A Practical Test for Candidate Measures
A proposed set function can look plausible and still fail to be a measure. Checking only that its values are nonnegative is not enough: its value at the empty set and its behavior on disjoint sets also matter. On a finite underlying set, the countable additivity condition becomes a finite calculation, so we can test candidates directly and describe exactly what valid ones look like.
Throughout, let \(X\) be a finite set and let \(\mathcal{M}\) be a sigma-algebra of subsets of \(X\). A candidate \(\mu\) has domain \(\mathcal{M}\) and takes values in \([0,\infty]\). The tests below do not require \(\mu(X)\) to be finite, and they make no assumption that the measure space is complete or sigma-finite. In the worked examples, the candidate values are finite. The underlying set being finite does not, by itself, force a measure to have finite total mass.
Here pairwise disjoint means \(E_i\cap E_j=\varnothing\) whenever \(i\ne j\). Since \(\mathcal{M}\) is a sigma-algebra, the union of any sequence in \(\mathcal{M}\) belongs to \(\mathcal{M}\), so the left side of the additivity requirement is defined. Countable additivity includes finite additivity: a finite disjoint family can be extended to a sequence by setting every remaining set equal to \(\varnothing\). On a finite \(X\), the converse also holds once the empty-set value is correct.
Proof. If \(\mu\) is a measure, its empty-set value is zero by definition. For a finite disjoint family, set \(E_n=\varnothing\) for \(n>k\). Countable additivity then gives finite additivity, since every added term is \(\mu(\varnothing)=0\).
For the reverse direction, assume \(\mu(\varnothing)=0\) and finite additivity. Let \((E_n)_{n\ge1}\) be a pairwise disjoint sequence in \(\mathcal{M}\). Each nonempty \(E_n\) contains at least one point of \(X\), and disjoint nonempty sets contain different points. There can therefore be at most \(|X|\) nonempty sets among the \(E_n\). In particular, there is an \(N\ge0\) such that \(E_n=\varnothing\) for every \(n>N\); if all sets are empty, take \(N=0\). Their union is \(\bigcup_{n=1}^N E_n\), and the series of their measures has only the first \(N\) terms possibly nonzero. Finite additivity gives \(\mu(\bigcup_{n\ge1}E_n)=\sum_{n=1}^N\mu(E_n)=\sum_{n\ge1}\mu(E_n)\). This proves countable additivity. If \(X=\varnothing\), then \(\mathcal{M}=\{\varnothing\}\); the condition \(\mu(\varnothing)=0\) is both necessary and sufficient, as the same argument with \(N=0\) shows. \(\square\)
Worked Checks on Explicit Sigma-Algebras
Worked Example: A Candidate on the Full Power Set
Take \(X=\{a,b,c\}\) and \(\mathcal{M}=\mathcal{P}(X)\), the collection of all subsets of \(X\). Define \(\mu(E)=2\chi_E(a)+0\chi_E(b)+3\chi_E(c)\) for every \(E\subseteq X\). Here \(\chi_E(x)\) is \(1\) when \(x\in E\) and \(0\) otherwise. Check the measure axioms and calculate \(\mu(\{a,c\})\).
Each indicator is either \(0\) or \(1\), so the three terms in the definition are nonnegative and \(\mu(E)\ge0\) for every \(E\). At \(E=\varnothing\), all three indicators are zero, giving \(\mu(\varnothing)=2(0)+0(0)+3(0)=0\). For \(E=\{a,c\}\), the indicators at \(a,b,c\) are respectively \(1,0,1\), so \(\mu(\{a,c\})=2(1)+0(0)+3(1)=5\).
To check additivity, let \(E_1,\ldots,E_k\) be any finite disjoint family. For each point \(x\in X\), disjointness implies that \(x\) belongs to at most one of these sets. Thus \(\sum_{j=1}^k\chi_{E_j}(x)=\chi_{\cup_jE_j}(x)\). Multiplying this identity at \(a,b,c\) by \(2,0,3\), respectively, and adding gives \(\sum_{j=1}^k\mu(E_j)=\mu(\cup_j E_j)\). The finite-space verification criterion now proves countable additivity.
Check: The disjoint partition \(X=\{a\}\cup\{b\}\cup\{c\}\) gives \(\mu(X)=2+0+3=5\). The set \(\{a,c\}\) has measure \(2+3=5\), matching the direct calculation above.
Worked Example: A Sigma-Algebra with Two Atoms
Let \(X=\{a,b,c,d\}\), \(A=\{a,b\}\), \(B=\{c,d\}\), and \(\mathcal{M}=\{\varnothing,A,B,X\}\). Define a candidate by \(\nu(\varnothing)=0\), \(\nu(A)=4\), \(\nu(B)=1\), and \(\nu(X)=5\). Verify that it is a measure on \((X,\mathcal{M})\).
All four listed values lie in \([0,\infty]\), and the empty set has value zero. To check finite additivity, consider disjoint measurable sets. A nonempty measurable set is \(A\), \(B\), or \(X\). If one set in a disjoint family is \(X\), every other set in that family must be empty. The union then has measure \(5\), while the sum of the values is \(5\) plus zeros. If no set is \(X\), the only possible nonempty disjoint sets are \(A\) and \(B\), and they cannot occur twice in a pairwise disjoint family. When both occur, their union is \(X\), and \(\nu(A)+\nu(B)=4+1=5=\nu(X)\). If just \(A\) occurs, the union has value \(4\); if just \(B\) occurs, it has value \(1\); if neither occurs, the union is empty and both sides are zero. These cases verify finite additivity, including families with empty sets.
The finite-space verification criterion gives countable additivity. For an explicit countable check, take \(E_1=A\), \(E_2=B\), and \(E_n=\varnothing\) for \(n\ge3\). These sets are pairwise disjoint, their union is \(X\), and \(\nu(\bigcup_{n\ge1}E_n)=\nu(X)=5=4+1+0+\cdots=\sum_{n\ge1}\nu(E_n)\).
Check: The only nontrivial split of \(X\) into disjoint nonempty measurable sets is \(A\) and \(B\). Its values add to \(4+1=5\), exactly the assigned value of \(X\).
Atoms Explain the Whole Test
The sets \(A\) and \(B\) in the previous example cannot be split into smaller nonempty members of \(\mathcal{M}\). Such minimal measurable pieces are called atoms. For a finite underlying set, every measurable set is a union of atoms, and a candidate measure is determined by its values on them.
Proof. First suppose \(X\ne\varnothing\). For each \(x\in X\), intersect all the measurable sets containing \(x\), and call the resulting set \(C_x\). This is a finite intersection because \(\mathcal{M}\) is finite: it is a subcollection of the finite power set \(\mathcal{P}(X)\). The intersection is measurable and contains \(x\). If \(y\in C_x\), then \(x\) and \(y\) belong to exactly the same measurable sets: membership of \(y\) in every measurable set containing \(x\) follows from the definition of \(C_x\), and if a measurable set contains \(y\) but not \(x\), its complement contains \(x\) but not \(y\), contradicting \(y\in C_x\). Consequently, \(C_x=C_y\). The distinct sets \(C_x\) partition \(X\). Each is an atom: a measurable subset \(D\subseteq C_x\) containing \(x\) must contain every point belonging to exactly the same measurable sets as \(x\), so \(D=C_x\); a nonempty measurable subset contains some \(y\) and then, by the same reasoning, contains all of \(C_y=C_x\). Also every measurable \(E\) is a union of these classes: if \(x\in E\), then \(C_x\subseteq E\) by the definition of \(C_x\). This proves the needed partition and union properties.
Now suppose \(\mu\) is a measure. Assign \(w_C=\mu(C)\) to each atom. Every measurable \(E\) is a union of finitely many disjoint atoms, so finite additivity, which follows from countable additivity, gives the displayed formula.
Conversely, suppose the displayed formula holds for nonnegative weights \(w_C\). The empty set contains no atoms, so \(\mu(\varnothing)=0\). Let \((E_n)_{n\ge1}\) be pairwise disjoint measurable sets. Each atom \(C\) is either contained in a given \(E_n\) or disjoint from it: since \(C\cap E_n\) is measurable, a nonempty intersection must equal \(C\) by the definition of an atom. Further, \(C\) can be contained in at most one \(E_n\), by disjointness. Thus the atoms contained in \(\bigcup_n E_n\) are exactly the atoms assigned to one of the sets \(E_n\), each at most once. There are finitely many atoms. Therefore the series \(\sum_n\mu(E_n)\) counts each such weight exactly once and no other weight; after the finitely many indices containing atoms have appeared, its partial sums have reached that finite sum, with the value \(+\infty\) allowed. Hence \(\mu(\bigcup_nE_n)=\sum_n\mu(E_n)\). If \(X=\varnothing\), there are no atoms and the formula gives only \(\mu(\varnothing)=0\), which is a measure by the definition. \(\square\)
Worked Example: A Nonnegative Candidate That Fails Additivity
On \(X=\{u,v\}\) with \(\mathcal{M}=\mathcal{P}(X)\), define \(\rho(E)=|E|^2\), where \(|E|\) is the number of points in \(E\). Test the axioms.
The values are \(\rho(\varnothing)=0^2=0\), \(\rho(\{u\})=1^2=1\), \(\rho(\{v\})=1^2=1\), and \(\rho(X)=2^2=4\). Thus the candidate is nonnegative and has the correct empty-set value. But \(\{u\}\) and \(\{v\}\) are disjoint, so additivity would require \(\rho(X)=\rho(\{u\})+\rho(\{v\})\). The left side is \(4\), while the right side is \(1+1=2\); they are unequal. Therefore \(\rho\) is not a measure.
Check: The atoms of this full power set are \(\{u\}\) and \(\{v\}\). If the candidate were a measure, the atom-weight formula would give \(\rho(X)=\rho(\{u\})+\rho(\{v\})=2\), confirming the contradiction with its assigned value \(4\).
Common Pitfalls and a Reliable Workflow
A common mistake is to check only nonnegativity and then assume that a formula defines a measure. For example, on \(X=\{u,v\}\), the candidate \(\eta(E)=1+|E|\) is nonnegative for every \(E\). But \(\eta(\varnothing)=1+0=1\ne0\), so it fails the empty-set axiom. It also fails additivity: the disjoint sets \(\varnothing\) and \(\{u\}\) have union \(\{u\}\), for which \(\eta(\{u\})=1+1=2\), whereas \(\eta(\varnothing)+\eta(\{u\})=1+2=3\). The correction is not to ignore the empty set in additivity; every disjoint sequence may include empty sets, and their measure must be zero.
Another error is to test only one partition of \(X\) and claim additivity for every measurable set. On a finite sigma-algebra, the atom-weight characterization gives a systematic alternative: list the atoms, record the candidate value on each, and check that every measurable set has the sum of the weights of its atoms. If even one set has a different assigned value, the candidate fails. If all values are nonnegative, the empty-set value is zero, and the formula holds for every measurable set, the characterization proves countable additivity.
For a direct verification, use this order:
- Check that the domain is exactly the stated sigma-algebra and that every candidate value belongs to \([0,\infty]\).
- Substitute \(E=\varnothing\) into the formula and verify that the result is zero.
- Identify the measurable atoms, or list the disjoint measurable sets that can occur.
- Check additivity on disjoint sets. On finite \(X\), a sequence of disjoint sets has only finitely many nonempty terms, so finite additivity suffices.
- Include the cases of empty sets and the empty underlying space rather than silently excluding them.
These checks separate three logically distinct requirements. Nonnegativity concerns the range of the candidate; the empty-set axiom fixes its value at one particular set; additivity constrains its values across disjoint sets. Passing two tests does not compensate for failing the third.
Check Your Understanding
For each question, state the measurable space and check the relevant axiom explicitly.
- On \(X=\{1,2,3\}\) with \(\mathcal{M}=\mathcal{P}(X)\), let \(\alpha(E)=2|E|\). Verify the empty-set value and additivity for a disjoint sequence, then compute \(\alpha(X)\).
- On \(X=\{a,b,c,d\}\) with \(\mathcal{M}=\{\varnothing,\{a,b\},\{c,d\},X\}\), what values on the two atoms are required for a measure with value \(7\) on \(X\)? Give one valid choice and check it.
- On a finite \(X\), explain why a pairwise disjoint sequence of measurable sets has only finitely many nonempty terms, and use this to justify the finite-space verification criterion.
- For \(X=\{u,v\}\), does \(\tau(E)=|E|\) define a measure on \(\mathcal{P}(X)\)? Check all disjoint cases, including cases involving the empty set.
- If the underlying set is empty, what is its sigma-algebra, and what value must a measure assign to its only member?