Countable Sets and Dense Null Examples

Countability Can Be Invisible to Lebesgue Measure

A set may meet every open interval and still have Lebesgue measure zero. The rational numbers provide the central example: they are dense in the real line, but their total length, in the sense of Lebesgue measure, is zero. The reason is not that the rationals are far apart. Rather, countability lets us assign each point a sufficiently short interval so that the sum of all the interval lengths is as small as we choose.

Throughout, the ambient measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but \(m(\mathbb{R})=\infty\), so it is not finite. A set \(A\subseteq\mathbb{R}\) is countable if it is finite or its elements can be listed as a sequence. A set is null if its Lebesgue outer measure is zero. We use the covering criterion for null sets: for every \(\varepsilon>0\), a null set can be covered by countably many open intervals whose total length is less than \(\varepsilon\). Conversely, such covers for every \(\varepsilon>0\) imply that the set is null. By the null-set criterion for Lebesgue measurability, a set of outer measure zero belongs to \(\mathcal{L}\) and has measure zero.

A Cover for Any Countable Set

Proof. If \(A=\varnothing\), the empty family of intervals covers \(A\) and has total length zero. If \(A\) has \(N\geq1\) elements, list them as \(a_1,\ldots,a_N\). For each \(j\in\{1,\ldots,N\}\), use the open interval centered at \(a_j\) with length \(\varepsilon/2^{j+1}\). These intervals cover \(A\), and their total length is

If \(A\) is countably infinite, list its distinct elements as \(a_1,a_2,\ldots\), and use the same interval around \(a_j\), now for every integer \(j\geq1\). The intervals cover every listed point, and the geometric series calculation above gives total length \(\varepsilon/2\). In each case the total length is less than \(\varepsilon\). The covering criterion gives outer measure zero, and the null-set criterion gives \(A\in\mathcal{L}\) and \(m(A)=0\). \(\square\)

The use of a geometric sequence is convenient because its sum can be controlled while every point receives a positive interval length. The intervals do not need to be disjoint, and the points need not be separated from one another. Only the sum of the covering lengths matters.

Ordering the Rationals and Covering a Bounded Part

To make a cover fully explicit, it helps to specify an enumeration rather than merely assert that the rationals can be listed. Write each rational uniquely as \(k/q\), where \(k\in\mathbb{Z}\), \(q\) is a positive integer, and \(\gcd(|k|,q)=1\). In particular, zero is represented as \(0/1\). Order these reduced pairs first by increasing \(q+|k|\), and within each fixed value of \(q+|k|\), by increasing \(k\). There are only finitely many pairs at each level, and every reduced pair occurs at a level, so this rule lists every rational exactly once.

The first levels illustrate why the ordering rule must be followed exactly. At \(q+|k|=1\), the only reduced pair is \((k,q)=(0,1)\), giving \(0/1\). At \(q+|k|=2\), the reduced pairs are \((-1,1)\) and \((1,1)\), giving \(-1/1\) and \(1/1\) in increasing \(k\). At \(q+|k|=3\), the reduced pairs in increasing \(k\) give \(-2/1,-1/2,1/2,2/1\). These are the first three groups of this enumeration.

Now retain from the enumeration only the rationals in \([0,1]\), including both endpoints, and call the resulting sequence \(r_1,r_2,\ldots\). Its first terms are \(0,1,1/2,1/3,1/4,2/3\): the terms \(1/2\) and \(1/3\) occur at levels \(3\) and \(4\), while \(1/4\) and \(2/3\) occur at level \(5\). For each \(j\geq1\), define the open interval

Its length is \(1/2^{j+1}\). The interval contains \(r_j\), so the family covers every rational in \([0,1]\), and its total length is the geometric series \(\sum_{j=1}^{\infty}1/2^{j+1}=1/2\). More generally, replacing \(1/2^{j+1}\) by \(\varepsilon/2^{j+1}\) produces total length \(\varepsilon/2\), for any prescribed \(\varepsilon>0\).

Dense and Null: The Rationals

Proof. The reduced-pair enumeration shows that \(\mathbb{Q}\) is countable, so the theorem gives \(\mathbb{Q}\in\mathcal{L}\) and \(m(\mathbb{Q})=0\). To prove density, let \((a,b)\) be any nonempty open interval, so \(a<b\). Choose a positive integer \(n\) with \(1/n<b-a\), and let \(k=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer not exceeding \(na\). Then \(na<k\leq na+1\), hence \(a<k/n\leq a+1/n<b\). Thus \(k/n\) is rational and belongs to \((a,b)\). Every nonempty open interval therefore meets \(\mathbb{Q}\), which proves that \(\mathbb{Q}\) is dense. \(\square\)

The same argument shows that \(\mathbb{Q}\cap(a,b)\) is dense in \((a,b)\) for every \(a<b\), and the countable-set theorem shows that it is null. Density is a topological property: it says that every open interval contains a point of the set. Measure zero is a different property: it says that the set can be covered by intervals of arbitrarily small total length. Neither property rules out the other.

A Common Covering Mistake

A tempting but invalid approach is to assign every point in an infinite enumeration an interval of the same fixed length, such as \(\varepsilon/4\). For the first \(N\) points, the total assigned length is \(N\varepsilon/4\); as \(N\) grows, this bound is not controlled by \(\varepsilon\). The mistake is to confuse short individual intervals with a small total length. The correction is to assign lengths from a summable sequence: for example, lengths \(\varepsilon/2^{j+1}\) have total \(\varepsilon/2\), no matter which countable set is being covered.

A second error is to infer from density that a set must have positive measure. The proposition supplies a counterexample with all details: every nonempty open interval contains a rational, yet \(\mathbb{Q}\) is covered by intervals with total length less than any prescribed \(\varepsilon>0\). Countability controls the covering sum; density does not prevent that sum from being arbitrarily small.