^N} =1-\frac1{2^N}, $$ so \(\sum_{n=1}^{\infty}\ell(I_n)=1\). The intervals are nested: \(I_n\subseteq I_1\) for every \(n\), and \(I_1\) itself is part of the union. Hence \(\bigcup_{n=1}^{\infty}I_n=I_1=[0,1/2]\), whose outer measure is \(1/2\). The theorem gives the valid, but non-sharp, estimate \(1/2\le1\). Check: the finite partial sums are \(1/2\), \(3/4\), and \(7/8\) for \(N=1,2,3\), respectively; the remainder after \(N\) is \(2^{-N}\), so each partial sum plus its remainder is \(1\).
Worked Example: Bounding a Union and Its Tail
For \(n\ge1\), let \(J_n=[n,n+2^{-n}]\). Bound the outer measure of the full union and the union with indices \(n>3\).
Here \(\ell(J_n)=2^{-n}\). The geometric sum gives $$ \sum_{n=1}^{\infty}\ell(J_n)=1,\qquad \sum_{n=4}^{\infty}\ell(J_n) =\frac1{16}+\frac1{32}+\cdots =\frac{1/16}{1-1/2} =\frac18. $$ The covering theorem therefore gives \(m^*(\bigcup_{n\ge1}J_n)\le1\), and the tail bound gives \(m^*(\bigcup_{n>3}J_n)\le1/8\). For a lower check on the first estimate, \(J_1=[1,3/2]\) is contained in the full union and has outer measure \(1/2\). Monotonicity gives \(1/2\le m^*(\bigcup_{n\ge1}J_n)\le1\). Check: the first three lengths sum to \(1/2+1/4+1/8=7/8\); adding the tail \(1/8\) recovers the full-series bound \(1\).
What the Bound Does—and Does Not—Say
The estimate is most useful when the sum of lengths is finite: it immediately proves that the union has finite outer measure, and it quantifies how little outer measure can remain after a finite stage. If the sum is infinite, the inequality remains true but supplies no finite upper bound. For example, if \(I_n=[n,n+1]\), then \(\sum_n\ell(I_n)=+\infty\), so the theorem yields only \(m^*(\bigcup_n I_n)\le+\infty\). An infinite right-hand side is not evidence that the union has infinite outer measure.
Common mistake: replacing the inequality by equality. In the nested-interval example, the sum of lengths is \(1\), but the union is \([0,1/2]\) and has outer measure \(1/2\). Thus the calculation \(m^*(\bigcup_n I_n)=\sum_n\ell(I_n)=1\) is wrong. The theorem proves only \(m^*(\bigcup_n I_n)\le\sum_n\ell(I_n)\); overlap can make the sum strictly larger. The telescoping example attains equality, but that does not make equality a general rule.
Endpoint conventions do not change these length bounds: changing whether an interval includes one or both endpoints leaves its length unchanged. Likewise, a singleton contributes length zero. The key information is the length series and how the intervals cover the union, not whether endpoints are open or closed.
Check Your Understanding
For each question, distinguish the bound supplied by the length series from any exact outer measure you can establish separately.
- Let \(I_n=[0,3^{-n}]\) for \(n\ge1\). Compute the sum of the lengths and the outer measure of the union.
- If \(\ell(I_n)=4^{-n}\), what upper bound does the theorem give for the outer measure of \(\bigcup_{n\ge1}I_n\)? What bound does it give for the union over \(n>2\)?
- For a family with \(\sum_{n\ge1}\ell(I_n)<\infty\), explain why \(m^*(U\setminus U_N)\le\sum_{n>N}\ell(I_n)\).
- Give a family of intervals for which the covering bound is strict, and identify the overlap responsible for the difference.
- What conclusion does the covering theorem give when \(\sum_{n\ge1}\ell(I_n)=+\infty\), and why is that conclusion not a finite estimate?