What a Measure Space Contains
A measure assigns a nonnegative size to each set in a specified collection of measurable sets. The collection matters: a measure is not, in general, defined on every subset of the underlying set. Its defining requirements are also distinct. The empty set must have measure zero, and every countable disjoint family of measurable sets must have a union whose measure is the sum of their measures. The sum and the measure are allowed to be infinite.
- \(\mu(\varnothing)=0\).
- If \(E_1,E_2,\ldots\in\mathcal{M}\) are pairwise disjoint, then \[ \mu\left(\bigcup_{j=1}^{\infty}E_j\right)=\sum_{j=1}^{\infty}\mu(E_j). \]
Pairwise disjoint means \(E_i\cap E_j=\varnothing\) whenever \(i\ne j\). The sets in the axiom must belong to \(\mathcal{M}\), and their union belongs to \(\mathcal{M}\) because \(\mathcal{M}\) is a sigma-algebra. Thus the axiom compares values that are all in the domain of \(\mu\). A measure is nonnegative and may take the value \(+\infty\); it is not a signed measure. In particular, an infinite value is not an error or an undefined quantity.
The sigma-algebra is part of the data. For example, if \(X\) has two elements, the collection \(\{\varnothing,X\}\) is a sigma-algebra, but it does not include either singleton. A measure on this space is only required to assign values to \(\varnothing\) and \(X\). By contrast, the power set \(\mathcal{P}(X)\) includes every subset of \(X\). Borel and Lebesgue measures are examples on particular sigma-algebras; the definition above does not identify those sigma-algebras or require a measure to be Lebesgue measure.
Reading the Countable-Additivity Axiom
For disjoint measurable sets, countable additivity says that the measure of the entire union is exactly the extended sum of the individual measures. The finite and infinite cases are handled by the same axiom: a finite list can be continued with empty sets, whose measure is zero. The order of the nonnegative terms does not change their extended sum. If one term is \(+\infty\), the sum is \(+\infty\); otherwise, it is the limit of finite partial sums.
Proof. By countable additivity, the measure on the left equals \(\sum_{j=1}^{\infty}\mu(E_j)\). For a sequence of nonnegative extended real numbers, its extended sum is defined as the limit of the nondecreasing partial sums, equivalently their supremum. Therefore the right side equals that same extended sum, proving the identity. If every term is zero, every partial sum and the supremum are zero; if some term is infinite or the partial sums are unbounded, the supremum is infinite. \(\square\)
Worked Example: A Measure on a Two-Set Space
Let \(X=\{a,b\}\), \(\mathcal{M}=\{\varnothing,X\}\), and define \(\mu(\varnothing)=0\), \(\mu(X)=3\). Check the measure axioms, including the countable condition.
The collection \(\mathcal{M}\) is a sigma-algebra: it contains \(X\), is closed under complements in \(X\), and any countable union of its members is either \(\varnothing\) (if all members are empty) or \(X\) (if at least one member is \(X\)). The assigned values are in \([0,\infty]\), and \(\mu(\varnothing)=0\).
For a pairwise disjoint sequence \((E_j)\) from \(\mathcal{M}\), there are two possibilities. If every \(E_j=\varnothing\), then the union has measure \(0\), and the sum is \(\sum_{j\ge1}0=0\). If some \(E_k=X\), every other \(E_j\) must be empty because it must be disjoint from \(X\). The union then has measure \(3\), while the sum has the one nonzero term \(\mu(E_k)=3\), so it also equals \(3\). This checks countable additivity in all possible cases. Check: the sequence \(E_1=X\), \(E_j=\varnothing\) for \(j\ge2\) gives \(\mu(\bigcup_jE_j)=3\) and \(\sum_j\mu(E_j)=3+0+0+\cdots=3\).
Why the Empty-Set Requirement Is Separate
With values in \([0,\infty]\), countable additivity alone does not force the measure of the empty set to be zero. The following proposition explains one circumstance in which countable additivity does force it: there must be at least one measurable set of finite measure.
Proof. Write \(c=\nu(\varnothing)\). Apply countable additivity to the sequence whose first set is \(A\) and whose remaining sets are all \(\varnothing\). This sequence is pairwise disjoint, and its union is \(A\). Hence \[ \nu(A)=\nu(A)+\sum_{j=2}^{\infty}c. \] If \(c>0\), the sum on the right is \(+\infty\): when \(c\) is finite and positive, its partial sums are unbounded, and when \(c=+\infty\), each partial sum containing a term is infinite. The displayed equality would then imply \(\nu(A)=+\infty\), contrary to the assumption \(\nu(A)<\infty\). Since \(c\in[0,\infty]\), the only remaining possibility is \(c=0\). \(\square\)
This extra hypothesis cannot be dropped. If all measurable sets have infinite assigned value, the equality in the countable-additivity axiom can hold even though the empty set has infinite value. Thus the empty-set condition in the definition is not redundant for general extended-valued measures.
Worked Example: Countable Additivity Without the Empty-Set Axiom
Take any nonempty set \(X\), let \(\mathcal{M}=\mathcal{P}(X)\), and define a set function \(\nu:\mathcal{M}\to[0,\infty]\) by \(\nu(E)=+\infty\) for every \(E\subseteq X\). This is a set function satisfying countable additivity, but it is not a measure.
Indeed, for any sequence \((E_j)\) of pairwise disjoint subsets, if its union is empty, then every \(E_j\) is empty, and \[ \nu\left(\bigcup_{j\ge1}E_j\right)=\nu(\varnothing)=+\infty \quad\text{and}\quad \sum_{j\ge1}\nu(E_j)=\sum_{j\ge1}(+\infty)=+\infty. \] If its union is nonempty, at least one \(E_j\) is nonempty. The union has value \(+\infty\), and the series has a term \(+\infty\), so both sides again equal \(+\infty\). Countable additivity holds in either case. But \(\nu(\varnothing)=+\infty\ne0\), so the empty-set axiom fails. Check: for \(E_j=\varnothing\) for all \(j\), both sides of countable additivity are \(+\infty\), not zero. This is why verifying only the countable-additivity equation does not establish that a set function is a measure.
A Countable Example with Finite Total Measure
Here is a measure whose values are obtained by adding weights assigned to points. The example also illustrates how a disjoint countable union can be checked by tracking each point exactly once.
Worked Example: Weights on the Positive Integers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and define \[ \mu(A)=\sum_{n\in A}2^{-n},\qquad A\subseteq\mathbb{N}. \] The sum over an arbitrary subset is the extended nonnegative sum of its terms. Show that this defines a measure and compute the measures of the even and odd integers.
Every summand is nonnegative. The empty set has no summands, so \(\mu(\varnothing)=0\). Now let \((A_j)_{j\ge1}\) be pairwise disjoint subsets of \(\mathbb{N}\). Each integer \(n\) belongs to at most one \(A_j\). Consequently, in the sum of the weights over all the \(A_j\), the term \(2^{-n}\) appears once if \(n\in\bigcup_jA_j\), and not at all otherwise. More formally, a nonnegative series is the supremum of the sums over finite selections of its terms. The finite selections from the disjoint sets together are exactly finite selections from their union. Taking these suprema gives \[ \sum_{j\ge1}\mu(A_j)=\sum_{n\in\bigcup_jA_j}2^{-n} =\mu\left(\bigcup_{j\ge1}A_j\right). \] Thus both measure axioms hold.
The total measure is the geometric series \[ \mu(\mathbb{N})=\sum_{n=1}^{\infty}2^{-n} =\frac{1/2}{1-1/2}=1. \] For the even integers, \[ \mu(\{2,4,6,\ldots\})=\sum_{k=1}^{\infty}2^{-2k} =\frac{1/4}{1-1/4}=\frac13. \] For the odd integers, \[ \mu(\{1,3,5,\ldots\})=\sum_{k=0}^{\infty}2^{-(2k+1)} =\frac{1/2}{1-1/4}=\frac23. \] The even and odd integers are disjoint and their union is \(\mathbb{N}\). Check: their measures sum to \(\frac13+\frac23=1=\mu(\mathbb{N})\), as countable additivity requires.
A Common Mistake: Checking Only One Axiom
A frequent error is to call any nonnegative set function satisfying the disjoint-union equation a measure. The all-infinite set function above is a counterexample: it satisfies that equation but assigns \(+\infty\) to the empty set. The correction is to check the two defining requirements separately. Conversely, the assignment \(\rho(A)=1\) for every nonempty subset \(A\) of \(X=\{a,b\}\), with \(\rho(\varnothing)=0\), does satisfy the empty-set requirement but fails additivity. The disjoint sets \(\{a\}\) and \(\{b\}\) have union \(X\), yet \[ \rho(\{a\}\cup\{b\})=1 \quad\text{while}\quad \rho(\{a\})+\rho(\{b\})=1+1=2. \] The sets must be disjoint for the countable-additivity axiom, and in this test they are; the unequal values show that \(\rho\) is not a measure. In later work, a candidate set function must be checked against the whole definition, on its stated sigma-algebra.
Check Your Understanding
For each question, distinguish the underlying set, the sigma-algebra, and the assigned set function where relevant.
- State the two axioms required of a measure \(\mu:\mathcal{M}\to[0,\infty]\).
- Why is it necessary for the sets in the countable-additivity axiom to be pairwise disjoint?
- For the all-infinite set function \(\nu\), which axiom fails, and what value does \(\nu\) assign to the empty set?
- In the weighted measure example, why does each point's weight occur at most once when the sets \(A_j\) are disjoint?
- What additional hypothesis in the proposition forces \(\nu(\varnothing)=0\) from countable additivity alone?