Measure Calculations On Finite Measurable Partitions

From Partition Cells to Measures of Unions

A finite measurable partition breaks a space into disjoint measurable pieces. Because the pieces do not overlap, countable additivity gives a direct calculation rule: the measure of a union of selected cells is the sum of their measures. The rule applies whether some cells have measure zero, whether some have infinite measure, and whether the whole space has finite measure. “Finite partition” refers to the number of cells, not to the measure of the space.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space: $\mathcal{M}$ is a sigma-algebra of subsets of $X$, and $\mu:\mathcal{M}\to[0,\infty]$ is a countably additive measure with $\mu(\varnothing)=0$. No finiteness, sigma-finiteness, or completeness assumption is imposed. All set equalities below are exact; the functions are not involved. A finite sum of values in $[0,\infty]$ is interpreted in the extended nonnegative reals: if any term is $+\infty$, the sum is $+\infty$, and the empty sum is $0$.

Definition: A finite measurable partition of $X$ is a finite family $E_1,\ldots,E_k$ of sets in $\mathcal{M}$ such that the cells are pairwise disjoint and $\bigcup_{i=1}^k E_i=X$. Empty cells may be allowed; they do not affect any of the formulas below. If $k=0$, this definition is applicable only when $X=\varnothing$.

For any subset $J\subseteq\{1,\ldots,k\}$, the selected union is $A_J=\bigcup_{i\in J}E_i$. It is measurable because a sigma-algebra is closed under finite unions. Its complement in $X$ is the union of exactly the unselected cells. These facts give two ways to calculate: sum the selected cell measures, or sum the unselected cell measures to obtain the complement. The complement calculation is especially important when the total measure is infinite.

The Finite Additivity Calculation

Theorem: Let $E_1,\ldots,E_k$ be a finite measurable partition of $X$ in a measure space $(X,\mathcal{M},\mu)$. For every $J\subseteq\{1,\ldots,k\}$, $$ \mu\left(\bigcup_{i\in J}E_i\right)=\sum_{i\in J}\mu(E_i). $$ This equality holds in $[0,\infty]$, with an empty sum equal to $0$. No assumption that $\mu(X)$ is finite is needed.

Proof. If $J$ is empty, its union is $\varnothing$, so both sides are $0$. If $J$ has one element, the equality states that $\mu(E_i)=\mu(E_i)$. For a nonempty $J$ with $r$ elements, list its indices as $i_1,\ldots,i_r$. The sets $E_{i_1},\ldots,E_{i_r}$ are measurable and pairwise disjoint, since they belong to a partition. Countable additivity applied to this finite disjoint family, extended by empty sets if necessary, gives $$ \mu\left(\bigcup_{j=1}^{r}E_{i_j}\right) =\sum_{j=1}^{r}\mu(E_{i_j}). $$ This is the claimed equality. The measure axiom permits the value $+\infty$, and a finite sum of nonnegative extended-real values is defined as stated above, so the argument also covers infinite cell measures. If $k=0$, then $X=\varnothing$; the only choice is $J=\varnothing$, already covered. $\square$

Corollary: For the same partition, $$ \mu(X)=\sum_{i=1}^{k}\mu(E_i), \qquad \mu(A_J^c)=\sum_{i\notin J}\mu(E_i). $$ In particular, $A_J$ has measure zero if and only if every selected cell has measure zero.

Proof. The first equality is the theorem with all indices selected. Since the cells are disjoint and cover $X$, a point lies outside $A_J$ precisely when it lies in a cell whose index is not in $J$. Thus $A_J^c=\bigcup_{i\notin J}E_i$, and the theorem gives the second equality. For the final assertion, a finite sum of nonnegative numbers is zero if every term is zero. Conversely, if the sum is zero, no term can be positive: a positive term would make the sum positive, including when that term is $+\infty$. Apply this to the selected cell measures. $\square$

Worked Calculations

Worked Example: A Finite Total with Zero-Measure Cells

Suppose $E_1,E_2,E_3,E_4$ partition $X$, and their measures are respectively $0,\frac32,0,7$. Find the measure of $A=E_1\cup E_3\cup E_4$, and then find the measure of its complement.

The selected index set is $J=\{1,3,4\}$. By the theorem,

\mu(A)=\mu(E_1)+\mu(E_3)+\mu(E_4) =0+0+7=7.

The complement consists of the one unselected cell, so $A^c=E_2$ and $\mu(A^c)=\frac32$. As a check, the measure of the whole space is $$ \mu(X)=0+\frac32+0+7=\frac{17}{2}, $$ and, because this total is finite, ordinary subtraction gives $\mu(X)-\mu(A)=\frac{17}{2}-7=\frac32$, agreeing with the direct complement calculation. The zero-measure cells $E_1$ and $E_3$ are still cells of the partition; zero measure does not mean that a set must be empty.

Worked Example: Infinite Total and Two Different Complements

Let $X=\mathbb{N}=\{1,2,\ldots\}$ with counting measure, for which the measure of a set is its number of elements if it is finite and $+\infty$ if it is infinite. Consider the partition $$ E_1=\{1\},\qquad E_2=\{2,4,6,\ldots\},\qquad E_3=\{3,5,7,\ldots\}. $$

These sets are disjoint and cover $\mathbb{N}$. Their measures are $\mu(E_1)=1$, $\mu(E_2)=+\infty$, and $\mu(E_3)=+\infty$, since each of the last two sets has infinitely many elements. For $A=E_1\cup E_3$, the sum rule gives $$ \mu(A)=1+(+\infty)=+\infty. $$ The complement is $E_2$, so its measure is also $+\infty$. Here $\mu(X)=+\infty$, and subtracting $\mu(A)=+\infty$ from $\mu(X)=+\infty$ is undefined. The direct complement formula is nonetheless exact: it gives $\mu(A^c)=\mu(E_2)=+\infty$.

Now select only $B=E_1$. Then $\mu(B)=1$, while the unselected cells give $\mu(B^c)=\mu(E_2)+\mu(E_3)=+\infty$. In this case $+\infty-1=+\infty$ agrees with the complement measure. This check illustrates the distinction: subtracting a finite selected measure from an infinite total can give the correct infinite complement, but subtracting an infinite selected measure from an infinite total is undefined. The sum over the unselected cells works in both cases.

Worked Example: A Union of Zero-Measure Cells

On $X=\mathbb{R}$, let $\delta_0$ be the Dirac measure at $0$, defined on Borel sets by $\delta_0(E)=1$ if $0\in E$, and $\delta_0(E)=0$ if $0\notin E$. Use the partition $E_1=\{0\}$, $E_2=\{1\}$, and $E_3=\mathbb{R}\setminus\{0,1\}$. Find the measure of $C=E_2\cup E_3$ and check it using the complement.

The cells are Borel, pairwise disjoint, and cover $\mathbb{R}$. The point $0$ lies in $E_1$, but not in $E_2$ or $E_3$, so $$ \delta_0(E_1)=1,\qquad \delta_0(E_2)=0,\qquad \delta_0(E_3)=0. $$ Therefore $\delta_0(C)=0+0=0$. The complement is $C^c=E_1$, whose measure is $1$. As an independent check, $C$ does not contain $0$, so the defining formula for $\delta_0$ gives $\delta_0(C)=0$ directly. Also $\delta_0(\mathbb{R})=1=0+0+1$, consistent with the sum over all partition cells.

Complements, Subtraction, and Common Mistakes

For every selection $J$, the reliable complement identity is $$ \mu(A_J^c)=\sum_{i\notin J}\mu(E_i). $$ It follows from the partition itself, so it is valid even when some terms are infinite. A subtraction formula requires more care. If $\mu(X)<\infty$, then every measurable subset has finite measure, and finite additivity applied to the disjoint sets $A_J$ and $A_J^c$ gives $$ \mu(A_J^c)=\mu(X)-\mu(A_J). $$ If $\mu(X)=+\infty$ but $\mu(A_J)<\infty$, the complement must have infinite measure: otherwise the sum of the two finite measures would make $\mu(X)$ finite. In this case the extended-real expression $+\infty-\mu(A_J)=+\infty$ matches the complement measure. If $\mu(X)=+\infty$ and $\mu(A_J)=+\infty$, the expression $+\infty-(+\infty)$ is undefined, so it cannot be used to calculate the complement. The unselected-cell sum still gives the answer.

Common mistake: subtracting two infinite values. In the counting-measure example, a tempting calculation is $$ \mu(E_2)=\mu(X)-\mu(E_1\cup E_3)=+\infty-(+\infty). $$ This is not a valid calculation: the right side is undefined. The correction is to use the unselected cell directly, giving $\mu(E_2)=+\infty$. By contrast, for $B=E_1$, the calculation $+\infty-1=+\infty$ is defined and agrees with the directly computed complement measure. The distinction is between an infinite total minus a finite quantity and an infinite total minus an infinite quantity.

Common mistake: counting overlapping sets as partition cells. Finite additivity in the theorem depends on disjointness. For example, on the three-point space $\{a,b,c\}$ with counting measure, the sets $F_1=\{a,b\}$ and $F_2=\{b,c\}$ overlap at $b$. Their measures sum to $2+2=4$, but their union has measure $3$. Thus they are not a partition, and the sum rule for partition cells does not apply to them. Before summing, verify both that the sets are measurable and that distinct cells are disjoint.

The rule matters because it reduces a set calculation to a finite list of cell measures. Keep the cases separate: zero terms contribute nothing; any selected infinite cell makes that union’s measure infinite; and an empty selection has measure zero. For complements, summing the unselected cells avoids any ambiguity about extended-real subtraction.

Check Your Understanding

Use disjointness and the sum over the relevant cells. For complement questions, identify the unselected cells before considering subtraction.

A partition has three cells of measures $2,0,5$. What is the measure of the union of the first and third cells, and what is the measure of its complement?
For a finite measurable partition, explain why the union over an empty index set has measure zero.
If one selected cell has measure $+\infty$, what is the measure of the selected union? State the reason using the finite-sum convention.
If $\mu(X)=+\infty$ and $\mu(A)=+\infty$, why can $\mu(X)-\mu(A)$ not be used to find $\mu(A^c)$? What formula should be used instead?
Give a reason that the measures of two overlapping measurable sets cannot in general be added to obtain the measure of their union.

Measure Calculations On Finite Measurable Partitions

What You'll Learn