Why Disjointize a Countable Family?
Overlapping sets are awkward to handle one at a time: a point that belongs to several of them may be counted repeatedly. A useful set-theoretic rearrangement assigns each point to the first set in the family that contains it. This produces disjoint pieces without changing the union. The key is that the construction of each piece uses only finitely many earlier sets, even when the original family is countable.
Throughout, let \(X\) be an arbitrary set and let \(\mu^*\) be an outer measure on \(X\), possibly taking the value \(+\infty\). Write \(\mathcal C\) for the Carathéodory measurable sets: \(E\in\mathcal C\) means that for every \(T\subseteq X\), \(\mu^*(T)=\mu^*(T\cap E)+\mu^*(T\cap E^c)\). No finiteness or sigma-finiteness assumption is imposed. The earlier results on complements and finite unions show that \(\mathcal C\) contains \(\varnothing\) and \(X\), is closed under complements, and is closed under finite unions. In particular, it is closed under differences, since \(E\setminus F=E\cap F^c\). We will use these finite closure properties, not countable-union closure. The latter is not being assumed here.
The First-Entry Construction
When every \(A_n\) is Carathéodory measurable, each \(D_n\) is measurable: \(D_1=A_1\), while for \(n\ge2\) the set \(\bigcup_{k=1}^{n-1}A_k\) is a finite union of measurable sets, and \(D_n\) is its difference from \(A_n\). This argument proves measurability of every individual piece; it does not require that the countable union of the \(A_n\) is already known to be measurable.
Proof. Measurability of each \(D_n\) follows from finite-union and difference closure as described above. To prove disjointness, take indices \(i<j\). By definition, \(D_i\subseteq A_i\). But \(D_j\) excludes every point in \(\bigcup_{k=1}^{j-1}A_k\), which includes \(A_i\) because \(i<j\). Hence \(D_i\cap D_j=\varnothing\).
For the equality of unions, first note that \(D_n\subseteq A_n\) for every \(n\). Therefore \(\bigcup_nD_n\subseteq\bigcup_nA_n\). Conversely, take any \(x\in\bigcup_nA_n\). The set of indices \(\{n\ge1:x\in A_n\}\) is nonempty, so it has a least element \(j\). Then \(x\in A_j\), and \(x\notin A_k\) for every \(k<j\). Consequently \(x\notin\bigcup_{k<j}A_k\), so \(x\in D_j\). This proves \(\bigcup_nA_n\subseteq\bigcup_nD_n\), and the two unions are equal. \(\square\)
What Happens at Each Finite Stage?
The construction can also be read one stage at a time. Set \(U_0=\varnothing\) and \(U_n=\bigcup_{k=1}^nA_k\) for \(n\ge1\). Each \(U_n\) is a finite union, so it is measurable when all the \(A_k\) are. The piece \(D_n\) is exactly the portion of \(U_n\) that was not already in \(U_{n-1}\).
Proof. For \(n=1\), \(U_1=A_1\) and \(U_0=\varnothing\), so \(U_1\setminus U_0=A_1=D_1\). For \(n\ge2\), \(U_n=U_{n-1}\cup A_n\). A point in \(U_n\setminus U_{n-1}\) must therefore belong to \(A_n\) and not to any \(A_k\) with \(k<n\). This is exactly the definition of \(D_n\). Thus \(D_n=U_n\setminus U_{n-1}\) for every \(n\).
We now prove the union identity by induction. It holds at \(n=1\), since \(U_1=D_1\). Suppose \(U_{n-1}=\bigcup_{k=1}^{n-1}D_k\). Since \(U_n=U_{n-1}\cup D_n\), substitution gives \(U_n=\bigcup_{k=1}^nD_k\). These pieces are pairwise disjoint by the theorem’s disjointness argument, which applies to any two indices in the finite range as well. The induction proves the identity for every \(n\). \(\square\)
Worked Example: Overlapping Subsets of a Finite Space
Let \(X=\{1,2,3,4,5\}\) with counting measure \(\#\), and let \(\mathcal C=\mathcal P(X)\), the collection of all subsets of \(X\). This is a finite measure space, and every set below is measurable. Take \(A_1=\{1,2,4\}\), \(A_2=\{2,3,4\}\), \(A_3=\{4,5\}\), and \(A_4=\{1,5\}\). Compute the first four pieces.
For the first index, \(D_1=A_1=\{1,2,4\}\). For the second, the preceding union is \(A_1=\{1,2,4\}\), so \(D_2=A_2\setminus A_1=\{3\}\). For the third, \(A_1\cup A_2=\{1,2,3,4\}\), so \(D_3=A_3\setminus(A_1\cup A_2)=\{5\}\). For the fourth, \(A_1\cup A_2\cup A_3=\{1,2,3,4,5\}=X\), so \(D_4=A_4\setminus(A_1\cup A_2\cup A_3)=\varnothing\).
Check: the pieces \(\{1,2,4\}\), \(\{3\}\), \(\{5\}\), and \(\varnothing\) are pairwise disjoint, and their union is \(\{1,2,3,4,5\}\). The original sets also have union \(X\): elements \(1,2,4\) occur in \(A_1\), element \(3\) occurs in \(A_2\), and element \(5\) occurs in \(A_3\). Counting points gives \(3+1+1+0=5=\#X\), in agreement with the union check.
Worked Example: Disjointizing Overlapping Intervals
On \(\mathbb R\) with Lebesgue measure \(m\), let \(A_1=[0,2]\), \(A_2=[1,3]\), and \(A_3=[2,4]\). These are Borel sets and hence Lebesgue measurable. The difference formula gives \(D_1=[0,2]\). Next, \(D_2=[1,3]\setminus[0,2]=(2,3]\): the point \(2\) is removed because it belongs to \(A_1\), while \(3\) remains because it does not belong to \(A_1\).
For the third piece, \(A_1\cup A_2=[0,3]\). Therefore \(D_3=[2,4]\setminus[0,3]=(3,4]\). The strict endpoint at \(3\) matters: \(3\) was already in \(A_2\), but \(4\) was not in either preceding set.
Check: \(D_1\cup D_2\cup D_3=[0,2]\cup(2,3]\cup(3,4]=[0,4]\), which is also \(A_1\cup A_2\cup A_3\). The intervals are disjoint because \(D_2\) starts strictly above \(2\), and \(D_3\) starts strictly above \(3\). Their lengths are \(2\), \(1\), and \(1\), respectively, adding to \(4\), the length of \([0,4]\).
Worked Example: A Countable Family on the Natural Numbers
Let \(X=\mathbb N=\{1,2,\ldots\}\), with its power-set sigma-algebra and counting measure, and define \(A_n=\{n,n+1\}\) for every \(n\ge1\). This is a sigma-finite measure space: \(X\) is the countable union of its singleton sets, each of counting measure \(1\). All subsets are measurable.
The first piece is \(D_1=A_1=\{1,2\}\). For \(n\ge2\), the union of the preceding sets is \(A_1\cup\cdots\cup A_{n-1}=\{1,2,\ldots,n\}\). Indeed, the formula holds for \(n=2\), when the preceding union is \(A_1=\{1,2\}\); if it holds at \(n\), adjoining \(A_n=\{n,n+1\}\) gives \(\{1,2,\ldots,n+1\}\), proving the formula at \(n+1\). Thus \(D_n=\{n,n+1\}\setminus\{1,2,\ldots,n\}=\{n+1\}\) for every \(n\ge2\).
Check: \(D_1=\{1,2\}\), \(D_2=\{3\}\), \(D_3=\{4\}\), and \(D_4=\{5\}\). In general, every integer \(j\ge3\) occurs in \(D_{j-1}\), while \(1\) and \(2\) occur in \(D_1\); hence the pieces have union \(\mathbb N\), as do the original sets. Distinct singleton pieces are disjoint, and none intersects \(D_1\).
A Common Pitfall: Subtracting Only the Previous Set
The set to remove from \(A_n\) is the union of all earlier sets, not just \(A_{n-1}\). For a concrete failure, take the measurable subsets \(A_1=\{a,b\}\), \(A_2=\{b,c\}\), and \(A_3=\{a,c,d\}\) of the finite space \(X=\{a,b,c,d\}\) with counting measure. The incorrect rule \(B_1=A_1\), \(B_n=A_n\setminus A_{n-1}\) gives \(B_1=\{a,b\}\), \(B_2=\{c\}\), and \(B_3=\{a,d\}\). Although \(B_2\) is disjoint from \(B_1\), \(B_3\cap B_1=\{a\}\), so the resulting sets are not pairwise disjoint.
The correction is to remove the full preceding union. Here \(D_1=\{a,b\}\), \(D_2=\{b,c\}\setminus\{a,b\}=\{c\}\), and \(D_3=\{a,c,d\}\setminus\{a,b,c\}=\{d\}\). These pieces are pairwise disjoint and their union is \(\{a,b,c,d\}\), exactly the union of the original three sets.
The construction is useful because it prepares a family for arguments that require disjoint sets, while keeping each piece measurable using only finite operations. At this stage, the proof establishes measurability of the pieces and equality of unions; it does not use countable additivity or claim that a countable union is measurable. Those questions require further results about countable families.
Check Your Understanding
Use the first-entry definition and justify each set operation in your answers.
- For \(A_1=\{1,3\}\), \(A_2=\{2,3,4\}\), and \(A_3=\{1,4,5\}\), compute \(D_1,D_2,D_3\).
- Explain why \(D_i\cap D_j=\varnothing\) whenever \(i<j\), using the definition of \(D_j\).
- If \(A_1\subseteq A_2\subseteq A_3\subseteq\cdots\), describe \(D_n\) for \(n\ge2\) and explain why the pieces are still a disjointization.
- Give a family for which the rule \(A_n\setminus A_{n-1}\) fails to produce pairwise disjoint sets.
- Why does proving that each \(D_n\) is measurable require only finite-union closure, even though the family has infinitely many sets?