Constructing the Jordan Parts from a Hahn Decomposition

From a Hahn Decomposition to Two Measures

A signed measure can assign both positive and negative values to measurable sets, so it is not itself a measure in the usual nonnegative sense. A Hahn decomposition separates the underlying space into a positive set and a negative set: on measurable subsets of the first, the signed measure is nonnegative; on measurable subsets of the second, it is nonpositive. Restricting to these two regions gives nonnegative set functions. The negative region requires a sign change, so that its restriction, too, takes nonnegative values.

Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a signed measure on it. We do not assume that \(\nu\) or the measures constructed below are finite or sigma-finite. A signed measure is countably additive and may take extended real values, but it does not take both \(+\infty\) and \(-\infty\). A (nonnegative) measure takes values in \([0,\infty]\). All equalities between set functions below hold for every \(E\in\mathcal{M}\).

The Hahn decomposition theorem guarantees such a pair for a signed measure; that theorem was established earlier. Here, fix any Hahn decomposition \((P,N)\). Define the positive and negative parts by restricting \(\nu\) to the corresponding sets, with a minus sign on the negative set.

Both values are nonnegative: \(E\cap P\) is a measurable subset of \(P\), and \(E\cap N\) is a measurable subset of \(N\). The negative sign in \(\nu^-\) therefore changes a nonpositive value into a nonnegative one. The construction does not claim that either part is finite: a value \(+\infty\) is allowed.

Why the Restrictions Are Measures

Proof. First, \(\nu^+(\varnothing)=\nu(\varnothing\cap P)=\nu(\varnothing)=0\), and \(\nu^+(E)\ge0\) for every \(E\in\mathcal{M}\) because \(E\cap P\subseteq P\) is measurable. Let \((E_k)_{k\ge1}\) be pairwise disjoint measurable sets. Then \((E_k\cap P)_{k\ge1}\) are pairwise disjoint, and their union is \((\bigcup_{k\ge1}E_k)\cap P\). Countable additivity of \(\nu\) gives \(\nu^+(\bigcup_{k\ge1}E_k)=\nu((\bigcup_{k\ge1}E_k)\cap P)=\sum_{k\ge1}\nu(E_k\cap P)=\sum_{k\ge1}\nu^+(E_k)\). Every term in this sum is nonnegative, so the sum is well-defined in \([0,\infty]\). Thus \(\nu^+\) is a measure.

For the negative part, \(\nu^-(\varnothing)=-\nu(\varnothing)=0\), and \(\nu^-(E)\ge0\) because \(\nu(E\cap N)\le0\). For the same disjoint sets, the sets \(E_k\cap N\) are disjoint and have union \((\bigcup_{k\ge1}E_k)\cap N\). Countable additivity gives \(\nu^-(\bigcup_{k\ge1}E_k)=-\nu((\bigcup_{k\ge1}E_k)\cap N)=-\sum_{k\ge1}\nu(E_k\cap N)=\sum_{k\ge1}\nu^-(E_k)\). The values \(-\nu(E_k\cap N)\) are nonnegative, so this series is well-defined in \([0,\infty]\). Hence \(\nu^-\) is a measure as well. \(\square\)

The restrictions also retain a useful location property. If \(E\subseteq N\), then \(E\cap P=\varnothing\), so \(\nu^+(E)=0\). If \(E\subseteq P\), then \(E\cap N=\varnothing\), so \(\nu^-(E)=0\). These identities follow directly from the definitions and \(\nu(\varnothing)=0\); they say where each part can have nonzero values.

Recovering the Signed Measure

Proof. Since \(P\) and \(N\) are disjoint and cover \(X\), the sets \(E\cap P\) and \(E\cap N\) are disjoint and their union is \(E\). Finite additivity of the signed measure therefore gives \(\nu(E)=\nu(E\cap P)+\nu(E\cap N)=\nu^+(E)-\nu^-(E)\). For the final expression to be meaningful, both parts cannot be infinite on \(E\). If they were, then \(\nu(E\cap P)=+\infty\) and \(\nu(E\cap N)=-\infty\), and additivity on the disjoint partition of \(E\) would require adding \(+\infty\) and \(-\infty\). That is excluded by the definition of a signed measure. Thus at most one part is infinite on \(E\), and the displayed difference is defined. \(\square\)

This reconstruction statement is about the given Hahn decomposition and its restrictions; it does not require the whole space to have finite measure. If both parts are finite on \(E\), their ordinary real difference gives \(\nu(E)\). If one is infinite, the other is finite, and the extended-real difference is still unambiguous.

Zero Values and a Sign Error to Avoid

A zero-valued region can be placed on either side of a Hahn decomposition without changing the values of the constructed parts there: if \(\nu\) is zero on all measurable subsets of that region, restricting to any such subset contributes zero. For a concrete instance, let \(X=\{0,1,2\}\) with counting measure as the underlying measure, and define the signed measure by \(\nu(\{0\})=1\), \(\nu(\{1\})=-1\), and \(\nu(\{2\})=0\). Both \(P_1=\{0,2\},N_1=\{1\}\) and \(P_2=\{0\},N_2=\{1,2\}\) are Hahn decompositions: every subset of each positive set has value \(0\) or \(1\), and every subset of each negative set has value \(0\) or \(-1\). In either construction, the positive part assigns mass \(1\) to \(\{0\}\) and \(0\) to \(\{1\}\) and \(\{2\}\); the negative part assigns mass \(1\) to \(\{1\}\) and \(0\) to the other singletons. For \(E=X\), the parts are \(1\) and \(1\), giving \(\nu(X)=1-1=0\), as also follows by adding the original atomic values.

Common mistake: defining the negative part as \(\nu(E\cap N)\) without changing its sign. In the density example, this incorrect rule would give \(\nu((1,3])=-2\cdot2=-4\), which is negative and therefore cannot be a measure. The correction is to define \(\nu^-(E)=-\nu(E\cap N)\); then \(\nu^-((1,3])=-(-4)=4\), a nonnegative value, and \(\nu(X)=3-4=-1\).

The construction turns a sign-changing set function into two ordinary measures concentrated on opposite sign regions. It is the starting point for the Jordan decomposition theorem, which states the decomposition and its uniqueness in a broader structural form. Here the essential work is establishing directly that the restrictions are measures and that their difference recovers \(\nu\).