Null Sets of an Outer Measure

What Does It Mean to Be Outer-Null?

An outer measure assigns a size to every subset of a space, including sets that have not been declared measurable. A set of outer measure zero is therefore meaningful before we have constructed a measure on a sigma-algebra. Such sets matter because they are stable under taking subsets and countable unions. Those closure properties make null sets useful both in constructing measures and in identifying changes that do not affect measure.

Fix a set \(X\) and an outer measure \(\mu^*\) on \(X\). This means that \(\mu^*\) is a function from the power set of \(X\) to \([0,\infty]\) such that \(\mu^*(\varnothing)=0\), \(\mu^*\) is monotone, and it is countably subadditive. In particular, for any sets \(A,A_1,A_2,\ldots\subseteq X\),

A\subseteq B\ \Longrightarrow\ \mu^*(A)\le\mu^*(B), \qquad \mu^*\!\left(\bigcup_{n=1}^{\infty}A_n\right) \le \sum_{n=1}^{\infty}\mu^*(A_n).

The value \(+\infty\) is allowed. Neither a sigma-algebra nor a measure space is required for these statements: \(\mu^*\) is defined on all subsets of \(X\).

Definition: A set \(N\subseteq X\) is outer-null with respect to \(\mu^*\) if \(\mu^*(N)=0\). The outer measure and the space are part of the meaning: a set may be outer-null for one outer measure and not for another.

Outer-null does not mean empty. For example, a nonempty set can have outer measure zero. Nor does “outer-null” automatically mean “of measure zero” for any measure one might also place on \(X\). That distinction will be made precise below.

Subsets and Countable Unions

Theorem: Let \(\mu^*\) be an outer measure on \(X\). Every subset of an outer-null set is outer-null. If \(N_1,N_2,\ldots\subseteq X\) are outer-null, then \(\bigcup_{n=1}^{\infty}N_n\) is outer-null.

Proof. Suppose \(N\subseteq X\) and \(\mu^*(N)=0\). If \(A\subseteq N\), monotonicity gives \(0\le\mu^*(A)\le\mu^*(N)=0\), hence \(\mu^*(A)=0\). Now suppose \(\mu^*(N_n)=0\) for every \(n\). Countable subadditivity gives

0\le \mu^*\!\left(\bigcup_{n=1}^{\infty}N_n\right) \le \sum_{n=1}^{\infty}\mu^*(N_n) =\sum_{n=1}^{\infty}0=0.

Thus the union is outer-null. This includes the empty set and finite unions: the empty set has outer measure zero by the outer-measure axiom, and a finite union can be written as a countable union by adjoining empty sets. \(\square\)

A family of sets closed under taking subsets and countable unions is often called a sigma-ideal. The theorem says that the outer-null sets of any outer measure form such a family. This is a closure property of the sets having outer measure zero; it does not assert that all subsets of \(X\) are measurable for an associated measure.

Worked Example: A Countable Set in the Real Line

Let \(\mu^*=m^*\) be Lebesgue outer measure on \(\mathbb{R}\), defined by covering a set with countably many intervals and taking the infimum of the sums of their lengths. Show directly that \(\mathbb{Q}\) is outer-null.

List the rationals as \(q_1,q_2,\ldots\). Given any \(\varepsilon>0\), cover \(q_n\) by the open interval centered at \(q_n\) of length \(\varepsilon/2^n\). The union of these intervals contains \(\mathbb{Q}\), and their total length is

\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n} =\varepsilon\sum_{n=1}^{\infty}\left(\frac12\right)^n =\varepsilon.

Consequently \(m^*(\mathbb{Q})\le\varepsilon\) for every \(\varepsilon>0\). Since \(m^*(\mathbb{Q})\ge0\), it follows that \(m^*(\mathbb{Q})=0\). The same covering argument applies to any countable subset of \(\mathbb{R}\), including a singleton. For a singleton \(\{x\}\), one can also use a single interval of length \(\varepsilon\) centered at \(x\), for every \(\varepsilon>0\). Check: the displayed geometric series has first term \(\varepsilon/2\) and ratio \(1/2\), so its sum is \((\varepsilon/2)/(1-1/2)=\varepsilon\), as required.

Worked Example: The Middle-Thirds Cantor Set

Let \(C\subseteq[0,1]\) be the middle-thirds Cantor set, obtained by repeatedly removing open middle thirds. At stage \(n\), the set remaining is covered by \(2^n\) closed intervals, each of length \(3^{-n}\). Therefore this cover has total length

2^n3^{-n}=\left(\frac23\right)^n.

Since the stage-\(n\) intervals contain \(C\), the definition of Lebesgue outer measure gives \(0\le m^*(C)\le(2/3)^n\) for every positive integer \(n\). The sequence \((2/3)^n\) tends to zero, so \(m^*(C)=0\). For a numerical check, at stage \(1\) the cover has \(2\) intervals of length \(1/3\), with total length \(2/3\); at stage \(2\) it has \(4\) intervals of length \(1/9\), with total length \(4/9=(2/3)^2\). Each stage gives the stated formula. This example shows why outer-null cannot mean empty: \(C\) contains \(0\) and \(1\), but has outer measure zero.

Outer-Null Sets and Caratheodory Measurability

Outer measure is defined on all subsets, but an associated measure is defined only on a chosen collection of measurable sets. For an outer measure \(\mu^*\) on \(X\), the Caratheodory test calls \(E\subseteq X\) measurable when, for every \(A\subseteq X\),

\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E).

The test compares the outer measure of any test set \(A\) with the two pieces formed by splitting \(A\) at \(E\). The following fact concerns only outer-null sets; it does not establish the broader closure properties of the Caratheodory-measurable sets.

Proposition: For any outer measure \(\mu^*\) on \(X\), every outer-null set \(N\subseteq X\) satisfies the Caratheodory measurability test.

Proof. Let \(N\subseteq X\) satisfy \(\mu^*(N)=0\), and take any \(A\subseteq X\). Since \(A\cap N\subseteq N\), monotonicity gives \(\mu^*(A\cap N)=0\). Since \(A\setminus N\subseteq A\), monotonicity gives \(\mu^*(A\setminus N)\le\mu^*(A)\). These inequalities give one direction of the required equality. For the other direction, \(A=(A\cap N)\cup(A\setminus N)\), so subadditivity gives

\mu^*(A) \le \mu^*(A\cap N)+\mu^*(A\setminus N) =0+\mu^*(A\setminus N) \le\mu^*(A).

All inequalities are therefore equalities, including when \(\mu^*(A)=+\infty\). Thus \(N\) satisfies the test for every \(A\), as required. \(\square\)

In the Caratheodory construction, the measure on the measurable sets is the restriction of the given outer measure: its value at a measurable \(E\) is \(\mu^*(E)\). Thus, when this restriction is being used as the associated measure, an outer-null measurable set has measure zero because its value is exactly its outer measure. This conclusion concerns the restriction of that same outer measure. It does not apply to an unrelated measure on \(X\).

Worked Example: Why the Measure Must Be the Restriction

Let \(X=\{a\}\), and define \(\mu^*(A)=0\) for every \(A\subseteq X\). This is an outer measure: it is zero on the empty set, monotone because every value is zero, and countably subadditive because both sides of the subadditivity inequality are zero. In particular, \(\mu^*(\{a\})=0\).

Now put the counting measure \(\nu\) on \(X\). Its sigma-algebra is the power set of \(X\), and \(\nu(A)\) is the number of elements of \(A\). Hence \(\nu(\{a\})=1\), not zero. There is no contradiction: \(\nu\) is not the restriction of \(\mu^*\); it is a different measure. In contrast, the restriction of \(\mu^*\) to its Caratheodory-measurable sets assigns \(\{a\}\) the value \(\mu^*(\{a\})=0\). Check: the two values are computed from their definitions: \(\mu^*(\{a\})=0\) by the constant-zero rule, whereas \(\nu(\{a\})=1\) because the set contains exactly one point.

A Common Mistake and Its Correction

A common mistake is to reason, “\(N\) has outer measure zero, so every measure on \(X\) must assign \(N\) measure zero.” The singleton example shows the error concretely: it yields the incorrect calculation \(\nu(\{a\})=\mu^*(\{a\})=0\), even though counting measure gives \(\nu(\{a\})=1\). The correction is to identify the measure under discussion. If \(\mu\) is the restriction of \(\mu^*\) to a collection of \(\mu^*\)-measurable sets, then \(\mu(E)=\mu^*(E)\) for each such \(E\), so an outer-null \(E\) has \(\mu(E)=0\). With a separate measure \(\nu\), outer-nullity for \(\mu^*\) alone says nothing about \(\nu(N)\).

There is a related distinction between belonging to a specified sigma-algebra and satisfying the Caratheodory test. The latter is defined from the outer measure itself, and the proposition proves that outer-null sets satisfy that test. Membership in some other specified sigma-algebra is a separate matter. In particular, “outer measure zero” is the appropriate statement before a relevant measurable structure or associated measure has been identified.

Takeaway: For a fixed outer measure, outer-null sets include all subsets of null sets and are closed under countable unions. They also satisfy the Caratheodory measurability test. A claim that a set has measure zero must specify a measure, and the direct conclusion from outer-nullity applies to the measure obtained by restricting that same outer measure.

Check Your Understanding

Use the definition of outer measure and the distinction between an outer measure and an unrelated measure in your answers.

If \(A\subseteq N\) and \(\mu^*(N)=0\), which outer-measure axiom gives \(\mu^*(A)=0\)?
Suppose \(\mu^*(N_n)=0\) for every positive integer \(n\). Write the inequality that proves \(\bigcup_{n=1}^{\infty}N_n\) is outer-null.
For Lebesgue outer measure, why does covering each rational \(q_n\) by an interval of length \(\varepsilon/2^n\) prove \(m^*(\mathbb{Q})=0\)?
In the Caratheodory test, explain why \(\mu^*(A\cap N)=0\) when \(\mu^*(N)=0\).
Give the precise condition under which an outer-null set has measure zero for a measure obtained from an outer measure.

What You'll Learn