What It Means for a Set to Be Null
Continuity from above describes how measures behave when sets decrease toward an intersection. A particularly important case occurs when the intersection has measure zero. Sets of measure zero can be negligible for measuring size while still containing points, or even being complicated. To use them correctly, we must also distinguish a measurable set of measure zero from an arbitrary subset of such a set.
Throughout, let \((X,\mathcal{F},\mu)\) be a measure space. The measure of a set is defined only when that set belongs to \(\mathcal{F}\). Thus the definition below applies to measurable sets; later we will explain the role of completeness when considering their subsets.
Every null set is measurable by definition. The second phrase, “contained in a null set,” does not itself assert that \(E\) is measurable. That distinction matters because a measure space need not include every subset of each of its measurable sets. In a complete measure space, however, every subset of a null set is measurable and has measure zero.
Basic Consequences of Zero Measure
A null set cannot contain a measurable subset of positive measure: if \(E\subseteq N\), with both sets measurable and \(\mu(N)=0\), monotonicity gives \(0\leq\mu(E)\leq\mu(N)=0\). Hence \(\mu(E)=0\). Also, the empty set is null, since every measure assigns it measure zero.
Countable unions of measurable null sets are null. This is the previously established Countable Union of Null Sets theorem: countable subadditivity gives a measure at most the sum of zeros, and the union is measurable by closure of the sigma-algebra under countable unions. This conclusion is powerful, but it does not say that an uncountable union of null sets is null. For instance, the real line is the union of its singleton sets, although the real line does not have measure zero.
Worked Example: A Singleton Has Lebesgue Measure Zero
Fix \(a\in\mathbb{R}\), and use Lebesgue measure on the Borel subsets of \(\mathbb{R}\). The singleton \(\{a\}\) is closed and therefore measurable. For every \(\varepsilon>0\), it is contained in the open interval $$ \left(a-\frac{\varepsilon}{2},a+\frac{\varepsilon}{2}\right), $$ whose length is \(\varepsilon\). Monotonicity therefore gives $$ 0\leq \mu(\{a\})\leq\varepsilon $$ for every \(\varepsilon>0\). The only nonnegative number bounded above by every positive \(\varepsilon\) is zero, so \(\mu(\{a\})=0\). This argument uses intervals of arbitrarily small length, not the assertion that a point has no size by visual inspection.
Proof. If \(E\) is finite, it is a finite union of closed singletons, so it is Borel measurable. Its measure is zero by finite additivity and the singleton result. Now suppose \(E\) is countably infinite, and list its elements as \(E=\{x_1,x_2,\ldots\}\). Each singleton is closed, so \(E=\bigcup_{n=1}^{\infty}\{x_n\}\) is Borel measurable. For any \(\varepsilon>0\), cover \(x_n\) by the open interval $$ I_n=\left(x_n-\frac{\varepsilon}{2^{n+2}},x_n+\frac{\varepsilon}{2^{n+2}}\right). $$ Its length is \(\varepsilon/2^{n+1}\), and $$ \sum_{n=1}^{\infty}\frac{\varepsilon}{2^{n+1}}=\frac{\varepsilon}{2}. $$ By countable subadditivity, \(\mu(E)\leq\mu(\bigcup_n I_n)\leq\varepsilon/2<\varepsilon\). Since this holds for every positive \(\varepsilon\), \(\mu(E)=0\). \(\square\)
Worked Example: The Rational Numbers in an Interval
The set \(\mathbb{Q}\cap[0,1]\) is countable, since the rational numbers are countable, and it is Borel measurable as a countable union of singletons. The theorem gives $$ \mu(\mathbb{Q}\cap[0,1])=0. $$ This does not mean that \([0,1]\) has measure zero: \(\mathbb{Q}\cap[0,1]\) is only part of the interval, and the interval has measure \(1\). In particular, a set can be dense in an interval and still have measure zero. Density describes which open intervals the set meets; measure describes its size in a different sense.
Completeness and Subsets of Null Sets
The countable-set theorem applies because countable subsets of \(\mathbb{R}\) are Borel measurable. In a general measure space, a subset of a measurable null set might not be measurable. Completeness is the condition that rules out this difficulty.
Proof. Let \(N\) be null and let \(E\subseteq N\). Completeness gives \(E\in\mathcal{F}\), so its measure is defined. By monotonicity, $$ 0\leq\mu(E)\leq\mu(N)=0. $$ Thus \(\mu(E)=0\), as required. \(\square\)
The measurability conclusion is the part supplied by completeness; monotonicity then gives the zero measure. In a space that is not complete, monotonicity cannot be applied to a nonmeasurable subset, because its measure has not been defined. One can often enlarge a measure space by adding all subsets of its measurable null sets and assigning them measure zero; this process is called completing the measure space.
Worked Example: A Null Set That Is Not Countable
Let \(C_0=[0,1]\). Construct \(C_n\) by removing the open middle third from each interval of \(C_{n-1}\). At stage \(n\), the set \(C_n\) consists of \(2^n\) closed intervals, each of length \(3^{-n}\). The intervals are disjoint, so $$ \mu(C_n)=2^n3^{-n}=\left(\frac{2}{3}\right)^n. $$ The sets decrease, and the Cantor set is \(C=\bigcap_{n=0}^{\infty}C_n\). Since \(\mu(C_0)=1<\infty\), Continuity from Above for Finite Measure applies and gives $$ \mu(C)=\lim_{n\to\infty}\mu(C_n) =\lim_{n\to\infty}\left(\frac{2}{3}\right)^n=0. $$ Thus \(C\) is a null set. Unlike the countable examples above, the construction shows how a nested sequence of finite unions of intervals can have a null intersection. The measure computation relies on the total length at each stage, not on the number of points in the final set.
Changing a Set on a Null Set
Null sets let us treat measurable sets as equivalent for purposes of measuring size when their difference is confined to a null set. The following result makes the claim precise. The symmetric difference \(A\mathbin{\triangle}B\) consists of points belonging to exactly one of \(A\) and \(B\).
Proof. The sets \(A\cap B\), \(A\setminus B\), and \(B\setminus A\) are measurable. Both \(A\setminus B\) and \(B\setminus A\) are contained in \(A\mathbin{\triangle}B\), so monotonicity gives $$ 0\leq\mu(A\setminus B)\leq\mu(A\mathbin{\triangle}B)=0, \qquad 0\leq\mu(B\setminus A)\leq\mu(A\mathbin{\triangle}B)=0. $$ Consequently, both differences have measure zero. The disjoint decompositions $$ A=(A\cap B)\cup(A\setminus B), \qquad B=(A\cap B)\cup(B\setminus A) $$ and finite additivity now yield $$ \mu(A)=\mu(A\cap B)+0=\mu(B). $$ These equalities remain valid if \(\mu(A\cap B)=\infty\); no subtraction of infinite quantities is involved. \(\square\)
Worked Example: Adding One Endpoint Does Not Change Length
Take \(A=(2,5)\) and \(B=[2,5]\) with Lebesgue measure. Both sets are Borel measurable, and $$ A\mathbin{\triangle}B=\{2,5\}. $$ This symmetric difference is a finite set, hence null by the countable-set theorem. Therefore \(\mu(A)=\mu(B)\). Directly, the interval lengths also give $$ \mu((2,5))=5-2=3=\mu([2,5]). $$ This illustrates how changing whether endpoints are included can alter the set without altering its measure. The theorem applies more generally; it does not require the sets to be intervals.
Why Null Sets Matter—and What They Do Not Say
A null set is negligible with respect to the measure, but it is not necessarily empty. The singleton example already shows this. Nor does a null set have to be countable: the Cantor construction gives a null set as an intersection of increasingly fine finite unions of intervals. These distinctions become important whenever a statement is said to hold “except on a null set.” The exceptional points may still exist, and their measurability must be handled according to the measure space.
There are two common pitfalls. First, being small in measure is not the same as being topologically small: the rational numbers in \([0,1]\) are dense there but have measure zero. Second, a subset of a null set is not automatically measurable in an arbitrary measure space. If a claim concerns such a subset, check that the space is complete or that measurability has been established by another argument.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What two conditions must a set satisfy to be a null set in a measure space?
- Why does a countable union of measurable null sets remain null, and why does that fact not imply that every uncountable union of null sets is null?
- What does completeness guarantee about a subset of a null set?
- How does the sequence of stage sets in the Cantor construction show that the Cantor set has Lebesgue measure zero?
- If two measurable sets have null symmetric difference, why are their measures equal even if the measure is infinite?