Countable Subadditivity Of Measures

From Disjoint Additivity to an Upper Bound

Countable additivity gives an exact formula for the measure of a countable union when the sets are pairwise disjoint. For sets that overlap, adding their measures counts some parts more than once. The general principle is to remove from each set the points already included in earlier sets. The resulting pieces are disjoint, have the same union as the original family, and each piece is contained in its corresponding original set. These facts yield countable subadditivity.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space: \(\mathcal{M}\) is a sigma-algebra on \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a countably additive measure with \(\mu(\varnothing)=0\). The sets in the statements below belong to \(\mathcal{M}\). Neither finiteness nor sigma-finiteness of \(\mu\) is assumed. Sums of nonnegative extended real numbers are interpreted as the limits of their partial sums, so \(+\infty\) is allowed.

Proposition (Monotonicity): If \(E,F\in\mathcal{M}\) and \(E\subseteq F\), then \(\mu(E)\le\mu(F)\), including when \(\mu(F)=+\infty\).

Proof. The sets \(E\) and \(F\setminus E\) are measurable, disjoint, and have union \(F\). Finite additivity, which follows from countable additivity by appending empty sets, gives \(\mu(F)=\mu(E)+\mu(F\setminus E)\). Since \(\mu(F\setminus E)\ge0\), the extended nonnegative sum is at least \(\mu(E)\). This proves the claim whether the measures are finite or infinite. \(\square\)

Disjointifying a Sequence

Given \(A_1,A_2,\ldots\in\mathcal{M}\), define the successive differences by retaining all of \(A_1\), then removing \(A_1\) from \(A_2\), then removing everything already used from \(A_3\), and continuing in this way.

D_1=A_1,\qquad D_n=A_n\setminus\bigcup_{k=1}^{n-1}A_k\quad(n\ge2).

Lemma (Disjointification): Each \(D_n\) is measurable, the sets \(D_n\) are pairwise disjoint, \(D_n\subseteq A_n\), and \(\bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}A_n\).

Proof. For each \(n\), the union \(\bigcup_{k=1}^{n-1}A_k\) is a finite union of measurable sets, so it is measurable. Closure of \(\mathcal{M}\) under set differences then shows that \(D_n\in\mathcal{M}\); also \(D_1=A_1\in\mathcal{M}\). The definition directly gives \(D_n\subseteq A_n\).

To show disjointness, take indices \(i<j\). Every point in \(D_i\) belongs to \(A_i\), and \(A_i\subseteq\bigcup_{k=1}^{j-1}A_k\). But \(D_j\) contains only points of \(A_j\) outside that earlier union. Thus \(D_i\cap D_j=\varnothing\).

Since \(D_n\subseteq A_n\), we have \(\bigcup_nD_n\subseteq\bigcup_nA_n\). Conversely, if \(x\in\bigcup_nA_n\), the set of indices \(n\) for which \(x\in A_n\) is a nonempty subset of the positive integers, so it has a least element \(j\). Then \(x\in A_j\) and \(x\notin A_k\) for every \(k<j\), which means \(x\in D_j\). Therefore \(\bigcup_nA_n\subseteq\bigcup_nD_n\), proving equality. \(\square\)

Countable Subadditivity

Theorem (Countable Subadditivity): Let \((X,\mathcal{M},\mu)\) be any measure space and let \(A_n\in\mathcal{M}\) for every positive integer \(n\). Then \[ \mu\left(\bigcup_{n=1}^{\infty}A_n\right)\le\sum_{n=1}^{\infty}\mu(A_n). \] Both sides may be \(+\infty\). No assumption that the sets are disjoint, or that the measure space is finite or sigma-finite, is needed.

Proof. Form the measurable, pairwise disjoint sets \(D_n\) from the lemma. They have the same union as the \(A_n\), so countable additivity gives \[ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) =\mu\left(\bigcup_{n=1}^{\infty}D_n\right) =\sum_{n=1}^{\infty}\mu(D_n). \] By \(D_n\subseteq A_n\) and monotonicity, \(\mu(D_n)\le\mu(A_n)\) for each \(n\). For every positive integer \(N\), summing these nonnegative inequalities gives \[ \sum_{n=1}^{N}\mu(D_n)\le\sum_{n=1}^{N}\mu(A_n). \] The infinite sums are the limits of these increasing partial sums. Taking their limits preserves the inequality, even if either limit is \(+\infty\). Combining the two displays proves countable subadditivity. \(\square\)

The proof separates two roles. Countable additivity applies to the disjoint pieces and supplies an equality. Containment and monotonicity compare each piece to its original set and supply the inequality. If the original sets are already pairwise disjoint, the disjointification leaves them unchanged, and countable additivity gives equality.

Corollary (Finite Subadditivity): For a positive integer \(N\) and measurable sets \(A_1,\ldots,A_N\) in any measure space \((X,\mathcal{M},\mu)\), \[ \mu\left(\bigcup_{n=1}^{N}A_n\right)\le\sum_{n=1}^{N}\mu(A_n). \]

Proof. Set \(A_n=\varnothing\) for every \(n>N\). The countable union is then \(\bigcup_{n=1}^{N}A_n\), and the infinite sum on the right has only the first \(N\) possibly nonzero terms because \(\mu(\varnothing)=0\). Apply countable subadditivity. \(\square\)

Worked Applications

Worked Example: Overlapping Intervals

In \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, let \(A_1=[0,2]\), \(A_2=[1,3]\), and \(A_3=[2,4]\). Compute the measure of their union and compare it with the sum of their individual measures.

The union is \([0,4]\), so \(m(A_1\cup A_2\cup A_3)=4-0=4\). Each original interval has length \(2\), giving \[ m(A_1)+m(A_2)+m(A_3)=2+2+2=6. \] Thus the subadditivity bound reads \(4\le6\).

To see the disjointification explicitly, \(D_1=[0,2]\). Removing \(A_1\) from \(A_2\) gives \(D_2=(2,3]\). Removing \(A_1\cup A_2=[0,3]\) from \(A_3\) gives \(D_3=(3,4]\). These pieces are disjoint, their union is \([0,4]\), and their measures are \(2\), \(1\), and \(1\), respectively. Their sum is \(2+1+1=4\), exactly the measure of the union. Check: the original sets each have length \(2\), whereas the disjoint pieces have lengths \(2\), \(3-2=1\), and \(4-3=1\); both calculations give the asserted totals.

Worked Example: Counting Measure on Overlapping Finite Sets

Let \(X=\{1,2,3,4,5,6\}\), \(\mathcal{M}=\mathcal{P}(X)\), and \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\). Take \(A_1=\{1,2,3\}\), \(A_2=\{3,4\}\), and \(A_3=\{2,5,6\}\).

The union contains \(1,2,3,4,5,6\), so its measure is \(6\). The individual measures are \(\mu(A_1)=3\), \(\mu(A_2)=2\), and \(\mu(A_3)=3\); their sum is \(3+2+3=8\). Thus \(6\le8\).

The successive differences are \(D_1=\{1,2,3\}\), \(D_2=\{4\}\), and \(D_3=\{5,6\}\). Their measures sum to \(3+1+2=6\), matching the measure of the union. The points \(2\) and \(3\) are not counted again after their first appearance. Check: the disjoint pieces contain \(3+1+2=6\) points altogether, and their union is exactly \(\{1,2,3,4,5,6\}\).

Worked Example: A Countable Disjoint Cover of an Interval

In \((\mathbb{R},\mathcal{L},m)\), define, for each \(n\ge1\), \[ A_n=\left(1-2^{-(n-1)},\,1-2^{-n}\right]. \] Find the union and compare its measure with the sum of the measures.

The first intervals are \(A_1=(0,\tfrac12]\), \(A_2=(\tfrac12,\tfrac34]\), and \(A_3=(\tfrac34,\tfrac78]\). In general, the right endpoint of \(A_n\) is the left endpoint of \(A_{n+1}\); it belongs to \(A_n\) and not to \(A_{n+1}\). Hence the intervals are pairwise disjoint. Their endpoints approach \(1\), and every \(x\in(0,1)\) lies in one of these intervals: choose the interval whose consecutive dyadic endpoints enclose \(x\), assigning a shared endpoint to the interval on its left. Thus \(\bigcup_{n=1}^{\infty}A_n=(0,1)\).

The length of \(A_n\) is \[ \left(1-2^{-n}\right)-\left(1-2^{-(n-1)}\right)=2^{-(n-1)}-2^{-n}=2^{-n}. \] Therefore \(\sum_{n=1}^{\infty}m(A_n)=\sum_{n=1}^{\infty}2^{-n}=1\), by the geometric series formula. Also \(m((0,1))=1-0=1\), so equality holds. Check: the first three lengths are \(\tfrac12,\tfrac14,\tfrac18\), with partial sum \(\tfrac78\); the remaining geometric tail is \(\tfrac18\), giving total \(1\), in agreement with the length of the union.

A Common Overcounting Error

A frequent mistake is to write \(\mu(A\cup B)=\mu(A)+\mu(B)\) without checking that \(A\) and \(B\) are disjoint. In the interval example, the incorrect calculation would give \(m([0,2]\cup[1,3])=2+2=4\). But the union is \([0,3]\), whose measure is \(3\), not \(4\). The overlap \([1,2]\) has been counted twice in the proposed sum.

The correction is to keep only the part of the second set not already in the first: \(A\) and \(B\setminus A=(2,3]\) are disjoint and have union \(A\cup B\). Their measures are \(2\) and \(1\), so countable additivity gives \(m(A\cup B)=2+1=3\). For a sequence, the sets \(D_n\) perform this correction at every stage. Countable subadditivity is useful precisely because it provides a valid upper bound even when the original sets overlap in complicated ways or there are infinitely many of them.

Check Your Understanding

Use disjointification and the hypotheses of countable subadditivity to check each answer.

For \(A_1=[0,1]\) and \(A_2=[\tfrac12,\tfrac32]\) in \((\mathbb{R},\mathcal{L},m)\), find the union measure and the sum of the individual measures. Which one is larger?
For a sequence \((A_n)\) of measurable sets, define \(D_n\) using successive differences. Explain why \(D_i\cap D_j=\varnothing\) whenever \(i<j\).
Why does the proof remain valid if \(\sum_n\mu(A_n)=+\infty\)? Identify the partial-sum inequality used in the proof.
Give a condition under which countable subadditivity becomes equality, and state the measure axiom that gives that equality.
For \(A_1=\{a,b\}\), \(A_2=\{b,c\}\) under counting measure, compute the measure of the union and the sum of the individual measures. Describe the disjoint successive differences.