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Lebesgue Outer Measure on R · Tutorial 102 of 798

Countable Sets Have Zero Outer Measure

Learn how a summable allocation of interval lengths proves that every countable subset of the real line has outer measure zero.

Intermediate 10 min read

What You'll Learn

  • Define outer measure using countable covers by intervals.
  • Construct intervals around the points of an enumeration with summably small lengths.
  • Handle finite, empty, and countably infinite sets in the proof.
  • Apply the construction to squares, integers, and a reciprocal sequence.
  • Identify why equal-length intervals do not give the needed estimate.

The Covering Idea

To show that a set has outer measure zero, it is not enough to cover each of its points separately with a short interval. The lengths of all the intervals must have a small total. For a countable set, enumeration lets us assign a different length to each point: the lengths can decrease quickly enough that their sum is as small as desired.

Throughout, we work on \(\mathbb{R}\) with Lebesgue outer measure \(m^*\). An interval cover of \(A\subseteq\mathbb{R}\) is a finite or countable family of intervals whose union contains \(A\). We use the definition

$$ m^*(A)=\inf\left\{\sum_{j} \ell(I_j): A\subseteq\bigcup_j I_j,\ I_j\text{ intervals}\right\}, $$

where \(\ell(I_j)\) denotes the length of the interval. In particular, if one interval cover has total length at most \(c\), then \(m^*(A)\le c\). Outer measure is nonnegative, and \(m^*(\varnothing)=0\), since the empty set has an empty cover.

Definition: A set is countable if it is finite or its elements can be listed in a sequence. A countably infinite set can be listed without repetitions as \(A=\{x_1,x_2,\ldots\}\). A finite set with \(N\) elements can be listed as \(A=\{x_1,\ldots,x_N\}\).

The lengths in the construction will come from a geometric series. For every \(\varepsilon>0\),

$$ \sum_{n=1}^{\infty}\frac{\varepsilon}{2^{n+1}} =\varepsilon\sum_{n=1}^{\infty}\frac{1}{2^{n+1}} =\varepsilon\left(\frac14+\frac18+\frac1{16}+\cdots\right) =\frac{\varepsilon}{2} <\varepsilon. $$

Thus the \(n\)th point can be covered by an interval of length \(\varepsilon/2^{n+1}\). The lengths are summable, and their total is smaller than the prescribed error.

The Countable-Set Theorem

Theorem: If \(A\subseteq\mathbb{R}\) is countable, then \(m^*(A)=0\).

Proof. If \(A=\varnothing\), then \(m^*(A)=0\) by the empty-cover convention. Now suppose \(A\) is nonempty. List its distinct elements as \(x_1,x_2,\ldots\) if \(A\) is infinite, or as \(x_1,\ldots,x_N\) if \(A\) is finite. Fix any \(\varepsilon>0\). For each index \(n\) in the list, take the open interval

$$ I_n=\left(x_n-\frac{\varepsilon}{2^{n+2}},\ x_n+\frac{\varepsilon}{2^{n+2}}\right). $$

The point \(x_n\) is the midpoint of \(I_n\), so \(x_n\in I_n\). Its length is

$$ \ell(I_n)=\left(x_n+\frac{\varepsilon}{2^{n+2}}\right) -\left(x_n-\frac{\varepsilon}{2^{n+2}}\right) =\frac{\varepsilon}{2^{n+1}}. $$

Consequently, the intervals cover every listed point, and hence cover \(A\). If \(A\) is infinite, their total length is \(\varepsilon/2\), by the geometric-series calculation above. If \(A\) has \(N\) points, their total length is the corresponding finite partial sum, which is at most \(\varepsilon/2\). In either case, the cover has total length at most \(\varepsilon/2<\varepsilon\). The definition of outer measure therefore gives \(0\le m^*(A)\le\varepsilon/2\) for every \(\varepsilon>0\). If \(m^*(A)\) were positive, choosing \(\varepsilon=m^*(A)\) would give \(m^*(A)\le m^*(A)/2\), a contradiction. Thus \(m^*(A)=0\). \(\square\)

The argument handles finite sets as well as countably infinite sets. For a finite set the construction simply stops after its last point; for the empty set no intervals are needed. No boundedness assumption is involved: the locations of the points affect the centers of the intervals, but not the sum of their lengths.

Corollary: Every subset of a countable set in \(\mathbb{R}\) has outer measure zero.

Proof. Let \(B\subseteq A\), where \(A\subseteq\mathbb{R}\) is countable. The theorem gives \(m^*(A)=0\). By monotonicity of outer measure, \(0\le m^*(B)\le m^*(A)=0\), so \(m^*(B)=0\). Monotonicity follows directly from the definition: any interval cover of \(A\) is also an interval cover of \(B\), so the infimum of cover lengths for \(B\) is no larger. \(\square\)

Worked Coverings

Worked Example: The Squares of the Positive Integers

Let \(A=\{n^2:n=1,2,\ldots\}\). Construct a cover with total length less than \(1\), then verify that the construction works for any positive error.

Use \(\varepsilon=1\) in the theorem. Around \(n^2\), take the interval with radius \(1/2^{n+2}\), so its length is \(1/2^{n+1}\). The first three intervals are

$$ I_1=\left(1-\frac18,1+\frac18\right)=\left(\frac78,\frac98\right),\qquad I_2=\left(4-\frac1{16},4+\frac1{16}\right)=\left(\frac{63}{16},\frac{65}{16}\right), $$
$$ I_3=\left(9-\frac1{32},9+\frac1{32}\right)=\left(\frac{287}{32},\frac{289}{32}\right). $$

Their lengths, computed from the endpoints, are \(9/8-7/8=1/4\), \(65/16-63/16=2/16=1/8\), and \(289/32-287/32=2/32=1/16\). In general, \(n^2\) is the midpoint of its assigned interval, so it is strictly inside the interval. The total length is

$$ \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} =\frac14+\frac18+\frac1{16}+\cdots =\frac12<1. $$

Check: the first three lengths sum to \(1/4+1/8+1/16=7/16\), and the remaining lengths sum to \(1/32+1/64+\cdots=1/16\); together they give \(7/16+1/16=1/2\), matching the geometric-series total. Replacing \(1\) by any \(\varepsilon>0\) scales every length by \(\varepsilon\), giving total length \(\varepsilon/2<\varepsilon\), as required.

Worked Example: All Integers

List the integers as \(z_1=0\), \(z_{2k}=k\), and \(z_{2k+1}=-k\) for positive integers \(k\). This lists zero and then each positive and negative integer exactly once. Use intervals of radius \(1/2^{j+1}\) around \(z_j\).

The first four centers are \(0,1,-1,2\). Their intervals and lengths are

$$ \left(-\frac14,\frac14\right)\quad(\ell=\tfrac12),\qquad \left(\frac78,\frac98\right)\quad(\ell=\tfrac14),\qquad \left(-\frac{17}{16},-\frac{15}{16}\right)\quad(\ell=\tfrac18),\qquad \left(\frac{63}{32},\frac{65}{32}\right)\quad(\ell=\tfrac1{16}). $$

For example, the third interval has center \(-1\) and radius \(1/16\), so its endpoints are \(-16/16-1/16=-17/16\) and \(-16/16+1/16=-15/16\); its length is \(2/16=1/8\). Each integer is the center of its own interval, hence belongs to the cover. The total length is

$$ \sum_{j=1}^{\infty}\frac{1}{2^j} =\frac12+\frac14+\frac18+\cdots=1. $$

To make the total arbitrarily small, fix any \(\eta>0\) and instead use radius \(\eta/2^{j+2}\) around \(z_j\). The \(j\)th length is \(\eta/2^{j+1}\), and the total is \(\eta/2<\eta\). Check: the first four lengths in this rescaled cover sum to \(\eta(1/4+1/8+1/16+1/32)=15\eta/32\); the remaining tail is \(\eta/32\), giving \(\eta/2\) altogether.

Worked Example: The Reciprocal Sequence

Let \(A=\{1/n:n=1,2,\ldots\}\). With \(\varepsilon=3/2\), assign to \(1/n\) an interval of radius \((3/2)/2^{n+2}=3/2^{n+3}\). Its length is \(3/2^{n+2}\).

For the first three points, the intervals are

$$ \left(\frac{13}{16},\frac{19}{16}\right),\qquad \left(\frac{13}{32},\frac{19}{32}\right),\qquad \left(\frac{55}{192},\frac{73}{192}\right). $$

Indeed, the first is \(1\pm3/16\); the second is \(1/2\pm3/32\); and the third is \(1/3\pm3/64=64/192\pm9/192\). Their lengths from the endpoints are \(6/16=3/8\), \(6/32=3/16\), and \(18/192=3/32\), respectively. The point \(1/n\) is the midpoint of its interval for every \(n\), so all points of \(A\) are covered. Summing the lengths gives

$$ \sum_{n=1}^{\infty}\frac{3}{2^{n+2}} =\frac38+\frac3{16}+\frac3{32}+\cdots =\frac34<\frac32. $$

Check: the first three terms sum to \(3/8+3/16+3/32=21/32\). The remaining terms sum to \(3/64+3/128+\cdots=3/32\), so the full sum is \(24/32=3/4\). For an arbitrary error \(\eta>0\), using radius \(\eta/2^{n+2}\) gives length \(\eta/2^{n+1}\) at index \(n\), and the total is \(\eta/2<\eta\). Thus the particular numerical cover illustrates the general construction.

Why the Lengths Must Be Summable

A tempting but ineffective choice is to give every point the same interval length. For a countably infinite listing and a fixed \(\varepsilon>0\), intervals each of length \(\varepsilon/4\) have total length

$$ \sum_{n=1}^{\infty}\frac{\varepsilon}{4}=+\infty. $$

This calculation does not show that the set has positive outer measure; it only shows that this particular cover gives no finite upper bound. The correction is to allocate a summable error, such as \(\varepsilon/2^{n+1}\), to the \(n\)th point. The resulting interval cover has total length \(\varepsilon/2\), and \(\varepsilon\) can be chosen as small as desired.

The conclusion concerns outer measure on \(\mathbb{R}\), not the size of the set in the sense of how many points it contains: a countable set may have infinitely many distinct points and still have outer measure zero. The proof depends on two ingredients working together: every point receives an interval, and the sum of all assigned lengths can be made arbitrarily small.

Check Your Understanding

For each question, identify the intervals and the total-length estimate that support the conclusion.

  1. For a listing \(A=\{x_1,x_2,\ldots\}\), what radius around \(x_n\) gives an interval of length \(\varepsilon/2^{n+1}\)?
  2. Why does the proof still work when \(A\) is finite, and what cover is used when \(A\) is empty?
  3. For the integer enumeration in the second example, calculate the length assigned to \(z_5\) in the unscaled cover.
  4. Explain why assigning every point an interval of length \(\varepsilon/4\) does not provide a finite total-length bound for an infinite set.
  5. If \(B\subseteq A\) and \(A\) is countable, which property of outer measure gives \(m^*(B)\le m^*(A)\)?

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