Countable Subadditivity

Bounding the Measure of a Union

The previous tutorial examined how inclusion orders measure: if $A\subseteq B$, then $\mu(A)\leq\mu(B)$. A related question is how to estimate the measure of a set assembled from many pieces. When those pieces are disjoint, countable additivity gives an exact formula. When they overlap, the measure of their union is at most the sum of their individual measures. This inequality is countable subadditivity.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. The Countable Subadditivity Theorem was established in “Definition of a Measure”; we will use it here rather than prove it again. For measurable sets $A_1,A_2,\ldots$, it says

\mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty}\mu(A_n).

The right side is an extended nonnegative sum, so it may be infinite. The inequality is useful even when it is not sharp: a point lying in several sets is counted several times on the right, but only once as a member of the union. Disjointness is therefore not required. In fact, if the sets are pairwise disjoint, countable additivity gives equality.

Key distinction: Countable additivity gives the exact measure of a countable disjoint union. Countable subadditivity gives an upper bound for a union that may overlap. Overlap can make the inequality strict.

A finite union can be handled by the same result: append empty sets to the finite list and apply countable subadditivity. The measure of each added empty set is zero. More generally, if $A_n\subseteq B_n$ for every $n$, then

\bigcup_{n=1}^{\infty} A_n\subseteq\bigcup_{n=1}^{\infty} B_n, \qquad \mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \mu\left(\bigcup_{n=1}^{\infty} B_n\right) \leq \sum_{n=1}^{\infty}\mu(B_n).

The first inequality uses monotonicity of measures; the second uses countable subadditivity. This is a common strategy: replace complicated sets by simpler measurable supersets whose measures are easier to estimate.

Worked Example: Overlapping Intervals

Take Lebesgue measure on the Borel subsets of $\mathbb{R}$, and set $A=[-2,1]$, $B=[0,4]$, and $C=\{5\}$. The individual measures are $$ \mu(A)=1-(-2)=3,\qquad \mu(B)=4-0=4,\qquad \mu(C)=0. $$ The intervals overlap on $[0,1]$, so adding their measures counts that part twice. Their union is $[-2,4]\cup\{5\}$, and therefore $$ \mu(A\cup B\cup C)=4-(-2)=6 \leq 3+4+0=7. $$ The estimate is valid but not an equality. The extra point contributes zero measure, while the overlap accounts for the gap between the union’s measure and the sum of the individual measures.

Countable Unions of Null Sets

Countable subadditivity immediately gives a useful stability property. A null set is a measurable set $N$ with $\mu(N)=0$. A countable union of null sets is still null, even if the sets overlap in complicated ways.

Theorem (Countable Union of Null Sets): If $N_n\in\mathcal{F}$ and $\mu(N_n)=0$ for every positive integer $n$, then $$ \mu\left(\bigcup_{n=1}^{\infty}N_n\right)=0. $$

Proof. Since $\mathcal{F}$ is a sigma-algebra, $\bigcup_{n=1}^{\infty}N_n$ is measurable. Countable subadditivity gives $$ 0\leq\mu\left(\bigcup_{n=1}^{\infty}N_n\right) \leq\sum_{n=1}^{\infty}\mu(N_n) =\sum_{n=1}^{\infty}0 =0. $$ The measure is therefore zero. $\square$

The theorem does not say that the union is empty, or even finite. It says that its measure remains zero. This distinction matters in Lebesgue measure, where a null set can contain infinitely many points and can be dense. Countable subadditivity controls the measure of the union without requiring a description of its points.

Worked Example: The Rational Points in the Unit Interval

Enumerate the rational numbers in $[0,1]$ as $q_1,q_2,\ldots$. Each singleton $\{q_n\}$ is Borel measurable and has Lebesgue measure zero. Since $$ \mathbb{Q}\cap[0,1]=\bigcup_{n=1}^{\infty}\{q_n\}, $$ countable subadditivity gives $$ 0\leq\mu(\mathbb{Q}\cap[0,1]) \leq\sum_{n=1}^{\infty}\mu(\{q_n\}) =\sum_{n=1}^{\infty}0=0. $$ Thus $\mathbb{Q}\cap[0,1]$ has measure zero, although it is a countably infinite set.

Summable Measures and Infinitely Many Occurrences

There is a stronger use of countable subadditivity. Given a sequence of sets, consider the points that belong to infinitely many of them. This set is the limit superior, $$ \limsup_{n\to\infty} E_n = \bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}E_n. $$ A point belongs to this intersection exactly when, no matter how far along the sequence one starts, it belongs to at least one set in that tail. Equivalently, it belongs to infinitely many of the $E_n$. The measurability of such set limits was established earlier in the course; in particular, this limit superior is measurable when every $E_n$ is measurable.

Theorem (First Borel–Cantelli Lemma): Let $E_n\in\mathcal{F}$ for every $n$. If $$ \sum_{n=1}^{\infty}\mu(E_n)<\infty, $$ then $$ \mu\left(\limsup_{n\to\infty}E_n\right)=0. $$

Proof. Fix a positive integer $N$. By the definition of the limit superior, $$ \limsup_{n\to\infty}E_n \subseteq\bigcup_{n=N}^{\infty}E_n. $$ Monotonicity and countable subadditivity therefore give $$ 0\leq\mu\left(\limsup_{n\to\infty}E_n\right) \leq\mu\left(\bigcup_{n=N}^{\infty}E_n\right) \leq\sum_{n=N}^{\infty}\mu(E_n). $$ This inequality holds for every $N$. Since the series of nonnegative terms $\sum_{n=1}^{\infty}\mu(E_n)$ converges, its tails $\sum_{n=N}^{\infty}\mu(E_n)$ tend to zero as $N$ tends to infinity. The fixed nonnegative number $\mu(\limsup E_n)$ is bounded above by every one of these tails, so it must be zero. $\square$

No independence assumption appears in this lemma. The conclusion follows solely from summability and the measure bound for each tail union. The hypothesis is sufficient, not necessary: a sequence can have a divergent sum of measures while its limit superior is still null.

Worked Example: A Sequence of Small Intervals

On $\mathbb{R}$ with Lebesgue measure, define, for $n\geq1$, $$ E_n=\bigcup_{k=0}^{2^n-1} \left[\frac{k}{2^n},\frac{k}{2^n}+2^{-3n}\right]. $$ For a fixed $n$, the intervals are disjoint: their starting points are spaced $2^{-n}$ apart, while each interval has length $2^{-3n}<2^{-n}$. There are $2^n$ intervals, each of length $2^{-3n}$, so $$ \mu(E_n)=2^n\,2^{-3n}=2^{-2n}=4^{-n}. $$ Consequently, $$ \sum_{n=1}^{\infty}\mu(E_n) =\sum_{n=1}^{\infty}4^{-n} =\frac{1/4}{1-1/4} =\frac{1}{3}<\infty. $$ The First Borel–Cantelli Lemma implies that the set of points lying in infinitely many $E_n$ has measure zero. For instance, $0$ belongs to every $E_n$, so the limit superior is not empty; the theorem says its measure is nevertheless zero.

What a Divergent Sum Does Not Tell Us

The First Borel–Cantelli Lemma has a one-way conclusion. If the sum of the measures is finite, the limit superior is null. If the sum diverges, this argument provides no conclusion about the measure of the limit superior. It would be an error to reverse the implication.

Worked Example: Divergent Total Measure but a Null Limit Superior

On $\mathbb{R}$ with Lebesgue measure, let $E_n=[0,1/n]$. Then $$ \mu(E_n)=\frac{1}{n}, $$ and the series $\sum_{n=1}^{\infty}1/n$ diverges. For example, each block of terms from $2^j+1$ through $2^{j+1}$ has $2^j$ terms, each at least $1/2^{j+1}$, so that block has sum at least $1/2$. Infinitely many such blocks force divergence.

Nevertheless, a point $x>0$ belongs to only finitely many $E_n$: once $n>1/x$, we have $1/n<x$, so $x\notin[0,1/n]$. The point $0$ belongs to every $E_n$. Hence $$ \limsup_{n\to\infty}E_n=\{0\}, \qquad \mu\left(\limsup_{n\to\infty}E_n\right)=0. $$ Thus divergence of the sum does not imply that the limit superior has positive measure.

Countable subadditivity is an upper-bound principle. It permits a useful conclusion when the total available bound is finite, as in the First Borel–Cantelli Lemma, but it cannot by itself supply a converse. To obtain a positive-measure conclusion from a divergent sum, additional hypotheses or a different argument would be needed.

A practical approach is to identify the union or tail union relevant to the question, estimate each set’s measure, and then sum those estimates. If the tail sums tend to zero, the points appearing infinitely often form a null set. If the total sum is infinite, do not infer the opposite conclusion: return to the actual sets and examine their overlaps or structure.

Check Your Understanding

Use countable subadditivity and the results proved above to answer the following questions.

What inequality does countable subadditivity give for $\mu(\bigcup_{n=1}^{\infty}A_n)$? Must the sets $A_n$ be disjoint?
Why does a countable union of measurable null sets have measure zero?
What does membership in $\limsup_{n\to\infty}E_n$ mean in terms of how often a point belongs to the sets $E_n$?
Which hypothesis in the First Borel–Cantelli Lemma ensures that the tail sums tend to zero?
Does divergence of $\sum_n\mu(E_n)$ imply that $\limsup E_n$ has positive measure? Give the example from this tutorial that supports your answer.

What You'll Learn