Disjointification for Countable Additivity Arguments

Why Disjointification Helps

A measure is additive on disjoint measurable sets. A cover, however, usually contains overlaps: adding the measures of its members may count the same point more than once. The standard repair is to keep the first set as it is, remove it from the second, remove the first two from the third, and continue. The resulting pieces are disjoint, have exactly the same union as the original sets, and can therefore be used with countable additivity.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space. No finiteness, sigma-finiteness, or completeness assumption is imposed; measures may take the value $+\infty$. All sets considered are in $\mathcal{M}$, and all set equalities below hold exactly, not merely almost everywhere. The only measure properties needed are countable additivity on pairwise disjoint measurable sets and its consequence, monotonicity: if $A\subseteq B$, then $\mu(A)\le\mu(B)$. Indeed, $B=A\cup(B\setminus A)$ is a disjoint union, so $\mu(B)=\mu(A)+\mu(B\setminus A)\ge\mu(A)$.

Definition (successive-difference disjointification): For a sequence $(E_n)_{n\ge1}$ of measurable sets, define $$ D_1=E_1,\qquad D_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\quad(n\ge2). $$ The sets $D_n$ are the new parts contributed by $E_n$ after all earlier sets have been counted.

Each $D_n$ is measurable: a finite union of measurable sets is measurable, and the difference of measurable sets is measurable. The construction respects the given order. A point in several of the $E_n$ is assigned to the first one containing it, rather than being counted in every one.

The Decomposition and Its Measure Consequences

Theorem (successive-difference decomposition): For any sequence $(E_n)_{n\ge1}$ in $\mathcal{M}$, its associated sets $D_n$ are pairwise disjoint and $$ \bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}E_n. $$ Consequently, for every measure $\mu$ on $(X,\mathcal{M})$, $$ \mu\left(\bigcup_{n=1}^{\infty}E_n\right)=\sum_{n=1}^{\infty}\mu(D_n) \le \sum_{n=1}^{\infty}\mu(E_n), $$ where sums and measures may equal $+\infty$.

Proof. If $i<j$, then $D_i\subseteq E_i$, while $D_j$ excludes every point in $\bigcup_{k<j}E_k$, including $E_i$. Thus $D_i\cap D_j=\varnothing$. To prove equality of the unions, first $D_n\subseteq E_n$ for every $n$, so $\bigcup_nD_n\subseteq\bigcup_nE_n$. Conversely, let $x\in\bigcup_nE_n$. There is a least positive integer $j$ such that $x\in E_j$. Then $x\notin E_k$ for every $k<j$, and hence $x\in E_j\setminus\bigcup_{k<j}E_k=D_j$. This proves the reverse inclusion. Countable additivity on the pairwise disjoint measurable sets $D_n$ gives the equality of measures and the series. Finally, $D_n\subseteq E_n$ implies $\mu(D_n)\le\mu(E_n)$ by monotonicity, so summing these inequalities gives the stated bound, including when a sum is infinite. $\square$

The inequality is countable subadditivity. A useful cover version follows immediately: if $A\in\mathcal{M}$ and $A\subseteq\bigcup_n E_n$, then $$ \mu(A)\le\sum_{n=1}^{\infty}\mu(E_n). $$ Indeed, monotonicity gives $\mu(A)\le\mu(\bigcup_nE_n)$, and the theorem bounds the latter by the sum. The hypothesis that the cover contains $A$ matters; no assumption that $A$ equals the union is required.

Proposition (disjoint pieces inside a covered set): If $A,E_1,E_2,\ldots\in\mathcal{M}$ and $A\subseteq\bigcup_nE_n$, define $F_1=A\cap E_1$ and, for $n\ge2$, define $$ F_n=(A\cap E_n)\setminus\bigcup_{k=1}^{n-1}E_k. $$ Then the $F_n$ are pairwise disjoint, their union is $A$, and $$ \mu(A)=\sum_{n=1}^{\infty}\mu(F_n)\le\sum_{n=1}^{\infty}\mu(E_n). $$

Proof. Each $F_n$ is measurable and is contained in $D_n$, so the pairwise-disjoint argument in the theorem shows that the $F_n$ are pairwise disjoint. If $x\in A$, the cover assumption places $x$ in at least one $E_n$. At the least such index $j$, $x\in A\cap E_j$ and $x\notin E_k$ for all $k<j$, so $x\in F_j$. Conversely, each $F_n\subseteq A$, proving $\bigcup_nF_n=A$. Countable additivity gives $\mu(A)=\sum_n\mu(F_n)$. Since $F_n\subseteq E_n$, monotonicity yields each termwise bound and hence the series inequality. $\square$

Worked Applications

Worked Example: Removing Overlaps from Three Intervals

On $(\mathbb{R},\mathcal{L},m)$, where $m$ is Lebesgue measure, take $E_1=[0,2]$, $E_2=[1,3]$, and $E_3=[2,4]$. Find the disjoint pieces and compute the measure of the union.

The first piece is $D_1=[0,2]$. The second removes all points already in $E_1$: $$ D_2=[1,3]\setminus[0,2]=(2,3]. $$ The earlier union is $E_1\cup E_2=[0,3]$, so $$ D_3=[2,4]\setminus[0,3]=(3,4]. $$ The endpoints follow from the set differences: $2$ is already in $E_1$, and $3$ is already in $E_2$. Thus $D_1,D_2,D_3$ are disjoint and their union is $[0,4]$. Their measures are $2$, $3-2=1$, and $4-3=1$, respectively. Countable additivity (here a finite disjoint sum) gives $m([0,4])=2+1+1=4$. By contrast, the original interval measures sum to $2+2+2=6$, which counts overlaps more than once. Check: the union is the interval $[0,4]$, whose Lebesgue measure is its length $4-0=4$, agreeing with the disjoint-piece calculation.

Worked Example: A Countable Cover with Redundant Sets

On $(\mathbb{R},\mathcal{L},m)$, define $$ E_1=[0,1/2],\qquad E_2=[1/4,3/4],\qquad E_3=(1/2,3/4]\cup\{1\}, $$ and, for $n\ge4$, let $E_n=(1-2^{-(n-2)},\,1-2^{-(n-1)}]$. Compute the union measure and compare it with the sum of the cover measures.

The disjoint pieces are $D_1=[0,1/2]$, $D_2=(1/2,3/4]$, and $D_3=\{1\}$: the interval part of $E_3$ was already covered by $E_2$. For $n\ge4$, the interval $E_n$ lies to the right of $3/4$, does not overlap earlier such intervals, and is therefore $D_n=E_n$. These pieces cover $[0,1]$. Their measures sum to $$ m([0,1/2]) + m((1/2,3/4]) + m(\{1\})+\sum_{n=4}^{\infty}m(E_n) =\frac12+\frac14+0+\sum_{n=4}^{\infty}2^{-(n-1)}. $$ The tail is a geometric series with first term $2^{-3}=1/8$ and ratio $1/2$, so it equals $(1/8)/(1-1/2)=1/4$. Thus the disjoint-piece sum is $1/2+1/4+0+1/4=1$. Independently, the union $[0,1]$ has Lebesgue measure $1-0=1$.

For the original cover, $m(E_1)=1/2$, $m(E_2)=3/4-1/4=1/2$, and $m(E_3)=3/4-1/2=1/4$, since a singleton has measure zero. For $n\ge4$, the interval length is $(1-2^{-(n-1)})-(1-2^{-(n-2)})=2^{-(n-1)}$, so the tail sum is $1/4$, as computed above. Therefore $\sum_nm(E_n)=1/2+1/2+1/4+1/4=3/2$. The cover inequality $1\le3/2$ holds; its excess reflects the overlap. In particular, $E_3$'s interval was already counted in $E_2$.

Worked Example: A Cover in Counting Measure

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}$ be its power set, and let $\mu$ be counting measure, so $\mu(E)$ is the number of elements of $E$, or $+\infty$ if $E$ is infinite. Set $A=\{1,2,3,4,5\}$, $E_1=\{1,2,3\}$, $E_2=\{2,3,4\}$, $E_3=\{4,5\}$, and $E_n=\varnothing$ for $n\ge4$. Find the disjoint pieces inside the cover and verify the measure bound.

The sets $E_1,E_2,E_3$ cover $A$. The pieces are $F_1=A\cap E_1=\{1,2,3\}$, $F_2=(A\cap E_2)\setminus E_1=\{4\}$, and $F_3=(A\cap E_3)\setminus(E_1\cup E_2)=\{5\}$; for $n\ge4$, $F_n=\varnothing$. Their cardinalities are $3,1,1,0,\ldots$, so $\mu(A)=3+1+1=5$. The cover-set measures are $3,3,2,0,\ldots$, whose sum is $8$. Hence $5\le8$. Directly, $A$ has five elements, confirming the disjoint-piece total, while the two shared points $2,3$ and the shared point $4$ account for the difference between the two totals.

Common Pitfalls and How to Avoid Them

Adding the cover measures as if the sets were disjoint. In the interval example, $m(E_1)+m(E_2)+m(E_3)=6$, but $m(E_1\cup E_2\cup E_3)=4$. The incorrect calculation is to assert equality between these quantities without checking disjointness. The correction is either to use countable subadditivity, which gives an upper bound, or to disjointify first and add the measures of the pieces.

Subtracting measures instead of subtracting sets. Set difference is always defined, but a difference of measures need not be. For example, on $(\mathbb{R},\mathcal{L},m)$, let $E_1=(0,\infty)$ and $E_2=(1,2)$. Then $D_2=E_2\setminus E_1=\varnothing$, so $m(D_2)=0$. The expression $m(E_2)-m(E_1)=1-\infty$ is not a valid real-number subtraction and does not compute $m(D_2)$. The formula $m(B\setminus A)=m(B)-m(A)$ is justified when $A\subseteq B$ and $m(A)<\infty$, by countable additivity applied to $B=A\cup(B\setminus A)$. Without those conditions, use the set difference itself and the measure properties that apply.

Disjointification is useful whenever a proof begins with a countable cover but needs an additive decomposition. Its exact union identity prevents gaps; its disjointness prevents double-counting; and its measure inequality remains valid even when some sets or sums have infinite measure. The order may change the individual pieces, but it cannot change their union or the resulting sum of their measures.

Check Your Understanding

For each question, keep the distinction between a set identity and an identity of measures in view.

For $E_1=[0,1]$, $E_2=[1/2,2]$, and $E_3=[3/2,3]$ in $(\mathbb{R},\mathcal{L},m)$, find all three disjoint pieces and compute the measure of their union.
Prove that if $A\subseteq\bigcup_nE_n$ with all sets measurable, then $A$ can be written as a countable disjoint union of measurable subsets $F_n\subseteq E_n$.
Give an example where $\sum_n\mu(E_n)$ is strictly larger than $\mu(\bigcup_nE_n)$, and identify which points are counted more than once in the original sum.
Why is the expression $+\infty-(+\infty)$ not a valid way to calculate the measure of a set difference?
Does changing the order of the sets change the disjoint pieces? Which properties of the pieces remain unchanged?

Disjointification for Countable Additivity Arguments

What You'll Learn