Finite Additivity From Countable Additivity

From Countable Additivity to a Finite Sum

A measure is defined by its values on measurable sets and by a countable-additivity requirement for disjoint sequences. The corresponding formula for a finite disjoint family is familiar, but it is worth deriving carefully: the finite family is not itself an infinite sequence until we extend it, and that extension must preserve disjointness and leave the union unchanged. Empty sets do exactly this.

Throughout, let $(X,\mathcal{M},\mu)$ be an arbitrary measure space. Thus $\mathcal{M}$ is a sigma-algebra on $X$, and $\mu:\mathcal{M}\to[0,\infty]$ satisfies $\mu(\varnothing)=0$ and countable additivity: for every sequence $(E_k)_{k\ge1}$ of pairwise disjoint sets in $\mathcal{M}$, $\mu(\bigcup_{k\ge1}E_k)=\sum_{k\ge1}\mu(E_k)$. No finiteness, sigma-finiteness, or completeness assumption is needed. All sets below are measurable, and all equalities of measures are exact, not almost-everywhere statements.

Theorem (Finite additivity): If $A_1,\ldots,A_n\in\mathcal{M}$ are pairwise disjoint, where $n\ge1$, then $$ \mu\left(\bigcup_{j=1}^{n}A_j\right)=\sum_{j=1}^{n}\mu(A_j). $$ The terms and the finite sum are allowed to take the value $+\infty$. For the empty family, the union is $\varnothing$ and its measure is $0$, corresponding to the empty sum.

The theorem asserts a consequence of the measure axioms, not an additional assumption. Its disjointness condition is essential. The proof does not require the measures of the sets to be finite: if even one term is $+\infty$, the finite sum on the right is $+\infty$, and countable additivity still gives the asserted equality.

Proof by Padding with Empty Sets

Proof. First suppose $n\ge1$. Define a sequence $(E_k)_{k\ge1}$ by setting $E_k=A_k$ when $1\le k\le n$, and $E_k=\varnothing$ when $k>n$. Every $E_k$ is measurable. The sequence is pairwise disjoint: the first $n$ sets are pairwise disjoint by hypothesis, and an empty set has empty intersection with every set. Also, $\bigcup_{k\ge1}E_k=\bigcup_{j=1}^{n}A_j$, since adding empty sets does not change a union. Countable additivity therefore gives $$ \mu\left(\bigcup_{j=1}^{n}A_j\right) =\sum_{k=1}^{\infty}\mu(E_k) =\sum_{j=1}^{n}\mu(A_j)+\sum_{k=n+1}^{\infty}\mu(\varnothing) =\sum_{j=1}^{n}\mu(A_j), $$ because $\mu(\varnothing)=0$. These equalities are valid in $[0,\infty]$, including when a term $\mu(A_j)$ is infinite. If the family is empty, its union is $\varnothing$, and $\mu(\varnothing)=0$ by the measure axiom; this is the value of the empty sum. $\square$

The padding step does not assert that an empty set contributes a new part of the union. It contributes only a zero term to the series. The argument also shows precisely which assumptions are used: each set must be measurable, the original finite family must be pairwise disjoint, and countable additivity and $\mu(\varnothing)=0$ must hold.

Worked Applications

Worked Example: A Disjoint Union of Intervals

Take Lebesgue measure $m$ on $(\mathbb{R},\mathcal{L},m)$, where $\mathcal{L}$ is the Lebesgue sigma-algebra. Let $A_1=[0,1]$, $A_2=(1,3]$, and $A_3=\{5\}$. These sets are Lebesgue measurable. Check disjointness: the endpoint $1$ belongs to $A_1$, not $A_2$, and $5$ is in neither interval, so every pair has empty intersection.

The measures are $m(A_1)=1-0=1$, $m(A_2)=3-1=2$, and $m(A_3)=0$, since a singleton has Lebesgue measure zero. Their union is $[0,3]\cup\{5\}$, whose measure is $3+0=3$. Finite additivity gives $$ m(A_1\cup A_2\cup A_3)=m(A_1)+m(A_2)+m(A_3)=1+2+0=3. $$ Check: The union consists of an interval of length $3-0=3$ and one singleton of measure $0$, so direct computation again gives $3$.

Worked Example: Counting Measure on a Finite Set

Let $X=\{a,b,c,d,e\}$, let $\mathcal{M}$ be the collection of all subsets of $X$, and let $\mu$ be counting measure, so $\mu(E)$ is the number of elements of $E$. Set $A_1=\{a,c\}$, $A_2=\{b,d,e\}$, and $A_3=\varnothing$. These sets are pairwise disjoint: the first two have no common elements, and the third has none.

Counting each set gives $\mu(A_1)=2$, $\mu(A_2)=3$, and $\mu(A_3)=0$. Their union is $\{a,b,c,d,e\}$, with five elements, so $$ \mu(A_1\cup A_2\cup A_3)=5=2+3+0=\mu(A_1)+\mu(A_2)+\mu(A_3). $$ Check: The elements $a,c$ contribute $2$, the distinct elements $b,d,e$ contribute $3$, and the empty set contributes $0$; the total is $2+3=5$, matching the count of the union.

Worked Example: A Finite Sum of Dirac Measures

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, define $\mu=2\delta_0+3\delta_1+\delta_2$. Here $\delta_x(E)=1$ if $x\in E$, and $\delta_x(E)=0$ if $x\notin E$; thus $\mu(E)=2\delta_0(E)+3\delta_1(E)+\delta_2(E)$. Consider $A_1=\{0\}$, $A_2=\{1,2\}$, and $A_3=(4,5)$. They are Borel measurable and pairwise disjoint.

Substitution into the defining formula gives $\mu(A_1)=2(1)+3(0)+1(0)=2$. For $A_2$, the indicators at $0,1,2$ are $0,1,1$, so $\mu(A_2)=2(0)+3(1)+1(1)=4$. For $A_3$, none of $0,1,2$ belongs to $(4,5)$, so $\mu(A_3)=2(0)+3(0)+1(0)=0$. Their union contains $0,1,2$ and no other atoms, hence $$ \mu(A_1\cup A_2\cup A_3)=2(1)+3(1)+1(1)=6=2+4+0. $$ Check: Adding the three individual values gives $2+4+0=6$, the same as evaluating the measure directly on the union.

Why Disjointness Cannot Be Dropped

The formula proved above applies to pairwise disjoint sets. If sets overlap, adding their measures can count the overlap more than once. For example, in $(\mathbb{R},\mathcal{L},m)$, take $B_1=[0,2]$ and $B_2=[1,3]$. Both have measure $2$, so the incorrect calculation $m(B_1\cup B_2)=m(B_1)+m(B_2)=2+2=4$ would result from using finite additivity without checking its hypothesis. But $B_1\cap B_2=[1,2]$, and the union is $[0,3]$, whose measure is $3$, not $4$.

One can see the overcount directly: the interval $[1,2]$ has length $2-1=1$ and belongs to both sets. The sum $2+2$ counts that length twice, whereas the union counts it once. Thus the correction in this example is $2+2-1=3$. The finite-additivity theorem makes no claim about overlapping sets; a formula for such unions requires additional reasoning and is not obtained by simply applying the theorem.

A second mistake is to treat the padding as if it justified adding arbitrary extra sets. It does not: only empty sets are appended, because they preserve both the union and disjointness. Appending a nonempty set could change the union, and appending a set that intersects one of the original sets could destroy the disjointness required by countable additivity.

What the Proof Gives

For every finite disjoint measurable family, the measure of the union is obtained by an ordinary finite sum of the individual measures. The result holds on every measure space, whether its total measure is finite or infinite, and it remains valid when some individual measures are infinite. In particular, it applies to Lebesgue measure, counting measure, Dirac measures, and any other set function that satisfies the measure axioms.

When applying the result, check the hypotheses before computing: the family must be finite, each member must belong to the sigma-algebra, and distinct members must have empty intersection. The empty-set terms introduced in the proof are harmless precisely because their measures are zero. These checks keep the passage from countable additivity to finite additivity exact, including in cases where the numerical values are $0$ or $+\infty$.

Check Your Understanding

For each question, state whether the finite-additivity theorem applies before carrying out any calculation.

In a measure space, $A_1,A_2,A_3$ are measurable and pairwise disjoint, with measures $4$, $0$, and $+\infty$. What is the measure of their union, and which part of the proof permits this value?
Explain why appending $E_k=\varnothing$ for every $k>n$ preserves the union of $A_1,\ldots,A_n$.
Under counting measure, compute the measure of the union of $\{1,3\}$, $\{2,4\}$, and $\varnothing$. Verify the finite sum by counting the union directly.
For Lebesgue measure, take $C_1=[0,1]$ and $C_2=[1,2]$. Does the finite-additivity theorem apply to this pair? Find the measure of the union and identify the overlap responsible for the issue.
Why is no assumption that $\mu(X)<\infty$ needed in the proof by padding?