The Covering Criterion for Null Sets

Small Covers Detect Zero Outer Measure

A set can be negligible for Lebesgue measure even when it is large in other senses: it may be infinite or dense, for example. The relevant measure-theoretic condition is not that the set fit inside one short interval, but that it fit inside a countable collection of intervals whose total length is as small as desired. The covering criterion makes this precise and gives a direct way to prove that subsets and countable unions remain negligible.

Throughout, the ambient measure space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. This measure is sigma-finite but not finite; no completeness result is needed here. For arbitrary \(E\subseteq\mathbb{R}\), including sets not yet known to be measurable, \(m^*(E)\) denotes Lebesgue outer measure. We use the interval-cover definition \(m^*(E)=\inf\{\sum_{j=1}^{\infty}|I_j|: E\subseteq\bigcup_{j=1}^{\infty}I_j,\ I_j\text{ bounded open intervals}\}\), where \(|I_j|\) is the length of \(I_j\). Finite covers are allowed by appending empty intervals, and the empty set has the empty cover of total length zero.

Proof. Suppose first that \(m^*(E)=0\). Fix \(\varepsilon>0\). By the definition of an infimum, for every positive \(\varepsilon\) there is an admissible interval cover whose total length is less than \(m^*(E)+\varepsilon=\varepsilon\). Thus the required cover exists. Conversely, suppose that such a cover exists for every \(\varepsilon>0\). The definition of outer measure gives \(0\le m^*(E)\le\sum_j|I_j|<\varepsilon\) for every \(\varepsilon>0\). A nonnegative number less than every positive \(\varepsilon\) must be zero: if \(m^*(E)>0\), choosing \(\varepsilon=m^*(E)/2\) contradicts the inequality. Hence \(m^*(E)=0\). \(\square\)

The criterion concerns outer measure, so it applies before measurability is known. When the set in question is measurable, it is also exactly a test for measure zero. The cover is allowed to depend on \(\varepsilon\): a single cover of small length is not required to work for every tolerance.

Subsets and Countable Unions

The criterion immediately gives two closure properties. Neither requires the sets themselves to be measurable: they are statements about outer measure.

Proof. For the subset assertion, fix \(\varepsilon>0\). Since \(A\) is outer-null, choose a countable interval cover of \(A\) with total length less than \(\varepsilon\). The same intervals cover \(B\), because \(B\subseteq A\). The covering criterion gives \(m^*(B)=0\).

For the union assertion, fix \(\varepsilon>0\). Since each \(A_n\) is outer-null, apply the criterion to \(A_n\) with tolerance \(\varepsilon/2^{n+1}\). This gives a cover of \(A_n\) with total length less than \(\varepsilon/2^{n+1}\). Put all of these covers together. They form a countable collection of intervals covering \(\bigcup_{n=1}^{\infty}A_n\), and their total length is at most \(\sum_{n=1}^{\infty}\varepsilon/2^{n+1}=\varepsilon/2<\varepsilon\). The covering criterion now gives \(m^*(\bigcup_{n=1}^{\infty}A_n)=0\). The argument includes empty sets, whose covers can be empty, and does not require the \(A_n\) to be disjoint. \(\square\)

For measurable sets, the same conclusions can be phrased using \(m\): a subset of a measurable null set, or a countable union of measurable null sets, has outer measure zero. The countable union is measurable because \(\mathcal{L}\) is a sigma-algebra. For a subset, the conclusion here is outer-nullity, not a claim of measurability; that distinction is essential before completeness has been established.

Positive-Length Intervals Fail the Criterion

The criterion is a useful way to prove zero outer measure, but it also rules out zero outer measure for sets that contain a genuine interval. We use the covering lower bound for intervals established earlier: if a countable collection of intervals covers \([a,b]\), where \(a<b\), the sum of their lengths is at least \(b-a\). This bound applies to open interval covers as well.

A Common Budgeting Mistake

A frequent error in the countable-union argument is to give every \(A_n\) a cover of total length less than the same \(\varepsilon\), and then conclude that the combined cover has total length less than \(\varepsilon\). For instance, with \(\varepsilon=0.1\), the resulting estimates are only “the \(n\)-th cover has length less than \(0.1\).” Adding these bounds gives an infinite series of \(0.1\)'s, not a finite upper bound below \(0.1\). This does not establish the desired conclusion.

The correction is to distribute the available length among the sets. With tolerance \(0.1\), choose for \(A_n\) a cover of length less than \(0.1/2^{n+1}\). The sum of these bounds is \(0.1/2=0.05<0.1\). More generally, the allocation \(\varepsilon/2^{n+1}\) has total budget \(\varepsilon/2\), leaving a strict margin below \(\varepsilon\). The essential point is that the budget depends on \(n\) and that the sum of all budgets is controlled.

The interval-cover criterion turns an abstract equality \(m^*(E)=0\) into a constructive test: for each tolerance, produce one suitable cover. Subsets inherit any such cover, while countable unions require a summable allocation of cover lengths. For measurable sets these arguments identify and preserve measure zero; for arbitrary subsets they establish outer-nullity without assuming measurability.