Interval Lengths and Covering Conventions

Why Interval Conventions Matter

Lebesgue outer measure on the real line is built by covering sets with intervals and adding the lengths of those intervals. To interpret that construction, one must specify both what “length” means for different endpoint conventions and which intervals count as admissible covers. Endpoint inclusion does not change the assigned length of a nondegenerate bounded interval, but it can change whether that interval is open and therefore admissible in the usual definition of outer measure.

Throughout, $m^*$ denotes Lebesgue outer measure on $\mathbb{R}$, not a measure on a sigma-algebra. In its interval-cover definition, an admissible cover of $E\subseteq\mathbb{R}$ is a finite or countable family of open intervals whose union contains $E$; its cost is the sum of their lengths. The infimum of these costs is $m^*(E)$. The empty family covers the empty set and has cost zero. One may use bounded open intervals in this definition; an unbounded open interval can instead be assigned infinite cost and cannot help produce a finite-cost cover.

Definition: For a bounded interval with finite endpoints $a<b$, assign length $b-a$, regardless of whether either endpoint is included. Thus $(a,b)$, $[a,b]$, $[a,b)$, and $(a,b]$ all have length $b-a$. The empty set has length $0$, and a singleton $\{a\}=[a,a]$ has length $0$. An interval unbounded to the left, to the right, or in both directions has length $+\infty$. In particular, $\mathbb{R}$ has length $+\infty$.

For degenerate endpoints, the notation must be interpreted as a set before assigning length. When both endpoints are $a$, the closed interval $[a,a]$ is $\{a\}$, while $(a,a)$, $[a,a)$, and $(a,a]$ are empty. They have length zero for different set-theoretic reasons. This convention does not make $[a,a]$ an open interval: a singleton is not open in $\mathbb{R}$.

An unbounded interval is also not a finite-length substitute for a bounded interval. The half-line $[a,\infty)$, for example, has length $+\infty$, whether or not its finite endpoint is included. If such intervals are allowed in a cover, any cover containing one has infinite cost. For questions about finite covering costs, only bounded intervals can contribute.

Endpoint Inclusion and Assigned Length

The length convention is deliberately insensitive to endpoint inclusion when $a<b$. This is an assignment used in the covering construction; it does not assert that all four intervals are the same set. In particular, the four intervals can differ at one or both endpoints even though their assigned lengths agree.

Worked Example: Comparing Endpoint Conventions

Find the lengths of $(-2,3)$, $[-2,3]$, $[-2,3)$, and $(-2,3]$, and compare them with the degenerate intervals $[-2,-2]$ and $(-2,-2]$.

The two finite endpoints of each of the first four intervals are $-2$ and $3$, with $-2<3$. The assigned length for each is therefore $3-(-2)=5$. The endpoint choices alter membership: for example, $-2$ belongs to $[-2,3]$ and $[-2,3)$, but not to $(-2,3)$ or $(-2,3]$. They do not alter the assigned length.

Next, $[-2,-2]=\{-2\}$ is a singleton, so its length is $0$. The interval $(-2,-2]$ is empty, so its length is also $0$. Check: the singleton contains one point and the other set contains none, but both have length zero by the degenerate-interval convention. Neither is a nonempty open interval suitable as a member of an open-interval cover.

These length conventions match the costs assigned to intervals in the construction. They do not, by themselves, compute the outer measure of any particular set. In particular, using a zero-length singleton as a member of a cover is not permitted when the definition requires open intervals.

Open Covers and Enlargement

Because the standard covering definition uses open intervals, a closed or half-open interval is not automatically an admissible member of a cover. A bounded interval can, however, be contained in a slightly larger open interval. The extra length can be calculated directly. For a nondegenerate interval with endpoints $a<b$, extending it by a distance $r>0$ at each end gives $(a-r,b+r)$, whose length is $(b+r)-(a-r)=(b-a)+2r$.

Proposition: Let $I$ be a nonempty bounded interval with finite endpoints, allowing $I$ to be a singleton. For every $\eta>0$, there is a bounded open interval $J$ containing $I$ whose length is at most the assigned length of $I$ plus $\eta$.

Proof. First suppose $I$ has distinct endpoints $a<b$, so its assigned length is $b-a$. Set $r=\eta/2$ and let $J=(a-r,b+r)$. Both endpoints of $I$ lie strictly inside $J$, so $I\subseteq J$, including when either endpoint belongs to $I$. Its length is $b-a+2r=b-a+\eta$, as required. Now suppose $I=\{a\}$, whose assigned length is zero. Set $J=(a-\eta/2,a+\eta/2)$. Then $a\in J$ and the length of $J$ is $\eta$. These cases include every nonempty bounded interval. $\square$

The proposition concerns one interval at a time. For a countable family, enlarging every member by the same positive amount need not keep the total added length finite. The following result gives a controlled enlargement for a fixed countable cover with finite total cost.

Theorem: Suppose a set $E\subseteq\mathbb{R}$ is covered by a finite or countable family of bounded intervals $I_j$ with finite total assigned length $S=\sum_j |I_j|<\infty$. For every $\varepsilon>0$, there is a finite or countable open-interval cover of $E$ whose total length is at most $S+\varepsilon$.

Proof. Omit any empty intervals from the family; removing them does not change its union or increase its cost. For each remaining interval $I_j$, apply the proposition with $\eta_j=\varepsilon 2^{-j}$, using indices $j=1,2,\ldots$ (for a finite family, use its existing indices). This gives a bounded open interval $J_j\supseteq I_j$ with $|J_j|\le |I_j|+\varepsilon 2^{-j}$. Since the $I_j$ cover $E$, the $J_j$ also cover $E$. For a countable family, summing the inequalities gives $$ \sum_j |J_j|\le \sum_j |I_j|+\varepsilon\sum_{j=1}^{\infty}2^{-j}=S+\varepsilon, $$ because the geometric series $\sum_{j=1}^{\infty}2^{-j}$ equals $1$. For a finite family the corresponding finite sum of $2^{-j}$ is at most $1$, so the same bound holds. If the family is empty, then $E$ is empty and the empty open cover has cost zero. This proves the claim in all cases. $\square$

This theorem explains how a fixed finite-cost cover made from bounded intervals of mixed endpoint types can be replaced by an open-interval cover without adding more than a prescribed amount to its cost. It does not say that a particular interval cover has the least possible cost, or that the infimum defining $m^*(E)$ is attained.

Worked Example: Making a Half-Open Interval Admissible

Give an open-interval cover of $[2,5)$ using one interval, and compute both lengths.

The interval $[2,5)$ is bounded and has assigned length $5-2=3$, but it is not open, so it is not an admissible member of an open-interval cover. Enlarge it by $1/10$ at each end: $$ [2,5)\subset (2-1/10,5+1/10)=(19/10,51/10). $$ The open interval has length $51/10-19/10=32/10=16/5$. Thus this one open interval covers the half-open interval, and the added length is $16/5-3=16/5-15/5=1/5$. Check: its left endpoint $19/10=1.9$ is less than $2$, and its right endpoint $51/10=5.1$ is greater than $5$, so it contains every point of $[2,5)$; the length increase $1/5$ is exactly $1/10$ from each side.

Worked Example: Enlarging a Countable Cover with a Length Budget

Consider the countable family $I_n=(-2^{-n},2^{-n})$, $n\ge1$, which covers the singleton $\{0\}$. Compute its total length and enlarge its members so that the total added length is at most $1/4$.

Each $I_n$ is already open, has length $2^{-n}-(-2^{-n})=2^{1-n}$, and contains $0$. The original total length is $$ \sum_{n=1}^{\infty}2^{1-n}=2\sum_{n=1}^{\infty}2^{-n}=2. $$ For the enlargement budget, assign $\eta_n=(1/4)2^{-n}=2^{-n-2}$. Enlarge each interval by $\eta_n/2$ at each end: $$ J_n=(-2^{-n}-2^{-n-3},\,2^{-n}+2^{-n-3}). $$ Its length is $2^{1-n}+2^{-n-2}$, since the two extensions together add $\eta_n=2^{-n-2}$. Hence $$ \sum_{n=1}^{\infty}|J_n| =\sum_{n=1}^{\infty}2^{1-n}+\sum_{n=1}^{\infty}2^{-n-2} =2+\frac14=\frac94. $$ Every $J_n$ contains $I_n$, so the enlarged family still covers $\{0\}$. Check: the added length is $\sum_{n=1}^{\infty}2^{-n-2}=(1/4)\sum_{n=1}^{\infty}2^{-n}=1/4$, exactly the assigned budget. This is a calculation of cover costs, not a computation of $m^*(\{0\})$.

Unbounded Intervals and Admissibility

The finite-endpoint length formula $b-a$ applies only when both endpoints are real numbers with $a\le b$. For example, $(-\infty,4]$ and $(4,\infty)$ are unbounded and each has assigned length $+\infty$. Including either in a cover makes its total cost infinite. Thus unbounded intervals may be named when listing interval conventions, but they are unnecessary for any cover whose cost is finite.

Conversely, the availability of bounded open intervals is not an obstacle to covering an unbounded subset of $\mathbb{R}$: the definition allows countably many intervals. For instance, the family $(-n,n)$, $n\ge1$, covers $\mathbb{R}$, since for every real $x$ there is an integer $n>|x|$. This particular family has infinite total cost: its $n$-th interval has length $n-(-n)=2n$, and the partial sums $\sum_{n=1}^{N}2n=N(N+1)$ are unbounded as $N$ increases. This example illustrates the distinction between being a cover and having finite cost.

A Common Covering Mistake

A common error is to treat every interval, including a singleton, as an admissible open interval and then use its assigned length as the covering cost. For example, one might claim that the one-member family $\{[3,3]\}$ is an open-interval cover of $\{3\}$ with cost $0$. The arithmetic $3-3=0$ agrees with the length convention for the singleton, but the cover is inadmissible: $[3,3]=\{3\}$ is not open.

The correction is to distinguish the assigned length of an interval from whether it belongs to the permitted cover class. For any $\eta>0$, the open interval $(3-\eta/2,3+\eta/2)$ contains $3$ and has length $\eta$. It is an admissible one-interval cover, with positive cost $(3+\eta/2)-(3-\eta/2)=\eta$. The infimum over such costs is a separate question from whether the zero-length singleton itself is allowed as a covering interval.

Keep three checks separate when constructing a cover: identify the set represented by the endpoint notation, verify that each covering interval is in the allowed class, and only then calculate its assigned length. Endpoint conventions determine the first and third checks; openness determines the second.

Check Your Understanding

For each question, distinguish the set being described, its assigned length, and whether it is an admissible member of an open-interval cover.

What are the assigned lengths of $(-1,4]$, $[-1,4)$, $[-1,-1]$, and $(-1,-1)$? Which of these sets are open intervals?
For $a<b$, show directly that enlarging $[a,b)$ to $(a-r,b+r)$, where $r>0$, adds $2r$ to its assigned length.
Why does a cover containing the unbounded interval $(0,\infty)$ have infinite cost under the stated convention?
Give a bounded open interval containing the singleton $\{6\}$ with length $1/5$, and verify both claims from its endpoints.
In the countable enlargement theorem, why must the added amounts be chosen with a finite sum rather than assigning the same positive amount to every interval?