Lebesgue Outer Measure from Interval Covers

From Interval Covers to Outer Measure

An interval cover gives a way to measure a set from the outside: cover it with intervals, add their lengths, and look for the smallest possible total. The definition is deliberately an infimum, not a claim that there is always one cover with the smallest cost. At this stage, the main tasks are to define that infimum carefully and to see what explicit covers do—and do not—tell us about bounded sets.

All sets considered here are subsets $E\subseteq\mathbb{R}$. For a bounded open interval $(a,b)$, where $a<b$, its length is $b-a$. A countable interval cover of $E$ is a finite or countable family of bounded open intervals whose union contains $E$. The empty family is permitted and covers the empty set. The cost of a cover is the sum of the lengths of its intervals; for a finite cover this is a finite sum, and for a countable cover it is a series of nonnegative terms, allowed to equal $+\infty$.

Definition: The Lebesgue outer measure $m^*(E)$ of $E\subseteq\mathbb{R}$ is the infimum of the costs of all countable interval covers of $E$: $$ m^*(E)=\inf\left\{\sum_{j=1}^{\infty}(b_j-a_j): E\subseteq\bigcup_{j=1}^{\infty}(a_j,b_j)\right\}. $$ Finite covers are included by allowing all remaining intervals to be empty, with length zero; the empty set also has the empty cover of cost zero. The value $m^*(E)$ lies in $[0,\infty]$.

Every $E\subseteq\mathbb{R}$ has at least one such cover: for example, the intervals $(-n,n)$, $n\ge1$, cover all of $\mathbb{R}$. Their total cost is infinite, which is still an admissible cost. Thus the collection whose infimum defines $m^*(E)$ is never empty. No measurability assumption on $E$ is part of this definition. Outer measure is not, by definition alone, a measure on a sigma-algebra; measurability and the measure properties require further results.

What an Infimum Says About Covers

A cover of cost $c$ immediately gives $m^*(E)\le c$. More generally, when the infimum is finite, it can be approached as closely as desired by cover costs. The precise statement below is often the useful way to work with the definition.

Proposition: Let $E\subseteq\mathbb{R}$ and let $L$ be a finite nonnegative real number. Then $m^*(E)\le L$ if and only if, for every $\varepsilon>0$, $E$ has a countable interval cover of cost less than $L+\varepsilon$.

Proof. Suppose first that $m^*(E)\le L$, and fix $\varepsilon>0$. By the definition of infimum, some cover has cost less than $m^*(E)+\varepsilon$; otherwise every cover cost would be at least $m^*(E)+\varepsilon$, contradicting that $m^*(E)$ is the infimum. This cost is at most $L+\varepsilon$, and in fact is strictly less than $L+\varepsilon$, as required. Conversely, suppose that for every $\varepsilon>0$ there is a cover of cost less than $L+\varepsilon$. Since $m^*(E)$ is no greater than the cost of any cover, $m^*(E)<L+\varepsilon$ for every $\varepsilon>0$. If $m^*(E)>L$, choosing $\varepsilon=(m^*(E)-L)/2>0$ contradicts that inequality. Therefore $m^*(E)\le L$. $\square$

Worked Example: The Empty Set

Compute $m^*(\varnothing)$ from the definition. The empty family is a cover of $\varnothing$, and its cost is the empty sum, which equals zero. Every cover has nonnegative cost because every interval length is nonnegative. Hence the infimum of all cover costs is both at most zero and at least zero:

0\le m^*(\varnothing)\le 0, \qquad\text{so}\qquad m^*(\varnothing)=0.

Check: the empty cover has cost $0$, and no cover can have negative cost. These two observations verify both bounds needed for the exact value.

Bounds from a Single Interval

A set contained in a bounded interval has an immediate outer-measure upper bound. This conclusion uses only one family of covers: slightly enlarge the containing interval and let the enlargement tend to zero. It does not, by itself, prove that the bound is exact, because it gives no lower bound on the costs of all possible covers.

Proposition: If $E\subseteq[a,b]\subseteq\mathbb{R}$, where $a\le b$, then $m^*(E)\le b-a$.

Proof. Fix $\varepsilon>0$. The open interval $(a-\varepsilon/2,b+\varepsilon/2)$ contains $[a,b]$, and therefore contains $E$. Its length, computed by subtracting its left endpoint from its right endpoint, is $$ (b+\varepsilon/2)-(a-\varepsilon/2)=b-a+\varepsilon. $$ Thus $E$ has a cover of cost $b-a+\varepsilon$ for every $\varepsilon>0$. The infimum characterization just proved, with $L=b-a$, gives $m^*(E)\le b-a$. This calculation also covers $a=b$: the covering interval then has length $\varepsilon$, so the conclusion is $m^*(E)\le0$. $\square$

Worked Example: A Set in a Length-Five Interval

Let $E$ be any subset of $[-2,3]$. For each $\varepsilon>0$, cover $E$ by the single interval $(-2-\varepsilon/2,3+\varepsilon/2)$. Its length is

\left(3+\frac{\varepsilon}{2}\right) -\left(-2-\frac{\varepsilon}{2}\right) =3+2+\varepsilon =5+\varepsilon.

For any target excess $\eta>0$, choose $\varepsilon=\eta$; the cover cost is then $5+\eta$. Since this works for every $\eta>0$, the definition gives $m^*(E)\le5$. Check: when $\eta=0.2$, the interval is $(-2.1,3.1)$, and its length is $3.1-(-2.1)=5.2=5+0.2$. This example establishes an upper bound for every $E\subseteq[-2,3]$, not an exact value for each such set.

Worked Example: A Symmetric Bounded Set

Let $E=\{x\in\mathbb{R}:x^2\le2\}$, so $E=[-\sqrt{2},\sqrt{2}]$. Fix $\eta>0$, and use the one-interval cover $(-\sqrt{2}-\eta/2,\sqrt{2}+\eta/2)$. The interval contains both endpoints of $E$, and its length is

\left(\sqrt{2}+\frac{\eta}{2}\right) -\left(-\sqrt{2}-\frac{\eta}{2}\right) =2\sqrt{2}+\eta.

So for every positive excess $\eta$, there is a cover of cost $2\sqrt{2}+\eta$, and consequently $m^*(E)\le2\sqrt{2}$. Check: taking $\eta=0.1$ gives endpoints $-\sqrt{2}-0.05$ and $\sqrt{2}+0.05$; subtracting the left endpoint from the right gives $2\sqrt{2}+0.1$. The calculation is an upper bound from the displayed covers; establishing equality would require ruling out every cover with smaller cost.

Several Intervals and the Cost of a Cover

A cover is a family of intervals, not a partition. Its members may overlap, and the cost is the sum of their individual lengths. Overlap does not reduce that sum. Choosing more than one interval can still be advantageous when a set lies in separated regions, but the resulting cost remains an upper bound unless a separate lower-bound argument is available.

Worked Example: Covering Two Separated Pieces

Let $E=[0,1]\cup[3,4]$. For $\delta>0$, cover the first piece by $(-\delta,1+\delta)$ and the second by $(3-\delta,4+\delta)$. The total cover cost is the sum of the two lengths:

\bigl((1+\delta)-(-\delta)\bigr) +\bigl((4+\delta)-(3-\delta)\bigr) =(1+2\delta)+(1+2\delta) =2+4\delta.

Given any $\eta>0$, set $\delta=\eta/4$. The cover cost becomes $2+4(\eta/4)=2+\eta$, so $m^*(E)\le2$. For comparison, one interval $(-\delta,4+\delta)$ also covers $E$, but its cost is $(4+\delta)-(-\delta)=4+2\delta$. The two-interval cover has lower cost whenever $2+4\delta<4+2\delta$, equivalently $2\delta<2$, or $\delta<1$. Check: with $\delta=1/4$, the two intervals have lengths $1.5$ and $1.5$, total $3$; the one-interval cover has length $4.5$. Both are valid covers, and their costs are computed by adding interval lengths, not by treating the cover as a partition.

A Common Mistake: Turning a Cover Cost into an Exact Value

Suppose someone covers $[0,1]$ by the two intervals $(-1,0.6)$ and $(0.4,2)$, then reports that the outer measure is $3$, the length of their union $(-1,2)$. This is not what the definition says. The cost of this particular cover is

\bigl(0.6-(-1)\bigr)+\bigl(2-0.4\bigr) =1.6+1.6 =3.2.

The intervals overlap, but their individual lengths are still both counted. This cover proves only $m^*([0,1])\le3.2$. A different cover, such as $(-\varepsilon,1+\varepsilon)$, has cost $1+2\varepsilon$ for each $\varepsilon>0$, giving the sharper upper bound $m^*([0,1])\le1$. Neither upper-bound calculation alone establishes equality: an exact computation must also show that every admissible cover has cost at least the proposed value.

The distinction is useful throughout the construction of Lebesgue measure. Explicit covers are often enough to prove an upper bound, including a bound of zero when the costs can be made arbitrarily small. Exact outer-measure calculations require attention to both directions: a good cover supplies the upper bound, while a lower-bound argument must control all covers. Finally, $m^*$ is outer measure on all subsets of $\mathbb{R}$; it should not be confused with Lebesgue measure $m$, which is the measure on Lebesgue-measurable sets.

Check Your Understanding

For each question, distinguish the cost of the stated cover from the infimum over all covers.

What is the cost of the single-interval cover $(-1-\varepsilon,2+\varepsilon)$ of $[-1,2]$, and what upper bound follows as $\varepsilon$ tends to zero?
Why does the empty cover imply $m^*(\varnothing)\le0$, and what fact gives the reverse inequality?
If $E\subseteq[4,7]$, use a cover with arbitrarily small excess to show $m^*(E)\le3$.
For a two-interval cover whose interval lengths are $2.4$ and $1.1$, what is its cost even if the intervals overlap?
State the additional kind of argument needed to turn an upper bound from a cover into an exact outer-measure value.