What Makes a Set Null?
A set is called null not simply because it is small in an informal sense, but because a particular measure assigns it value zero. The measure and the measurable space matter: the same underlying set can be null for one measure and non-null for another. There is also a separate question of measurability. A measure is defined only on its sigma-algebra, so a subset of a measurable null set is not automatically a set to which the measure can be applied.
Throughout the general statements, \((X,\mathcal{M},\mu)\) is a measure space: \(\mathcal{M}\) is a sigma-algebra on \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a measure. No assumption that \(\mu\) is finite, sigma-finite, or complete is made unless stated. The term “measurable” means membership in \(\mathcal{M}\).
The phrase “measure zero” is sometimes used informally for a set whose outer measure is zero, even when measurability has not yet been established. Here, “null set” always means a measurable set of measure zero. If \(A\notin\mathcal{M}\), then \(\mu(A)\) is not defined by the measure \(\mu\), even if \(A\) is contained in a measurable null set.
Basic Consequences of the Definition
Proof. Since \(A,E\in\mathcal{M}\) and \(A\subseteq E\), the set \(E\setminus A\) belongs to \(\mathcal{M}\). The disjoint union \(E=A\cup(E\setminus A)\) and finite additivity give \(\mu(E)=\mu(A)+\mu(E\setminus A)\). Both terms on the right are nonnegative, so \(0\le\mu(A)\le\mu(E)=0\). Thus \(\mu(A)=0\), and \(A\in\mathcal{M}\) by assumption. Therefore \(A\) is a measurable null set. \(\square\)
The measurability assumption on \(A\) is essential to this conclusion. The proposition does not assert that every subset of a measurable null set belongs to \(\mathcal{M}\); it says that if the subset is measurable, then its measure must be zero.
Dirac Measures: Null Sets Avoid the Point Mass
Let \((X,\mathcal{M})\) be a measurable space and fix \(x_0\in X\). The Dirac measure at \(x_0\), denoted \(\delta_{x_0}\), is the measure on \(\mathcal{M}\) defined by \(\delta_{x_0}(E)=1\) when \(x_0\in E\), and \(\delta_{x_0}(E)=0\) when \(x_0\notin E\). It is a finite measure, with total mass \(\delta_{x_0}(X)=1\). Its null-set test depends only on whether the measurable set contains \(x_0\).
Proof. Suppose first that \(E\in\mathcal{M}\) and \(x_0\notin E\). By the defining formula for the Dirac measure, \(\delta_{x_0}(E)=0\), so \(E\) is null. Conversely, suppose \(E\in\mathcal{M}\) is \(\delta_{x_0}\)-null. If \(x_0\in E\), the defining formula gives \(\delta_{x_0}(E)=1\), contradicting \(\delta_{x_0}(E)=0\). Hence \(x_0\notin E\). This proves both directions. \(\square\)
Worked Example: Testing Sets Under a Dirac Measure
Take \(X=\{a,b,c\}\), \(\mathcal{M}=\mathcal{P}(X)\), and the Dirac measure \(\delta_b\). This is a finite, complete, sigma-finite measure space: every subset is measurable, the total measure is \(1\), and the space can be covered by a single set of finite measure. Calculate the measure of \(E=\{a,c\}\), \(F=\{a,b\}\), and \(G=\varnothing\).
The point carrying mass is \(b\). Since \(b\notin E\), \(\delta_b(E)=0\). Since \(b\in F\), \(\delta_b(F)=1\). Since \(b\notin G\), \(\delta_b(G)=0\). Thus \(E\) and \(G\) are null, while \(F\) is not. The strict membership test here is simply whether \(b\in E\): containing \(b\) gives measure \(1\), and not containing \(b\) gives measure \(0\).
Check: the partition \(X=E\cup\{b\}\) is disjoint, and its measures add to \(\delta_b(X)=0+1=1\). Also \(X=F\cup\{c\}\) is disjoint, and its measures add to \(1+0=1\). These calculations agree with the defining formula.
Counting Measure: Only the Empty Set Is Null
For any set \(X\), counting measure is defined on the full sigma-algebra \(\mathcal{P}(X)\) by \(\#E\), the number of elements of \(E\) if \(E\) is finite, and \(+\infty\) if \(E\) is infinite. This measure is complete because every subset of a measurable set is measurable when the sigma-algebra is \(\mathcal{P}(X)\). It is finite when \(X\) is finite; for infinite \(X\), its total mass is infinite. On an uncountable \(X\), it is not sigma-finite, whereas on a countable \(X\), it is sigma-finite.
Proof. If \(E=\varnothing\), its number of elements is \(0\), so its counting measure is zero. If \(E\ne\varnothing\), it contains at least one element. If it contains finitely many elements, its counting measure is a positive integer; if it is infinite, its counting measure is \(+\infty\). In either case its measure is not zero. Every \(E\subseteq X\) is measurable under \(\mathcal{P}(X)\), so the only null set is the empty set. \(\square\)
Worked Example: Counting Measure on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\), let \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and let \(\mu\) be counting measure. This measure space is infinite, complete, and sigma-finite, since \(\mathbb{N}=\bigcup_{k=1}^{\infty}\{k\}\) and each singleton has measure \(1\). Compare \(A=\{2,5\}\), \(B=\{2,5,8\}\), and \(C=\varnothing\).
The set \(A\) has two elements, so \(\mu(A)=2\). The set \(B\) has three elements, so \(\mu(B)=3\). The empty set has zero elements, so \(\mu(C)=0\). Thus \(C\) is null, while \(A\) and \(B\) are not. For the infinite set \(D=\{2,4,6,\ldots\}\), the counting measure is \(\mu(D)=+\infty\), since \(D\) has infinitely many elements.
Check: the disjoint decomposition \(B=A\cup\{8\}\) gives \(\mu(B)=\mu(A)+\mu(\{8\})=2+1=3\). Also, \(D\) contains the \(n\) distinct elements \(2,4,\ldots,2n\) for every positive integer \(n\), so its cardinality is not finite and its counting measure is \(+\infty\).
Measurability Is a Separate Test
The counting-measure example has every subset measurable because its sigma-algebra is the full power set. That feature is not automatic for a general measure space. The distinction can be seen in a finite example, where measurability can be checked by listing the entire sigma-algebra.
Worked Example: A Subset of a Null Set That Is Not Measurable
Let \(X=\{a,b,c\}\) and let \(\mathcal{M}=\{\varnothing,\{a,b\},\{c\},X\}\). This is a sigma-algebra: its members occur in complementary pairs, and on a finite set it is closed under countable unions. Define \(\mu(\varnothing)=0\), \(\mu(\{a,b\})=0\), \(\mu(\{c\})=1\), and \(\mu(X)=1\). This is a finite, sigma-finite measure space, but not every subset of \(X\) is measurable.
To verify the measure, note that \(\mathcal{M}\) consists of unions of the two disjoint pieces \(\{a,b\}\) and \(\{c\}\), assigned masses \(0\) and \(1\). For disjoint measurable sets, each piece contributes its assigned mass at most once, so the mass of their union is the sum of their masses; this also verifies countable additivity. The set \(E=\{a,b\}\) is measurable and \(\mu(E)=0\). Its subset \(A=\{a\}\) is not in the displayed list of members of \(\mathcal{M}\). Therefore \(\mu(A)\) is not defined by this measure. In particular, it is not correct to call \(A\) a measurable null set.
Check: \(E\) itself is in \(\mathcal{M}\), so its measure is defined and equals \(0\). The list of \(\mathcal{M}\) contains \(\varnothing\) and \(\{a,b\}\), but not \(\{a\}\), so the subset fails the measurability test before any measure calculation can be made.
A Common Mistake and a Reliable Test
A common mistake is to infer that every subset of a measurable null set is measurable and has measure zero. In the finite example, the incorrect calculation would be: “\(A=\{a\}\subseteq E=\{a,b\}\), and \(\mu(E)=0\), so \(\mu(A)=0\).” The conclusion does not follow, because \(A\notin\mathcal{M}\), and \(\mu(A)\) is not defined. The valid proposition applies only after measurability of \(A\) has been established.
For any proposed null set \(E\), use two separate checks. First ask whether \(E\in\mathcal{M}\). If not, it is not a measurable null set under the definition used here. If it is measurable, then calculate \(\mu(E)\) and check whether the value is exactly zero. The Dirac and counting examples show why the measure-specific calculation matters: for \(\delta_{x_0}\), measurable sets avoiding \(x_0\) are null; for counting measure on a power set, only the empty set is null.
These examples also show that nullness is not the same as being empty. Under a Dirac measure, a set can contain many points and still have measure zero, provided it avoids the point mass. Under counting measure, every nonempty set has positive measure or infinite measure. In all cases, “null” is a relation between a set, a sigma-algebra that makes it measurable, and a measure defined on that sigma-algebra.
Check Your Understanding
For each question, keep the measurability test separate from the measure calculation.
- For \(\delta_{x_0}\) on a measurable space \((X,\mathcal{M})\), characterize all measurable null sets.
- For counting measure on \(\mathcal{P}(X)\), what is the measure of a nonempty finite set with exactly \(k\) elements, where \(k\ge1\)?
- In the finite example with \(\mathcal{M}=\{\varnothing,\{a,b\},\{c\},X\}\), why is it invalid to write \(\mu(\{a\})=0\)?
- If \(A,E\in\mathcal{M}\), \(A\subseteq E\), and \(\mu(E)=0\), which measure property shows that \(\mu(A)=0\)?
- Can a nonempty set be null for a Dirac measure? State the exact condition on the point \(x_0\).