Why Countably Many Exceptions Can Still Be Negligible
In measure theory, a set of measure zero is negligible for many purposes. But arguments often produce a different exceptional set at each stage: one set for each index \(n\). To control all stages at once, we need to know what happens to the union of those sets. The essential point is that countable additivity controls this union, while the sigma-algebra property guarantees that it remains measurable.
Throughout the main result, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space. No finiteness, sigma-finiteness, or completeness assumption is imposed. A measurable null set means a set \(E\in\mathcal{M}\) with \(\mu(E)=0\). Measures take values in \([0,\infty]\). The countable subadditivity property of a measure, established from countable additivity, states that for measurable sets \(E_n\), \(\mu(\bigcup_{n=1}^{\infty}E_n)\le\sum_{n=1}^{\infty}\mu(E_n)\).
Proof. Since \(\mathcal{M}\) is a sigma-algebra and each \(E_n\) belongs to \(\mathcal{M}\), their countable union \(E=\bigcup_{n=1}^{\infty}E_n\) belongs to \(\mathcal{M}\). Countable subadditivity therefore applies and gives \(\mu(E)\le\sum_{n=1}^{\infty}\mu(E_n)=\sum_{n=1}^{\infty}0=0\). Measures are nonnegative, so also \(\mu(E)\ge0\). Hence \(\mu(E)=0\), as required. \(\square\)
This proof uses two different properties for two different purposes. Closure of \(\mathcal{M}\) under countable unions gives measurability; countable subadditivity gives measure zero. Neither step asks the whole space to have finite measure. The empty union is \(\varnothing\), which belongs to every sigma-algebra and has measure zero, so the same conclusion also holds if there are no sets in the collection.
Proof. By the theorem, \(N=\bigcup_{n=1}^{\infty}N_n\) is measurable and \(\mu(N)=0\). If \(x\notin N\), then \(x\notin N_n\) for each \(n\), because \(N_n\subseteq N\). The hypothesis for \(P_n\) therefore gives \(P_n(x)\) for every \(n\). Thus all statements hold simultaneously outside \(N\). \(\square\)
This is a set-theoretic way to manage a countable collection of exceptions: collect them into one measurable null set. It does not assert that there is a single exception for an uncountable collection of statements. Nor does it say that a countable union is null merely because it is countable; each set in the theorem must itself be null.
Worked Example: The Rationals in the Unit Interval
Use the theorem to show that \(Q\cap[0,1]\) has Lebesgue measure zero. Work in \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete, sigma-finite, and has infinite total measure.
The set \(Q\cap[0,1]\) is countable, so list its distinct elements as \(q_1,q_2,\ldots\). Each singleton \(\{q_j\}\) is Lebesgue measurable. For every \(\varepsilon>0\), it lies inside the interval \((q_j-\varepsilon/2,q_j+\varepsilon/2)\), whose length is \(\varepsilon\). Monotonicity of Lebesgue measure gives \(0\le m(\{q_j\})\le\varepsilon\) for every \(\varepsilon>0\), and hence \(m(\{q_j\})=0\). The theorem now gives \(m(Q\cap[0,1])=m(\bigcup_{j=1}^{\infty}\{q_j\})=0\).
Check by a direct cover: Fix \(\varepsilon>0\), and cover \(q_j\) by the interval centered at \(q_j\) with radius \(\varepsilon/2^{j+2}\). Its length is \(\varepsilon/2^{j+1}\). The total length of these intervals is \(\sum_{j=1}^{\infty}\varepsilon/2^{j+1}=\varepsilon/2\). Thus the Lebesgue outer measure of \(Q\cap[0,1]\) is at most \(\varepsilon/2\) for every \(\varepsilon>0\), so it is zero. Since the set is measurable, its Lebesgue measure is zero as well.
Worked Example: Collecting Dyadic Boundary Exceptions
For each positive integer \(n\), define \(D_n=\{k/2^n:k=0,1,\ldots,2^n\}\) in \((\mathbb{R},\mathcal{L},m)\). These are the grid points that can occur as boundaries in a partition of \([0,1]\) into intervals of length \(2^{-n}\). Show that all such boundary points, across every level \(n\), form a null set.
Each \(D_n\) is finite and measurable. A singleton has Lebesgue measure zero, as shown in the previous example, so finite subadditivity gives \(m(D_n)=0\). The theorem applies to the sequence \((D_n)\): the union \(D=\bigcup_{n=1}^{\infty}D_n\), consisting of all dyadic rationals in \([0,1]\), is measurable and satisfies \(m(D)=0\).
Check: There are only countably many pairs \((n,k)\) with \(n\ge1\) and \(0\le k\le2^n\), so \(D\) is countable. Enumerate its distinct points as \(d_1,d_2,\ldots\). For any \(\varepsilon>0\), intervals centered at \(d_j\) with radius \(\varepsilon/2^{j+2}\) cover \(D\), and their total length is \(\sum_{j=1}^{\infty}\varepsilon/2^{j+1}=\varepsilon/2\). This independently confirms that \(D\) has Lebesgue measure zero.
The same reasoning applies whenever each stage contributes a measurable null set of possible boundary or exceptional points: their countable collection is still contained in one measurable null set.
Worked Example: A Countable Union for Dirac Measure
Consider \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)\), where \(\delta_0(A)=1\) if \(0\in A\) and \(\delta_0(A)=0\) if \(0\notin A\). This is a finite, sigma-finite measure space on the Borel sigma-algebra; it is not complete. For \(n\ge1\), let \(E_n=\{1/n\}\).
Each \(E_n\) is Borel and does not contain \(0\), so the defining formula for Dirac measure gives \(\delta_0(E_n)=0\). Their union is \(E=\{1/n:n\ge1\}\), which also does not contain \(0\). Therefore \(\delta_0(E)=0\), in agreement with the theorem. In this case the union’s measure can also be checked directly from the definition of \(\delta_0\).
For contrast, the measurable singleton \(\{0\}\) has \(\delta_0(\{0\})=1\), not zero. Thus a set being a singleton—or a union being countable—does not by itself imply that its measure is zero. The measure and the nullity of each set both matter.
What the Result Does Not Say
A common error is to replace “a countable union of null sets” with “a countable union of sets.” For a concrete correction, take counting measure \(\#\) on \((\mathbb{N},\mathcal{P}(\mathbb{N}),\#)\), where \(\mathbb{N}=\{1,2,\ldots\}\). This measure space is infinite, sigma-finite, and complete. Each singleton \(E_n=\{n\}\) has \(\#(E_n)=1\), so these sets are not null. Their union is \(\mathbb{N}\), and \(\#(\mathbb{N})=\infty\). The incorrect calculation “each set is a singleton, so each has measure zero” fails because counting measure assigns measure one to every singleton. The theorem is not contradicted: its hypothesis \(\#(E_n)=0\) is absent.
There is also an important measurability boundary. The theorem starts with measurable sets \(E_n\); it then obtains a measurable union from the sigma-algebra axiom. It does not claim that every subset of a measurable null set is measurable in an arbitrary measure space. That further property is part of completeness, which is not assumed here. For the theorem itself, completeness is unnecessary: measurability of the union follows directly from countable closure of \(\mathcal{M}\).
The result is useful whenever separate arguments produce countably many measurable null exceptional sets. The proof needs no estimates beyond countable subadditivity, and it does not require a common finite-measure region. Its precise scope is simple: in any measure space, countably many measurable sets of measure zero can be gathered into one measurable set of measure zero.
Check Your Understanding
For each question, identify the measure space and verify whether the null-set hypothesis is satisfied before applying the theorem.
- In \((\mathbb{R},\mathcal{L},m)\), let \(E_n=\{2^{-n}\}\). What is the measure of \(\bigcup_{n=1}^{\infty}E_n\), and which theorem justifies the conclusion?
- In a general measure space, why are both sigma-algebra closure and countable subadditivity needed in the proof?
- For Dirac measure \(\delta_0\) on \(\mathbb{R}\), does the theorem imply that \(\bigcup_{n=1}^{\infty}\{n\}\) is null? Check the measure of each individual set.
- Suppose \(P_n(x)\) can fail only on a measurable null set \(N_n\). State a measurable set outside which every \(P_n\) holds simultaneously.
- Does the countable-union theorem require \(\mu(X)<\infty\), or that \((X,\mathcal{M},\mu)\) be complete? Explain which steps in the proof use its actual assumptions.