Outer Measure of a Singleton

Why a Point Has Zero Outer Measure

Lebesgue outer measure is defined through interval covers. A cover of a set may have positive total length even when the set itself consists of just one point. The key is that outer measure takes an infimum over all allowed covers: a point can be covered by intervals whose lengths are as small as desired. To determine the outer measure exactly, we need both an upper bound from such covers and a lower bound that applies to every cover.

We work in \(\mathbb{R}\). For an interval \(I\), its length \(|I|\) is its usual nonnegative length; an unbounded interval has length \(+\infty\). A countable interval cover of \(A\subseteq\mathbb{R}\) is a finite or countable family \((I_j)\) of intervals such that \(A\subseteq\bigcup_j I_j\). Its total length is the sum \(\sum_j |I_j|\), interpreted as a nonnegative extended real number. The Lebesgue outer measure \(m^*(A)\) is the infimum of these total lengths over all countable interval covers of \(A\). This definition concerns outer measure; it does not by itself assert that \(A\) is Lebesgue measurable.

Proof. Fix \(x\in\mathbb{R}\). First consider any countable interval cover \((I_j)\) of \(\{x\}\). Each interval length satisfies \(|I_j|\ge 0\), so the total length \(\sum_j|I_j|\) is nonnegative, including when the sum is \(+\infty\). Thus every total length over which the infimum is taken is at least zero. Since there is at least one cover, the infimum is defined, and this gives \(m^*(\{x\})\ge 0\).

For the other bound, let \(\varepsilon>0\). The open interval \(I_\varepsilon=(x-\varepsilon/2,x+\varepsilon/2)\) contains \(x\), so it is a one-interval cover of \(\{x\}\). Its length is \((x+\varepsilon/2)-(x-\varepsilon/2)=\varepsilon\). Consequently \(m^*(\{x\})\le\varepsilon\) for every \(\varepsilon>0\). If \(m^*(\{x\})\) were positive, choosing \(\varepsilon=m^*(\{x\})/2\) would contradict this inequality. Hence \(m^*(\{x\})\le0\). Combining the bounds gives \(m^*(\{x\})=0\). \(\square\)

The lower bound uses only nonnegativity: no interval cover can have negative total length. The upper bound is the part that uses the special geometry of a point: an interval centered at that point can be made arbitrarily short. Neither argument claims that a cover has length exactly zero. Rather, the infimum of the cover lengths is zero.

Using the Covering Infimum

The same reasoning can be expressed as a useful criterion for this particular set. For a singleton, having outer measure zero is equivalent to being coverable by intervals of total length below every prescribed positive tolerance.

Proof. Suppose first that \(m^*(\{x\})=0\). The collection of total lengths of interval covers is nonempty, is bounded below by zero, and has infimum zero. By the defining property of an infimum, for every \(\delta>0\) there must be a cover with total length less than \(0+\delta=\delta\). Otherwise every cover would have total length at least \(\delta\), making \(\delta\) a lower bound and contradicting that the infimum is zero.

Conversely, suppose that for every \(\delta>0\) there is a cover with total length less than \(\delta\). Since all cover lengths are nonnegative, their infimum satisfies \(m^*(\{x\})\ge0\). Given any \(\delta>0\), the assumed cover also gives \(m^*(\{x\})<\delta\). A nonnegative number less than every positive \(\delta\) must be zero: if it were \(c>0\), taking \(\delta=c/2\) would be a contradiction. Therefore \(m^*(\{x\})=0\). \(\square\)

A Common Mistake: Confusing One Cover with the Infimum

Suppose someone covers \(\{4\}\) by the interval \([3,5]\), calculates its length as \(5-3=2\), and concludes that \(m^*(\{4\})=2\). The length calculation is correct, but the conclusion is not: it computes the cost of one cover, not the infimum over all covers. The definition gives only \(m^*(\{4\})\le2\) from this interval.

For a direct correction, use the intervals \((4-1/n,4+1/n)\) for integers \(n\ge1\). Each contains \(4\), and its length is \((4+1/n)-(4-1/n)=2/n\). For example, at \(n=1000\) the length is \(2/1000=1/500\), already less than \(2\), and increasing \(n\) makes the cover length arbitrarily small. Meanwhile every cover has nonnegative total length. Thus the infimum is zero, not the length of the first interval chosen.

This distinction is central to outer measure: an individual cover supplies an upper bound, while a lower bound must apply to every cover. Here the universal lower bound comes from nonnegative lengths, and the arbitrarily short centered intervals supply the matching upper bound. The conclusion is specifically about Lebesgue outer measure. Whether and how the resulting null sets are used as measurable sets is a separate question.