Set Functions on Every Subset
A measure is defined on a sigma-algebra, so its value is only assigned to sets in that sigma-algebra. An outer measure has a broader domain: it assigns a value to every subset of the underlying space. This broader domain is useful because it allows set sizes to be discussed before selecting which sets will be measurable. The defining requirements constrain those values without requiring additivity on disjoint sets.
Let \(X\) be any set, and write \(\mathcal{P}(X)\) for its power set, the collection of all subsets of \(X\). No sigma-algebra or measure space is assumed here. The values of an outer measure may be infinite; in particular, the sum of a sequence of nonnegative extended real numbers is interpreted as the limit of its increasing partial sums, and may equal \(+\infty\).
- Empty set: \(\mu^*(\varnothing)=0\).
- Monotonicity: if \(A\subseteq B\subseteq X\), then \(\mu^*(A)\le\mu^*(B)\).
- Countable subadditivity: for every sequence \((A_n)_{n\ge1}\) of subsets of \(X\), \(\displaystyle \mu^*\!\left(\bigcup_{n=1}^{\infty}A_n\right)\le\sum_{n=1}^{\infty}\mu^*(A_n)\).
These conditions apply to every subset, not just to sets that are already known to be measurable. Monotonicity says that enlarging a set cannot decrease its assigned size. Countable subadditivity says that covering a union by sets gives an upper bound for the size of the union: overlaps may be counted more than once, and equality is not required. Countable subadditivity also covers finite unions, by taking all later sets to be empty.
What the Axioms Immediately Imply
Proof. Extend the finite list to a sequence by setting \(A_j=\varnothing\) for every \(j>r\). Its countable union is still \(\bigcup_{j=1}^{r}A_j\). Countable subadditivity gives an upper bound by \(\sum_{j=1}^{\infty}\mu^*(A_j)\). The terms after \(r\) are all \(\mu^*(\varnothing)=0\), so this sum is exactly \(\sum_{j=1}^{r}\mu^*(A_j)\). This proves the inequality. \(\square\)
The empty-set condition is also needed for this formulation to give the intended zero contribution when a finite union is padded with empty sets. The three requirements are stated separately: in particular, monotonicity should not be silently inferred from countable subadditivity.
Worked Example: A Weighted Outer Measure on a Finite Set
Let \(X=\{a,b,c\}\), and assign weights \(w(a)=2\), \(w(b)=3\), and \(w(c)=5\). Define \(\mu^*(A)=\sum_{x\in A}w(x)\) for every \(A\subseteq X\). For example, \(\mu^*(\{a,c\})=2+5=7\), and \(\mu^*(\varnothing)=0\).
If \(A\subseteq B\), every weight included in the sum for \(A\) also occurs in the sum for \(B\), and all weights are nonnegative. Thus \(\mu^*(A)\le\mu^*(B)\). For a sequence \((A_n)\), each \(x\in\bigcup_n A_n\) belongs to at least one \(A_n\), so \(\chi_{\cup_n A_n}(x)\le\sum_n\chi_{A_n}(x)\). Multiplying by \(w(x)\) and summing over the three points gives \[ \mu^*\!\left(\bigcup_n A_n\right) =\sum_{x\in X}w(x)\chi_{\cup_n A_n}(x) \le\sum_{x\in X}w(x)\sum_n\chi_{A_n}(x) =\sum_n\mu^*(A_n). \] The last equality follows by taking partial sums in \(n\): there are only three points, so each partial sum is the corresponding finite sum of the three pointwise partial sums, and taking their increasing limits gives the equality. The example satisfies all three requirements. As a direct check of subadditivity, take \(A_1=\{a,b\}\), \(A_2=\{b,c\}\), and \(A_n=\varnothing\) for \(n\ge3\). The union has value \(2+3+5=10\), while the sum of the set values is \((2+3)+(3+5)=13\), so \(10\le13\).
Worked Example: An Outer Measure Concentrated at One Point
Let \(X=\{p,q,r\}\), and define \(\mu^*(A)=1\) if \(p\in A\), and \(\mu^*(A)=0\) if \(p\notin A\). The empty set does not contain \(p\), so its value is \(0\). If \(A\subseteq B\) and \(p\in A\), then \(p\in B\), so both values are \(1\); if \(p\notin A\), then \(\mu^*(A)=0\le\mu^*(B)\). This verifies monotonicity.
For countable subadditivity, if \(p\notin\bigcup_n A_n\), the union has value \(0\), and the inequality holds because the right side is nonnegative. If \(p\in\bigcup_n A_n\), at least one \(A_k\) contains \(p\), so \(\sum_n\mu^*(A_n)\ge\mu^*(A_k)=1\), which is the value of the union. For instance, with \(A_1=\{p,q\}\), \(A_2=\{p,r\}\), and \(A_n=\varnothing\) for \(n\ge3\), the union has value \(1\) and the sum is \(1+1=2\). The inequality is strict here; subadditivity does not assert additivity.
Worked Example: A Fixed Cost for Every Nonempty Set
Fix a real number \(c>0\) and define \(\mu^*(\varnothing)=0\) and \(\mu^*(A)=c\) for every nonempty \(A\subseteq X\). This function is monotone: if \(A\subseteq B\), then either \(A=\varnothing\), whose value is \(0\), or both sets are nonempty, whose values are both \(c\).
If \(\bigcup_n A_n=\varnothing\), every \(A_n\) is empty, so both sides of countable subadditivity are \(0\). If the union is nonempty, some \(A_k\) is nonempty. The sum on the right is therefore at least \(\mu^*(A_k)=c\), while the union has value \(c\). Thus the inequality holds. With \(c=4\), \(A_1=\{u\}\), \(A_2=\{v\}\), and all later sets empty, the union has value \(4\) and the sum is \(4+4=8\). This example is an outer measure, but its values are not additive on disjoint nonempty sets.
Why Each Requirement Is Separate
It is possible to satisfy the empty-set condition and countable subadditivity while failing monotonicity. This shows why monotonicity is included explicitly in the definition.
Worked Example: Subadditivity Does Not Force Monotonicity
On \(X=\{a,b\}\), define \(q(\varnothing)=0\), \(q(\{a\})=2\), \(q(\{b\})=0\), and \(q(X)=1\). Then \(\{b\}\subseteq X\), but \(q(\{b\})=0\le1=q(X)\) does not fail; instead, \(\{a\}\subseteq X\) gives \(q(\{a\})=2>1=q(X)\), so monotonicity fails.
To check countable subadditivity, consider any sequence \((A_n)\). If its union is empty or \(\{b\}\), the union has value \(0\), so the inequality holds. If its union is \(\{a\}\), at least one \(A_n=\{a\}\), and the sum of the values is at least \(2=q(\{a\})\). If its union is \(X\), either some \(A_n=X\), contributing \(1\), or the sequence contains both \(\{a\}\) and \(\{b\}\), contributing at least \(2\). In either case the sum is at least \(1=q(X)\). Thus \(q\) is countably subadditive despite not being monotone.
Conversely, monotonicity and the empty-set condition do not imply countable subadditivity. On \(X=\mathbb{N}\), define \(s(A)=0\) when \(A\) is finite and \(s(A)=1\) when \(A\) is infinite. This function is monotone: a superset of an infinite set is infinite, and the value of any finite set is \(0\le s(B)\). Also \(s(\varnothing)=0\). But for \(A_n=\{n\}\), every \(s(A_n)=0\), whereas \(\bigcup_{n=1}^{\infty}A_n=\mathbb{N}\) has value \(1\). The required inequality would be \(1\le\sum_n0=0\), which is false.
Outer Measure Is Not Additivity
An outer measure is not, merely by its definition, a measure on a sigma-algebra. Its domain is the entire power set, and its union axiom is an inequality rather than an equality. Even when two sets are disjoint, countable subadditivity only gives an upper bound. The fixed-cost example exhibited a strict inequality for disjoint sets.
A common mistake is to replace the defining inequality by an equality for every union. For the point-concentrated example, taking \(A=\{p,q\}\) and \(B=\{p,r\}\) gives \(\mu^*(A\cup B)=1\), but \(\mu^*(A)+\mu^*(B)=2\). The correct statement is \(1\le2\), not \(1=2\). Another mistake is to check only finite unions: countable subadditivity must hold for every sequence, including sequences of infinitely many sets, as the singleton sequence in the example on \(\mathbb{N}\) demonstrates.
The power-set domain makes the definition easy to state, but it does not make every subset measurable in the later sense of Carathéodory. At this stage, the object is simply a set function satisfying the three conditions above. Those axioms supply the basic control needed to investigate which sets have stronger splitting properties, without assuming that every union has an exactly additive value.
Check Your Understanding
For each question, identify which defining requirement is being used or tested.
- Why does countable subadditivity imply finite subadditivity when \(\mu^*(\varnothing)=0\)?
- Give subsets \(A\subseteq B\) for which the function \(q\) in the example fails monotonicity.
- For the fixed-cost outer measure with \(c=4\), compute the values of \(\varnothing\), \(\{u\}\), and \(\{u,v\}\).
- For \(s(A)=0\) on finite subsets of \(\mathbb{N}\) and \(s(A)=1\) on infinite subsets, compute both sides of countable subadditivity for \(A_n=\{n\}\).
- Does countable subadditivity require equality for disjoint sets? Explain using the point-concentrated example.