From a Sign Partition to Two Positive Measures
A signed measure can assign positive values to some measurable sets and negative values to others. The Jordan decomposition separates these contributions into two positive measures. The separation is not made by taking the positive or negative part of the number \(\nu(E)\) independently for each set \(E\). Instead, it uses a measurable partition of the space into regions on which the signed measure has a fixed sign.
Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a finite signed measure: \(\nu\) is real-valued and countably additive, and there is a finite constant \(C\) such that, for every finite measurable partition \(E_1,\ldots,E_k\) of \(X\), \(\sum_{j=1}^k|\nu(E_j)|\le C\). No completeness or sigma-finiteness assumption is needed. A positive measure is a countably additive set function with values in \([0,\infty]\). Two positive measures \(\alpha,\beta\) are mutually singular, written \(\alpha\perp\beta\), if some \(A\in\mathcal{M}\) satisfies \(\alpha(X\setminus A)=0\) and \(\beta(A)=0\).
Fix a Hahn decomposition \((P,N)\). The positive part will retain \(\nu\) on \(P\) and assign zero to the negative region. The negative part will reverse the sign of \(\nu\) on \(N\) and assign zero to \(P\). These definitions make sense on every measurable set, not just on the two pieces of the partition.
The Jordan Decomposition Theorem
Proof. We verify each assertion. First, \(\nu^+(\varnothing)=\nu(\varnothing)=0\). If \(E\in\mathcal{M}\), then \(E\cap P\subseteq P\), so the defining property of a positive set gives \(\nu^+(E)=\nu(E\cap P)\ge0\). For disjoint measurable sets \(E_1,E_2,\ldots\), the sets \(E_j\cap P\) are disjoint and their union is \((\bigcup_j E_j)\cap P\). Countable additivity of \(\nu\) therefore gives $$ \nu^+\left(\bigcup_{j=1}^{\infty}E_j\right) =\nu\left(\bigcup_{j=1}^{\infty}(E_j\cap P)\right) =\sum_{j=1}^{\infty}\nu(E_j\cap P) =\sum_{j=1}^{\infty}\nu^+(E_j). $$ All terms in this series are nonnegative. Thus \(\nu^+\) is a positive measure. Likewise, \(\nu^-(\varnothing)=0\), and \(E\cap N\subseteq N\) implies \(\nu^-(E)=-\nu(E\cap N)\ge0\). Countable additivity of \(\nu\) on the disjoint sets \(E_j\cap N\), followed by multiplication by \(-1\), gives \(\nu^-(\bigcup_j E_j)=\sum_j\nu^-(E_j)\). Hence \(\nu^-\) is also a positive measure.
To check finiteness, apply the bound in the definition of a finite signed measure to the partition \(P,N\), omitting an empty piece if necessary. It gives $$ \nu^+(X)+\nu^-(X)=\nu(P)-\nu(N)=|\nu(P)|+|\nu(N)|\le C. $$ Here \(\nu(P)\ge0\) and \(\nu(N)\le0\) by the Hahn properties. In particular, both positive measures have finite total mass.
For singularity, take \(A=P\). Since \(P\cap N=\varnothing\), \(\nu^-(P)=-\nu(\varnothing)=0\); since \(X\setminus P=N\), \(\nu^+(X\setminus P)=\nu(N\cap P)=0\). Thus \(\nu^+\perp\nu^-\). Finally, \(E\cap P\) and \(E\cap N\) are a disjoint measurable partition of \(E\). Finite additivity of \(\nu\) gives $$ \nu(E)=\nu(E\cap P)+\nu(E\cap N)=\nu^+(E)-\nu^-(E). $$ This holds for every \(E\in\mathcal{M}\), as required. The proof also covers \(P=\varnothing\) or \(N=\varnothing\): the corresponding part is the zero measure. \(\square\)
The theorem’s conclusion has three separate ingredients: the parts are positive measures, they are concentrated on disjoint measurable regions, and their difference agrees with \(\nu\) on every measurable set. Finiteness here means each part has finite mass; it does not mean that \(\nu(E)\) is nonnegative or finite only on sets in one side of the partition.
Worked Example: An Atomic Signed Measure
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and define \(\nu(\{a\})=3\), \(\nu(\{b\})=-2\), and \(\nu(\{c\})=0\), extending by additivity. Then \(P=\{a,c\}\) and \(N=\{b\}\) form a Hahn decomposition: every subset of \(P\) has measure \(0\) or \(3\), and every subset of \(N\) has measure \(0\) or \(-2\).
The Jordan parts are \(\nu^+(E)=\nu(E\cap\{a,c\})\) and \(\nu^-(E)=-\nu(E\cap\{b\})\). Their values on all subsets are:
| \(E\) | \(\nu(E)\) | \(\nu^+(E)\) | \(\nu^-(E)\) |
|---|---|---|---|
| \(\varnothing\) | \(0\) | \(0\) | \(0\) |
| \(\{a\}\) | \(3\) | \(3\) | \(0\) |
| \(\{b\}\) | \(-2\) | \(0\) | \(2\) |
| \(\{c\}\) | \(0\) | \(0\) | \(0\) |
| \(\{a,b\}\) | \(1\) | \(3\) | \(2\) |
| \(\{a,c\}\) | \(3\) | \(3\) | \(0\) |
| \(\{b,c\}\) | \(-2\) | \(0\) | \(2\) |
| \(X\) | \(1\) | \(3\) | \(2\) |
For example, on \(E=\{a,b\}\), \(\nu(E)=3+(-2)=1\), while \(\nu^+(E)-\nu^-(E)=3-2=1\). The parts are supported on the disjoint sets \(\{a,c\}\) and \(\{b\}\), respectively, so they are mutually singular. Check: on the whole space, \(\nu(X)=3-2+0=1\) and \(\nu^+(X)-\nu^-(X)=3-2=1\); also \(\nu^+(X)+\nu^-(X)=5\), the sum of the absolute values of the three atomic masses.
Worked Example: A Signed Measure Defined by a Density
On \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), define \(\nu(E)=\int_E f\,dm\), where \(f=2\) on \([0,1/3]\) and \(f=-3\) on \((1/3,1]\). This is a finite signed measure because \(f\) is integrable: \(\int_{[0,1]}|f|\,dm=2(1/3)+3(2/3)=8/3\).
Take \(P=[0,1/3]\) and \(N=(1/3,1]\). Every measurable subset of \(P\) has nonnegative \(\nu\)-measure, since its density is \(2\); every measurable subset of \(N\) has nonpositive \(\nu\)-measure, since its density is \(-3\). The resulting measures are $$ \nu^+(E)=2m(E\cap[0,1/3]),\qquad \nu^-(E)=3m(E\cap(1/3,1]). $$ In particular, \(\nu^+(X)=2(1/3)=2/3\) and \(\nu^-(X)=3(2/3)=2\). Their difference on \(X\) is \(2/3-2=-4/3\), which agrees with the direct computation \(\nu(X)=2(1/3)-3(2/3)=2/3-2=-4/3\). For any measurable \(E\), splitting the integral over \(E\cap P\) and \(E\cap N\) gives \(\nu(E)=2m(E\cap P)-3m(E\cap N)=\nu^+(E)-\nu^-(E)\). Since \(\nu^+\) vanishes on \(N\) and \(\nu^-\) vanishes on \(P\), the measures are mutually singular.
Check: the boundary point \(1/3\) belongs to \(P\), not \(N\). It has Lebesgue measure zero, but the stated partition is still disjoint and covers all of \(X\), so the formulas apply without leaving an endpoint unassigned.
Worked Example: A Point Mass and a Diffuse Negative Part
On \(X=[0,1]\), with its Borel sigma-algebra, let \(\delta_0\) be the Dirac measure at \(0\), so \(\delta_0(E)=1\) if \(0\in E\) and \(0\) otherwise. Define \(\nu(E)=\delta_0(E)-2m(E)\). This is a finite signed measure: for any finite measurable partition \(E_1,\ldots,E_k\), the triangle inequality gives $$ \sum_{j=1}^k|\nu(E_j)| \le \sum_{j=1}^k\delta_0(E_j)+2\sum_{j=1}^k m(E_j) =1+2=3. $$ Set \(P=\{0\}\) and \(N=(0,1]\). For \(A\subseteq P\), \(\nu(A)=\delta_0(A)\ge0\). For measurable \(A\subseteq N\), \(\delta_0(A)=0\), so \(\nu(A)=-2m(A)\le0\). Thus \((P,N)\) is a Hahn decomposition.
The parts are \(\nu^+(E)=\delta_0(E\cap P)=\delta_0(E)\) and \(\nu^-(E)=2m(E\cap N)=2m(E)\), since \(m(\{0\})=0\). They are mutually singular: \(\nu^+\) is concentrated on \(\{0\}\), while \(\nu^-(\{0\})=0\). On \(X\), \(\nu(X)=1-2m([0,1])=1-2=-1\), and \(\nu^+(X)-\nu^-(X)=1-2=-1\). For an arbitrary measurable \(E\), the identity is \(\nu(E)=\delta_0(E)-2m(E)=\nu^+(E)-\nu^-(E)\). This example shows that the positive and negative measures need not both come from densities with respect to Lebesgue measure.
A Common Mistake: Taking Positive Values Set by Set
It is tempting to define a candidate positive part by \(\widetilde{\nu}^+(E)=\max(\nu(E),0)\). This operation on real numbers does not in general preserve additivity. For a concrete example, let \(X=\{a,b\}\), with \(\nu(\{a\})=1\) and \(\nu(\{b\})=-1\). Then \(\widetilde{\nu}^+(\{a\})=1\) and \(\widetilde{\nu}^+(\{b\})=0\), but \(\widetilde{\nu}^+(X)=\max(1-1,0)=0\). Additivity would require \(\widetilde{\nu}^+(X)=1+0=1\), a contradiction.
The correction is to use the sign regions of a Hahn decomposition, not the sign of each set’s total value. Here \(P=\{a\}\), \(N=\{b\}\), so the correct parts satisfy \(\nu^+(X)=\nu(P)=1\), \(\nu^-(X)=-\nu(N)=1\), and \(\nu(X)=1-1=0\). Each part is additive because it is defined by restricting the original measure to one fixed measurable region.
What the Decomposition Gives
The Jordan decomposition turns a finite signed measure into a difference of positive measures with disjoint measurable supports. The Hahn decomposition theorem supplies the sign partition; restricting \(\nu\) to its positive region and reversing its sign on its negative region supplies the two parts. Their difference recovers \(\nu\) on every measurable set, while their mutual singularity records that positive and negative mass lie on separate regions. The theorem proved here establishes these properties for parts constructed from a Hahn decomposition; uniqueness of the Jordan decomposition is a separate result.
Check Your Understanding
Use the definitions and verify each claim on measurable sets, not only on the whole space.
- For a Hahn decomposition \((P,N)\), explain why \(\nu^+(E)=\nu(E\cap P)\) is nonnegative for every measurable \(E\).
- Show that the set \(P\) witnesses \(\nu^+\perp\nu^-\).
- For the density example, compute \(\nu^+([0,1/6])\) and \(\nu^-([1/2,1])\).
- For \(\nu=\delta_0-2m\) on \([0,1]\), compute \(\nu([1/4,3/4])\) and compare it with \(\nu^+([1/4,3/4])-\nu^-([1/4,3/4])\).
- Give a reason the set function \(E\mapsto\max(\nu(E),0)\) need not be additive.