What Uniqueness Means
A Hahn decomposition divides a measurable space into a positive set and a negative set for a signed measure. The division need not be unique point by point: a part of the space on which every measurable subset has signed measure zero can be assigned to either side. The precise uniqueness statement is therefore not that two positive regions must be equal, but that their difference is null in this stronger sense.
Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu:\mathcal{M}\to\mathbb{R}\) is a finite-valued, countably additive signed measure. All sets discussed are in \(\mathcal{M}\), and all set equalities are exact. A measurable set \(P\) is positive if \(\nu(E)\ge 0\) for every measurable \(E\subseteq P\). A measurable set \(N\) is negative if \(\nu(E)\le 0\) for every measurable \(E\subseteq N\). These conditions apply to every measurable subset, not just to \(P\) or \(N\) themselves.
The uniqueness theorem compares two decompositions of the same signed measure on the same measurable space. It does not assert that either component is uniquely determined as an exact set. Instead, it identifies the only possible ambiguity: a \(\nu\)-null set can be transferred between the positive and negative regions.
The Uniqueness Theorem
Proof. First let \(E\) be any measurable subset of \(P\cap N'\). Since \(P\) is positive, \(\nu(E)\ge 0\). Since \(N'\) is negative, \(\nu(E)\le 0\). Thus \(\nu(E)=0\). This holds for every measurable \(E\subseteq P\cap N'\), so \(P\cap N'\) is \(\nu\)-null. Now let \(E\) be any measurable subset of \(N\cap P'\). Negativity of \(N\) gives \(\nu(E)\le 0\), while positivity of \(P'\) gives \(\nu(E)\ge 0\). Hence \(\nu(E)=0\) for every such \(E\), and \(N\cap P'\) is \(\nu\)-null.
Because \(P\cup N=X\) and \(P'\cup N'=X\), with each pair disjoint, the symmetric difference \(P\,\triangle\,P'\) is the disjoint union \((P\cap N')\cup(N\cap P')\). To check that this union is \(\nu\)-null, take any measurable \(E\subseteq P\,\triangle\,P'\). Write \(E_1=E\cap(P\cap N')\) and \(E_2=E\cap(N\cap P')\). These are disjoint measurable sets, each is a subset of a \(\nu\)-null set, and \(E=E_1\cup E_2\). Therefore \(\nu(E)=\nu(E_1)+\nu(E_2)=0+0=0\). This proves that \(P\,\triangle\,P'\) is \(\nu\)-null. Interchanging the roles of positive and negative regions gives the same conclusion for \(N\,\triangle\,N'\). \(\square\)
The proof relies on the defining inequalities for positive and negative sets. On their overlap, each measurable subset must have signed measure both at least zero and at most zero, so it must have signed measure exactly zero. The conclusion is stronger than merely showing that the overlap itself has signed measure zero.
Finite Examples
Worked Example: Moving a Zero-Value Point
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the power set of \(X\), and assign the singleton values \(\nu(\{a\})=4\), \(\nu(\{b\})=-3\), and \(\nu(\{c\})=0\). For each subset \(E\), \(\nu(E)\) is the sum of the singleton values in \(E\). Verify two Hahn decompositions and compare their positive regions.
The values of all measurable sets are \(\nu(\varnothing)=0\), \(\nu(\{a\})=4\), \(\nu(\{b\})=-3\), \(\nu(\{c\})=0\), \(\nu(\{a,b\})=4-3=1\), \(\nu(\{a,c\})=4+0=4\), \(\nu(\{b,c\})=-3+0=-3\), and \(\nu(X)=4-3+0=1\).
Take \(P=\{a,c\}\) and \(N=\{b\}\). Every subset of \(P\) has value either \(0\) or \(4\), so \(P\) is positive; the subsets of \(N\) have values \(0\) and \(-3\), so \(N\) is negative. Another decomposition is \(P'=\{a\}\) and \(N'=\{b,c\}\). The subsets of \(P'\) have values \(0\) and \(4\); the subsets of \(N'\) have values \(0\), \(-3\), and \(-3+0=-3\). Thus \(P'\) is positive and \(N'\) is negative. The two positive regions differ by \(\{c\}\). Its only measurable subsets are \(\varnothing\) and \(\{c\}\), of values \(0\) and \(0\), so the difference is \(\nu\)-null. Check: the common positive part is \(\{a\}\), and the common negative part is \(\{b\}\); only the zero-value point changes sides.
Worked Example: A Null Measurable Block
Let \(X=\{a,z_1,z_2,b\}\), and let \(\mathcal{M}\) consist of all unions of the three blocks \(A=\{a\}\), \(Z=\{z_1,z_2\}\), and \(B=\{b\}\). Set \(\nu(A)=2\), \(\nu(Z)=0\), and \(\nu(B)=-5\), extending \(\nu\) additively to \(\mathcal{M}\). Compare the decompositions that assign \(Z\) to opposite sides.
The measurable sets are the eight unions of \(A,Z,B\). Their values are \(\nu(\varnothing)=0\), \(\nu(A)=2\), \(\nu(Z)=0\), \(\nu(B)=-5\), \(\nu(A\cup Z)=2+0=2\), \(\nu(A\cup B)=2-5=-3\), \(\nu(Z\cup B)=0-5=-5\), and \(\nu(X)=2+0-5=-3\).
For \(P=A\cup Z\) and \(N=B\), the measurable subsets of \(P\) are \(\varnothing,A,Z,A\cup Z\), with values \(0,2,0,2\); all are nonnegative. The subsets of \(N\) have values \(0,-5\), so all are nonpositive. For \(P'=A\) and \(N'=Z\cup B\), the subsets of \(P'\) have values \(0,2\), and those of \(N'\) have values \(0,0,-5,-5\); these have the required signs as well. The symmetric difference of the positive regions is \(Z\). Its measurable subsets are \(\varnothing\) and \(Z\), since \(z_1\) and \(z_2\) cannot be separated by a set in \(\mathcal{M}\); their values are both zero. Thus \(Z\) is \(\nu\)-null even though the individual singletons are not measurable. Check: the two displayed decompositions each cover \(X\), have disjoint components, and differ only on \(Z\).
Worked Example: When the Decomposition Is Exact
Let \(X=\{r,s\}\), let \(\mathcal{M}\) be its power set, and set \(\nu(\{r\})=3\) and \(\nu(\{s\})=-2\). Verify the Hahn decomposition and show directly that no alternative positive region is possible.
The four set values are \(\nu(\varnothing)=0\), \(\nu(\{r\})=3\), \(\nu(\{s\})=-2\), and \(\nu(X)=3-2=1\). Thus \(P=\{r\}\) is positive, because its subsets have values \(0,3\), and \(N=\{s\}\) is negative, because its subsets have values \(0,-2\). If a Hahn decomposition placed \(r\) in its negative region, that region would contain the measurable subset \(\{r\}\) of value \(3>0\), contradicting negativity. If it placed \(s\) in its positive region, that region would contain the measurable subset \(\{s\}\) of value \(-2<0\), contradicting positivity. Therefore every Hahn decomposition has positive region \(\{r\}\) and negative region \(\{s\}\). Check: there is no nonempty \(\nu\)-null set here: each nonempty measurable set contains either \(\{r\}\) or \(\{s\}\), whose signed measure is nonzero.
A Common Mistake: Zero Total Is Not Null
It is incorrect to infer that a measurable set is \(\nu\)-null from the single calculation \(\nu(E)=0\). For a concrete example, let \(X=\{u,v\}\), let \(\mathcal{M}\) be the power set, and set \(\nu(\{u\})=1\), \(\nu(\{v\})=-1\). Then \(\nu(X)=1-1=0\), but \(X\) is not \(\nu\)-null: its measurable subset \(\{u\}\) has value \(1\), and its measurable subset \(\{v\}\) has value \(-1\). The corrected test is to check \(\nu(E)=0\) for every measurable \(E\subseteq X\). In the uniqueness theorem, that stronger property follows because every such subset lies in both a positive set and a negative set.
The conclusion is useful whenever a Hahn decomposition is constructed in more than one way. One may compare the resulting regions without demanding exact equality: any discrepancy is invisible to the signed measure on every measurable subset. The theorem does not say that an arbitrary set of signed measure zero can be reassigned; reassignment is justified only for a set that is \(\nu\)-null in the stated sense.
Check Your Understanding
Use the definitions and the theorem to answer these questions. When testing whether a set is null, consider every measurable subset, not only the set itself.
- If \(E\subseteq P\cap N'\) is measurable, which two inequalities apply to \(\nu(E)\), and what do they imply?
- Why does \(\nu(P\cap N')=0\) alone not suffice to show that \(P\cap N'\) is \(\nu\)-null?
- In the first worked example, list the measurable subsets of the symmetric difference and their signed measures.
- In the second worked example, why are \(\{z_1\}\) and \(\{z_2\}\) not among the measurable subsets that need to be tested?
- Give the set identity that expresses \(P\,\triangle\,P'\) as a union of intersections involving the two decompositions.