Countable Unions via Disjointization

From Overlapping Sets to Disjoint Pieces

Closure under countable disjoint unions does not immediately give closure under arbitrary countable unions: the sets in an arbitrary family may overlap. The key is to assign every point in the union to its first set in the sequence. This produces disjoint pieces without changing the union. Once the pieces are shown to be measurable, the result for disjoint unions applies.

Work on a set $X$ equipped with an outer measure $\mu^*:\mathcal{P}(X)\to[0,\infty]$. Here $\mathcal{P}(X)$ is the power set of $X$. An outer measure is monotone, vanishes on the empty set, and is countably subadditive. We do not assume $X$ has a measure space structure or that $\mu^*(X)$ is finite.

Definition: A set $E\subseteq X$ is Caratheodory measurable with respect to $\mu^*$ if, for every $T\subseteq X$, $$ \mu^*(T)=\mu^*(T\cap E)+\mu^*(T\setminus E). $$ The equality is in $[0,\infty]$. Write $\mathcal{M}_{\mu^*}$ for the collection of such sets.

The finite closure results established earlier in the material on Caratheodory measurability show that $\mathcal{M}_{\mu^*}$ contains $X$, is closed under complements, and is closed under finite unions. In particular, if $A_1,\ldots,A_{n-1}$ are measurable, then their finite union is measurable, and its complement is measurable. Intersecting that complement with a measurable set also gives a measurable set, since an intersection is the complement of a finite union of complements.

The Disjointization Construction

Given any sequence $(A_n)_{n\ge1}$ of subsets of $X$, define its disjointized pieces by $$ D_1=A_1,\qquad D_n=A_n\setminus\bigcup_{j=1}^{n-1}A_j\quad(n\ge2). $$ Thus $D_n$ consists of the points that belong to $A_n$ but to none of the earlier sets. The definition uses only finite unions at each stage; it does not require forming or subtracting an infinite union.

Lemma: For every sequence $(A_n)_{n\ge1}$ of subsets of $X$, the sets $D_n$ defined above are pairwise disjoint and $$ \bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}A_n. $$ If every $A_n$ is Caratheodory measurable, then every $D_n$ is Caratheodory measurable.

Proof. First take indices $i<j$. By definition, $D_i\subseteq A_i$. But $D_j$ excludes every point in $\bigcup_{k<j}A_k$, which includes $A_i$. Therefore $D_i\cap D_j=\varnothing$, proving pairwise disjointness.

Since $D_n\subseteq A_n$, we have $\bigcup_nD_n\subseteq\bigcup_nA_n$. For the reverse inclusion, take $x\in\bigcup_nA_n$. The set of indices $n\ge1$ for which $x\in A_n$ is nonempty, so it has a least element $k$. Then $x\in A_k$, and $x\notin A_j$ for every $j<k$. Hence $x\in D_k$. This proves $\bigcup_nA_n\subseteq\bigcup_nD_n$, and the two unions are equal.

Now suppose every $A_n$ is Caratheodory measurable. The set $D_1=A_1$ is measurable. For $n\ge2$, the finite union $F_{n-1}=\bigcup_{j=1}^{n-1}A_j$ is measurable by finite-union closure. Its complement $F_{n-1}^c$ is measurable, and $$ D_n=A_n\cap F_{n-1}^c. $$ The right side is a finite intersection of measurable sets, hence is measurable by complement and finite-union closure. Thus every $D_n$ is measurable. $\square$

Closure Under Arbitrary Countable Unions

The earlier result on countable disjoint unions of measurable sets states that, for an outer measure $\mu^*$ on $X$, a countable pairwise disjoint family of Caratheodory measurable sets has a Caratheodory measurable union. This result requires disjointness; the construction above supplies it for an arbitrary family.

Theorem: If $(A_n)_{n\ge1}$ is any sequence of Caratheodory measurable subsets of $X$, then $\bigcup_{n=1}^{\infty}A_n$ is Caratheodory measurable. No finiteness assumption on $\mu^*(X)$ is required.

Proof. Define $D_n=A_n\setminus\bigcup_{j<n}A_j$, with $D_1=A_1$. By the lemma, the $D_n$ are pairwise disjoint, each is Caratheodory measurable, and $\bigcup_nD_n=\bigcup_nA_n$. Apply the countable disjoint-union result to $(D_n)$. It follows that $\bigcup_nD_n$, and therefore $\bigcup_nA_n$, is Caratheodory measurable. $\square$

The theorem establishes the countable-union part of the closure properties needed for a sigma-algebra. It does not assert countable additivity of the outer measure on these sets; that is a distinct claim, requiring its own argument.

Worked Example: Disjointizing a Finite Family with Overlap

Let $X=\{1,2,3,4,5,6,7,8\}$, and define $$ A_1=\{1,2,5\},\quad A_2=\{2,3,5,7\},\quad A_3=\{1,4,7,8\},\quad A_n=\varnothing\ (n\ge4). $$ For a concrete outer measure, take counting measure $\mu^*(E)=|E|$ on $X$. Every subset of $X$ is Caratheodory measurable: for every $T,E\subseteq X$, the two sets $T\cap E$ and $T\setminus E$ partition $T$, so $|T|=|T\cap E|+|T\setminus E|$.

The first piece is $D_1=A_1=\{1,2,5\}$. For the second, remove the elements of $A_1$ from $A_2$: $$ D_2=A_2\setminus A_1=\{2,3,5,7\}\setminus\{1,2,5\}=\{3,7\}. $$ The union of the first two original sets is $A_1\cup A_2=\{1,2,3,5,7\}$. Therefore $$ D_3=A_3\setminus(A_1\cup A_2) =\{1,4,7,8\}\setminus\{1,2,3,5,7\} =\{4,8\}. $$ For $n\ge4$, $D_n=\varnothing$, since $A_n=\varnothing$.

The pieces $\{1,2,5\}$, $\{3,7\}$, and $\{4,8\}$ are pairwise disjoint. Their union is $\{1,2,3,4,5,7,8\}$, which is also $A_1\cup A_2\cup A_3$; the point $6$ belongs to neither union. As a check using counting measure, the piece sizes sum to $3+2+2=7$, and the original union also has seven elements.

Worked Example: An Infinite Overlapping Sequence

Take $X=\mathbb{N}=\{1,2,\ldots\}$ with counting outer measure, and let $A_n=\{n,n+1\}$ for each $n\ge1$. Every subset of $\mathbb{N}$ is Caratheodory measurable: for any $T,E\subseteq\mathbb{N}$, the disjoint partition $T=(T\cap E)\cup(T\setminus E)$ gives the counting-measure splitting identity, including when one or more counts are infinite.

We have $D_1=A_1=\{1,2\}$. For $n\ge2$, the earlier sets have union $$ \bigcup_{j=1}^{n-1}A_j=\{1,2,\ldots,n\}. $$ Indeed, the first set gives $1,2$, and each subsequent set $A_j=\{j,j+1\}$, up to $j=n-1$, adds the next integer through $n$. Thus $$ D_n=\{n,n+1\}\setminus\{1,2,\ldots,n\}=\{n+1\}\quad(n\ge2). $$ The strict membership condition matters here: $n$ is removed because it belongs to an earlier set, while $n+1$ does not.

Consequently the disjoint pieces are $\{1,2\},\{3\},\{4\},\{5\},\ldots$. Their union is all of $\mathbb{N}$, as is the union of the original sets. Under counting outer measure both unions have outer measure $+\infty$, since each contains infinitely many points. The disjointization changes the representation, not the union.

Worked Example: Repeated Sets and Empty Pieces

Let $X=\{a,b,c\}$, use counting outer measure, and set $A_n=\{a,c\}$ for every $n\ge1$. All subsets are measurable by the partition argument in the preceding example. The first piece is $D_1=A_1=\{a,c\}$. For every $n\ge2$, the earlier union is already $\{a,c\}$, so $$ D_n=A_n\setminus\bigcup_{j<n}A_j=\{a,c\}\setminus\{a,c\}=\varnothing. $$ The pieces are pairwise disjoint, and their union is $\{a,c\}$, exactly the union of all the $A_n$. The counting measure of the original union is $2$; the disjoint pieces contribute $2$ from $D_1$ and $0$ from each later piece. This also shows why empty pieces are permitted and must not cause an exception in the construction.

A Common Mistake: Treating Overlapping Sets as Disjoint

It is not valid to apply the disjoint-union result directly to an arbitrary family. For example, in $X=\{1,2,3\}$ with counting measure, let $A_1=\{1,2\}$ and $A_2=\{2,3\}$. Both sets are measurable, but they overlap at $2$. Their measures sum to $2+2=4$, whereas the measure of their union is $|\{1,2,3\}|=3$. The overlap has been counted twice, so the family is not disjoint and the disjoint-union result does not apply.

The correction is to replace the family by $D_1=A_1=\{1,2\}$ and $D_2=A_2\setminus A_1=\{3\}$. These pieces are disjoint, have the same union as the original sets, and have measures $2$ and $1$. This calculation illustrates the purpose of disjointization. The closure proof itself needs only that the pieces are measurable and disjoint; it does not infer countable additivity from the example.

When using the construction, check three points: each piece excludes all earlier sets, so the pieces are disjoint; each point in the original union has a least index of membership, so it appears in a piece; and each piece is formed using only finite unions and complements, so finite closure preserves measurability. These checks turn the disjoint-union theorem into closure under arbitrary countable unions.

Check Your Understanding

Use the disjointization formula and state where finite closure or disjoint-union closure is needed.

For $A_1=\{a,b\}$, $A_2=\{b,c\}$, and $A_3=\{a,d\}$, compute $D_1,D_2,D_3$ and their union.
Why does the least index $k$ for which $x\in A_k$ guarantee that $x\in D_k$?
If all $A_n$ are Caratheodory measurable, explain why $A_n\setminus\bigcup_{j<n}A_j$ is measurable without assuming closure under countable unions.
For $A_n=\{n,n+1\}$ on $\mathbb{N}$, compute $D_1,D_2,D_3$ and identify the general form of $D_n$.
Give an example of two measurable sets whose overlap makes it invalid to apply the countable disjoint-union result directly.

Countable Unions via Disjointization

What You'll Learn