Finite Signed Measures as Lists of Weights
On a finite space, a signed measure can be calculated without evaluating a limit or an integral: each measurable piece contributes one real weight, and the value of a set is the sum of the weights it contains. The measurable pieces that cannot be split further are the relevant atoms. Keeping track of these atoms also makes an important distinction visible: a set can have signed measure zero because positive and negative weights cancel, even though some of its subsets have nonzero signed measure.
Throughout this tutorial, \((X,\mathcal{M})\) is a finite measurable space: \(X\) is finite and \(\mathcal{M}\) is a sigma-algebra of subsets of \(X\). The measure \(\nu\) is a finite, real-valued signed measure on \(\mathcal{M}\), so it is countably additive and \(\nu(A)\in\mathbb{R}\) for every \(A\in\mathcal{M}\). In particular, \(\nu(\varnothing)=0\) and \(\nu\) is finitely additive on disjoint measurable sets. No assumption that \(\mathcal{M}=\mathcal{P}(X)\) is made.
Every measurable set is a union of atoms. Thus, if the atoms are \(C_1,\ldots,C_k\) and \(A\in\mathcal{M}\) is the union of the atoms indexed by \(I\subseteq\{1,\ldots,k\}\), finite additivity gives
When \(\mathcal{M}=\mathcal{P}(X)\), each atom is a singleton \(\{x\}\), and the weights are the point weights \(w_x=\nu(\{x\})\). If the sigma-algebra is smaller, a point need not itself be measurable. In that case, weights belong to the atoms, not to individual points inside them.
Positive, Negative, and Zero-Value Sets
The tests for positive and negative sets concern every measurable subset, not just the value of the entire set. In a finite space, it suffices to check the weights of the atoms it contains.
Proof. Suppose first that \(P\) is positive. If \(C_j\subseteq P\), then \(C_j\) is a measurable subset of \(P\). The definition of a positive set gives \(w(C_j)=\nu(C_j)\ge 0\). Conversely, suppose every atom contained in \(P\) has nonnegative weight. Any measurable \(B\subseteq P\) is a union of some of those atoms. Finite additivity writes \(\nu(B)\) as a finite sum of their weights, so \(\nu(B)\ge 0\). This proves both directions for positive sets. For negative sets, if \(N\) is negative, applying its definition to each atom \(C_j\subseteq N\) gives \(w(C_j)\le 0\). Conversely, if all such weights are nonpositive, each measurable \(B\subseteq N\) is a union of atoms with nonpositive weights, and their finite sum satisfies \(\nu(B)\le 0\). Finally, every measurable \(A\) is a disjoint union of its atoms, so finite additivity gives \(\nu(A)=\sum_{C_j\subseteq A}w(C_j)\). This value is zero exactly when that sum is zero. If \(X=\varnothing\), there are no atoms, the only measurable set is \(\varnothing\), and the assertions hold with the empty sum equal to zero. \(\square\)
The inequalities in these tests are inclusive: an atom of weight zero may belong to a positive set and to a negative set. The zero-value test, by contrast, asks whether the total sum is exactly zero. It does not require each weight to be zero.
Worked Example: Testing Sign and Zero Value on a Five-Point Space
Let \(X=\{a,b,c,d,e\}\), \(\mathcal{M}=\mathcal{P}(X)\), and assign the point weights \(w_a=4\), \(w_b=-3\), \(w_c=0\), \(w_d=2\), and \(w_e=-1\). For each \(A\subseteq X\), \(\nu(A)\) is the sum of the weights of its points.
Consider \(P=\{a,c,d\}\). Its weights are \(4,0,2\), all at least zero. Every measurable subset of \(P\) has a sum of some of those three weights, so every such sum is nonnegative; in particular, \(\nu(P)=4+0+2=6\), and \(P\) is positive. Consider \(N=\{b,c,e\}\). Its weights are \(-3,0,-1\), all at most zero. Every measurable subset has a nonpositive sum, and \(\nu(N)=-3+0-1=-4\), so \(N\) is negative. The point \(c\), whose weight is zero, is allowed in both sign tests.
The set \(A=\{a,b,e\}\) has value \(4+(-3)+(-1)=0\). It is a zero-value set, but it is neither positive nor negative: the subset \(\{b\}\subseteq A\) has value \(-3<0\), and the subset \(\{a\}\subseteq A\) has value \(4>0\). Check: adding the weights of all five points gives \(4+(-3)+0+2+(-1)=2\). Also, \(P\) contains only nonnegative weights and \(N\) only nonpositive weights, as required by the theorem.
Atoms May Contain More Than One Point
When a finite sigma-algebra does not contain every subset, it is the atom weights—not hypothetical point weights—that determine the measure of every measurable set. The next example lists all measurable sets so that the finite-sum rule can be checked directly.
Worked Example: Computing Values on a Coarse Sigma-Algebra
Let \(X=\{a,b,c,d,e\}\), and let \(\mathcal{M}\) consist of all unions of \(C_1=\{a,b\}\), \(C_2=\{c\}\), and \(C_3=\{d,e\}\). Assign the atom weights \(\nu(C_1)=5\), \(\nu(C_2)=-2\), and \(\nu(C_3)=-3\). There are \(2^3=8\) measurable sets, since each is formed by including or excluding each of the three atoms.
Their values, computed by adding the included atom weights, are
- \(\nu(\varnothing)=0\).
- \(\nu(C_1)=5\), \(\nu(C_2)=-2\), and \(\nu(C_3)=-3\).
- \(\nu(C_1\cup C_2)=5+(-2)=3\).
- \(\nu(C_1\cup C_3)=5+(-3)=2\).
- \(\nu(C_2\cup C_3)=(-2)+(-3)=-5\).
- \(\nu(X)=5+(-2)+(-3)=0\).
The positive sets are exactly \(\varnothing\) and \(C_1\): a positive set cannot contain either negative-weight atom, and either allowed choice has nonnegative weight. The negative sets are exactly \(\varnothing\), \(C_2\), \(C_3\), and \(C_2\cup C_3\): these are precisely the unions made only from the two nonpositive atoms. The zero-value sets are \(\varnothing\) and \(X\). Check: the table accounts for all eight unions, and \(X\) has value \(5-2-3=0\) even though its individual atom weights are not all zero. No set such as \(\{a\}\) can be tested separately, because \(\{a\}\notin\mathcal{M}\).
Cancellation and Sets of Measure Zero
A zero-value set need not have zero value on all its measurable subsets. To avoid conflating these notions, call a measurable set \(A\) a null set for \(\nu\) if \(\nu(B)=0\) for every measurable \(B\subseteq A\). In a finite measurable space, the atomic criterion makes this stronger condition exact: \(A\) is a null set if and only if every atom it contains has weight zero. Indeed, testing \(B\) equal to each included atom proves necessity; if all included atoms have weight zero, every measurable subset of \(A\) is a union of zero-weight atoms, proving sufficiency.
Worked Example: Enumerating Zero-Value Sets
Let \(X=\{p,q,r,s,z\}\), \(\mathcal{M}=\mathcal{P}(X)\), and give the points weights \(w_p=2\), \(w_q=-2\), \(w_r=1\), \(w_s=-1\), and \(w_z=0\). A subset has value zero exactly when it includes either both \(p,q\) or neither, and either both \(r,s\) or neither; the point \(z\) may be included or omitted independently.
For the pair \(\{p,q\}\), the possible contributions are \(0\) when neither is selected and \(2+(-2)=0\) when both are selected. Selecting exactly one contributes either \(2\) or \(-2\), not zero. For \(\{r,s\}\), selecting neither contributes \(0\), selecting both contributes \(1+(-1)=0\), and selecting exactly one contributes either \(1\) or \(-1\), not zero. The point \(z\) contributes \(0\) in either case. Thus the eight zero-value sets are \(\varnothing\), \(\{z\}\), \(\{p,q\}\), \(\{p,q,z\}\), \(\{r,s\}\), \(\{r,s,z\}\), \(\{p,q,r,s\}\), and \(\{p,q,r,s,z\}\).
Check: the choices for the first pair are “neither” or “both,” giving two options; the second pair also has two options; and \(z\) has two options. The total is \(2\cdot2\cdot2=8\), matching the list. The set \(\{p,q\}\) has value \(2+(-2)=0\), but its subset \(\{p\}\) has value \(2\), so \(\{p,q\}\) is not a null set. By contrast, \(\{z\}\) is a null set: its only subsets are \(\varnothing\) and \(\{z\}\), both of value zero.
A Common Sign-Test Mistake
A common mistake is to decide that a set is positive merely because its total signed measure is nonnegative. For example, take \(X=\{u,v\}\) with the full sigma-algebra and weights \(w_u=4\) and \(w_v=-3\). The calculation \(\nu(\{u,v\})=4+(-3)=1\ge0\) is correct, but it does not show that \(\{u,v\}\) is positive. Its measurable subset \(\{v\}\) has value \(-3<0\), which violates the definition. The correction is to inspect every included atom: a positive set can contain no negative-weight atom, and a negative set can contain no positive-weight atom. A total value of zero likewise does not by itself imply that all subsets have value zero.
These finite calculations provide a direct way to inspect sign structure before more general results about signed measures are needed. The method is always the same: find the measurable atoms, record their real weights, and use finite sums. It applies even when some weights vanish, when cancellations produce zero-value sets, and when the sigma-algebra is too coarse to measure individual points.
Check Your Understanding
For each question, use the atoms of the stated finite measurable space and apply the inclusive sign inequalities from the definitions.
- On a finite space with full sigma-algebra and weights \(3,-1,0\), list all positive singleton atoms and all negative singleton atoms.
- If a measurable set is a union of atoms with weights \(2,-2,5\), what condition on the chosen atoms makes its value zero?
- In a finite measurable space, why must every positive set contain no atom of strictly negative weight?
- Suppose an atom contains two points but is the smallest nonempty measurable set. Can the weight of either point be inferred as the value of a measurable singleton?
- Give a zero-value set that is not a null set using weights \(4\) and \(-4\), and identify a measurable subset witnessing the distinction.