Hahn Decompositions for Atomic Measures

Separating the Signs of Atomic Weights

On a countable discrete space, every singleton is measurable, so a signed measure is controlled by its values on individual points. This makes Hahn decompositions concrete: put the points with positive weight on the positive side and the points with negative weight on the negative side. Points of weight zero contribute nothing and can go on either side. The details matter for infinite spaces, where absolute summability ensures that sums over arbitrary subsets are well-defined and that the resulting set function is countably additive.

Throughout, let \(S\) be a finite or countably infinite set, equipped with the sigma-algebra \(2^S\) of all subsets and counting measure \(\kappa\). Enumerate \(S\) as \(\{s_i:i\in I\}\), where \(I\) is finite or \(I=\mathbb{N}\). We consider real-valued signed measures \(\nu\) given by weights \(a_i\in\mathbb{R}\) satisfying \(\sum_{i\in I}|a_i|<\infty\). This condition is automatic when \(S\) is finite; on an infinite space, it is the summability condition used here. For \(A\subseteq S\), define

\nu(A)=\sum_{\{i\in I:s_i\in A\}}a_i.

The series converges absolutely for every \(A\), since the sum of the absolute values over \(A\) is at most \(\sum_{i\in I}|a_i|\). In particular, the value does not depend on the order in which the points are listed. This formula includes the zero measure, the empty space, and finite spaces.

Definition: A Hahn decomposition of the measurable space \((S,2^S)\) for the signed measure \(\nu\) is a pair of disjoint sets \(P,N\subseteq S\) with \(P\cup N=S\), such that \(\nu(E)\ge0\) for every \(E\subseteq P\) and \(\nu(F)\le0\) for every \(F\subseteq N\). Thus \(P\) is a positive set and \(N\) is a negative set. The inequalities include equality.

The Sign-Splitting Construction

Theorem: Let \(S\), \(2^S\), and \(\nu\) be as above. Define \(P_+=\{s_i:a_i>0\}\), \(N_-=\{s_i:a_i<0\}\), and \(Z=\{s_i:a_i=0\}\). Then \((P_+\cup Z,N_-)\) is a Hahn decomposition. More generally, for every \(Z_+\subseteq Z\), the pair \((P_+\cup Z_+,N_-\cup(Z\setminus Z_+))\) is a Hahn decomposition.

Proof. Fix any subset \(Z_+\subseteq Z\), and set \(P=P_+\cup Z_+\) and \(N=N_-\cup(Z\setminus Z_+)\). These sets are disjoint and their union is \(S\): each weight is positive, negative, or zero, and each zero-weight point is assigned to exactly one side.

Let \(E\subseteq P\). Every weight \(a_i\) with \(s_i\in E\) is nonnegative. The series defining \(\nu(E)\) is absolutely convergent, and each of its terms is nonnegative, so \(\nu(E)\ge0\). This also holds when \(E=\varnothing\), since the empty sum is zero. If \(F\subseteq N\), every weight in the absolutely convergent series defining \(\nu(F)\) is nonpositive, so \(\nu(F)\le0\); again the empty set gives value zero. Therefore \((P,N)\) is a Hahn decomposition. \(\square\)

Absolute convergence is what allows the sign argument to apply to every subset, including infinite subsets. The result concerns the signed measure on \((S,2^S)\); no claim about a different sigma-algebra or about a non-atomic part of a measure is needed.

Worked Example: A Finite Space with Positive, Negative, and Zero Weights

Let \(S=\{a,b,c,d,e\}\), with all subsets measurable, and give the points weights \(a_a=4\), \(a_b=-3\), \(a_c=0\), \(a_d=-1\), and \(a_e=2\). Construct a Hahn decomposition and calculate the signed measures of its two sides.

The strictly positive weights occur at \(a,e\), the strictly negative weights at \(b,d\), and the zero weight at \(c\). Assign \(c\) to the positive side. Then \(P=\{a,c,e\}\) and \(N=\{b,d\}\). The measure of \(P\) is \(4+0+2=6\), while the measure of \(N\) is \(-3+(-1)=-4\).

To check the defining conditions, any subset \(E\subseteq P\) has measure equal to a sum chosen from \(4,0,2\), so \(\nu(E)\ge0\). In particular, the possible values are \(0,2,4,6\). Any subset \(F\subseteq N\) has measure equal to a sum chosen from \(-3,-1\), so \(\nu(F)\le0\); the possible values are \(0,-1,-3,-4\). The two sets partition \(S\), as required.

Check: Directly summing all five weights gives \(4-3+0-1+2=2\). Summing the two side values gives \(6+(-4)=2\), the same total.

Which Atoms Can Belong to Each Side?

The construction above gives a decomposition, and the singleton sets show that it also describes every possible decomposition. This is a useful distinction: the positive and negative weights have forced locations, while only the zero-weight atoms allow a choice.

Proposition: For the signed measure \(\nu\) on \((S,2^S)\), a partition \((P,N)\) is a Hahn decomposition if and only if every point with positive weight belongs to \(P\), every point with negative weight belongs to \(N\), and every zero-weight point is assigned to exactly one of \(P,N\).

Proof. First suppose \((P,N)\) is a Hahn decomposition. If \(a_i>0\) but \(s_i\in N\), then the subset \(\{s_i\}\subseteq N\) has measure \(\nu(\{s_i\})=a_i>0\). This contradicts the requirement that every subset of \(N\) have measure at most zero. Hence every positive-weight point lies in \(P\). If \(a_i<0\) but \(s_i\in P\), the singleton \(\{s_i\}\subseteq P\) has measure \(a_i<0\), contradicting the requirement that every subset of \(P\) have measure at least zero. Thus every negative-weight point lies in \(N\). Since \((P,N)\) is a partition, each zero-weight point lies in exactly one side.

Conversely, suppose the stated assignments hold. Every weight of a point in \(P\) is nonnegative, and every weight of a point in \(N\) is nonpositive. For any \(E\subseteq P\), the absolutely convergent sum defining \(\nu(E)\) is a sum of nonnegative terms and is therefore nonnegative. For any \(F\subseteq N\), the sum defining \(\nu(F)\) is a sum of nonpositive terms and is therefore nonpositive. Thus \((P,N)\) is a Hahn decomposition. If \(S=\varnothing\), both parts are empty and the same conditions and conclusions hold vacuously. \(\square\)

Worked Example: A Countably Infinite Atomic Space

Let \(S=\mathbb{N}\), with sigma-algebra \(2^{\mathbb{N}}\) and counting measure \(\kappa\). Define the weights by \(a_n=(-1)^{n+1}2^{-n}\). Verify summability, construct a Hahn decomposition, and compute the measure of each side and of the whole space.

The absolute weights sum to \(\sum_{n=1}^{\infty}2^{-n}=1\), so the required absolute summability holds. The odd indices have positive weights and the even indices have negative weights. Therefore take \(P=\{1,3,5,\ldots\}\) and \(N=\{2,4,6,\ldots\}\). The positive-side measure is

\nu(P)=\sum_{k=0}^{\infty}2^{-(2k+1)} =\frac{1/2}{1-1/4} =\frac{2}{3}.

The negative-side measure is

\nu(N)=-\sum_{k=1}^{\infty}2^{-2k} =-\frac{1/4}{1-1/4} =-\frac{1}{3}.

Every subset of \(P\) has a sum of positive terms and hence nonnegative measure; every subset of \(N\) has a sum of negative terms and hence nonpositive measure. To check the total independently, the full series is geometric with first term \(1/2\) and ratio \(-1/2\), so

\nu(S)=\sum_{n=1}^{\infty}(-1)^{n+1}2^{-n} =\frac{1/2}{1-(-1/2)} =\frac{1}{3} =\frac{2}{3}-\frac{1}{3}.

Thus the whole-space value agrees with the sum of the two side values, and the decomposition satisfies the required sign conditions on all subsets, not only on \(P\) and \(N\) themselves.

Worked Example: Zero-Weight Atoms and All Possible Decompositions

Let \(S=\{u,v,w,r,t\}\), with weights \(a_u=3\), \(a_v=-1\), \(a_w=-2\), \(a_r=0\), and \(a_t=0\). Find all Hahn decompositions and verify the side values.

The positive-weight point \(u\) must lie in \(P\), while the negative-weight points \(v,w\) must lie in \(N\). Each of \(r,t\) can be assigned independently. Consequently, the four possible positive sides are \(\{u\}\), \(\{u,r\}\), \(\{u,t\}\), and \(\{u,r,t\}\); in each case \(N\) is the complement in \(S\). The value of each positive side is \(3\), since adding either zero-weight point leaves the sum unchanged. The value of each negative side is \(-1-2=-3\), also unchanged by any zero-weight assignments.

For any subset of any such \(P\), the available weights are \(3,0,0\), so its measure is either \(0\) or \(3\). For any subset of \(N\), the available weights are \(-1,-2\), so its measure is \(0\), \(-1\), \(-2\), or \(-3\). This verifies the sign conditions for all four partitions. The total is \(3-1-2+0+0=0\), and the two side values sum to \(3+(-3)=0\).

Common Pitfalls and Why Summability Matters

Pitfall: deciding the sign of a whole space from its total value. For weights \(3,-2,-2\) on a three-point space, the total is \(3-2-2=-1\). It would be incorrect to call the whole space a negative set: its subset consisting of the point of weight \(3\) has positive measure. The correction is to check the sign of every subset, or, in this atomic setting, to separate the signs of the individual weights. Here the positive point belongs in \(P\), and the two negative points belong in \(N\).

Pitfall: treating a zero-weight atom as forced onto one side. A zero-weight singleton has measure zero, so it satisfies both the nonnegative and nonpositive inequalities. As the finite example shows, it can be put in either part without changing any subset’s sign. By contrast, a positive-weight atom cannot be placed in \(N\), and a negative-weight atom cannot be placed in \(P\); the singleton tests prove both restrictions.

Pitfall: omitting absolute summability in the countable case. A convergent series in one chosen order is not enough to define a finite signed measure on all subsets. For instance, the weights \(a_n=(-1)^{n+1}/n\) have a convergent alternating sum in the displayed order, but the positive weights over the odd indices have divergent sum, and the negative weights over the even indices have divergent sum in magnitude. The proposed values on those subsets are not finite real numbers. The absolute-summability hypothesis prevents this problem and makes every subset sum well-defined, independently of its enumeration.

For countable atomic signed measures, the Hahn decomposition is therefore a direct consequence of the individual weights, but the subset conditions and the convergence assumption are essential parts of the argument. This sign split identifies the positive and negative regions; it does not by itself construct the positive and negative measures considered in the Jordan decomposition.

Check Your Understanding

Use the definition of a Hahn decomposition, and test the sign of arbitrary subsets rather than only the whole-space value.

On a four-point space with weights \(2,-5,0,1\), give one Hahn decomposition and calculate the measure of each side.
For a countable atomic signed measure with absolutely summable weights, explain why every subset of the strictly positive-weight atoms has nonnegative measure.
Prove that a point of negative weight cannot belong to the positive side of any Hahn decomposition.
If a finite space has three zero-weight atoms, how many assignments of those atoms to the two sides are possible once all nonzero weights have been assigned correctly?
Why does convergence of the full series in one order fail to replace absolute summability in the countable construction?

Hahn Decompositions for Atomic Measures

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