Uniqueness of the Jordan Decomposition

What Uniqueness Must Show

A Jordan decomposition writes a signed measure as the difference of two positive measures that are mutually singular. The decomposition is useful only if the two positive measures are determined by the signed measure itself, rather than by a choice made during construction. The key is to compare the two decompositions on regions where their concentration sets overlap: on each such region, the sign of the signed measure forces the candidate measures to agree or to vanish.

Throughout, let $(X,\mathcal{M})$ be a measurable space. A finite signed measure $\nu$ here is a countably additive, real-valued set function on $\mathcal{M}$ with finite total variation; in particular, it takes finite values on measurable sets. All positive measures considered in the theorem are finite. Equalities of measures below mean equality on every $E\in\mathcal{M}$, not merely almost everywhere. No completeness or sigma-finiteness assumption on the measurable space is needed.

Definition: Positive measures $\alpha$ and $\beta$ on $(X,\mathcal{M})$ are mutually singular, written $\alpha\perp\beta$, if there is a measurable set $A\in\mathcal{M}$ such that $\alpha(A^c)=0$ and $\beta(A)=0$. Thus $\alpha$ is concentrated on $A$, while $\beta$ is concentrated on the disjoint set $A^c$. A Jordan decomposition of $\nu$ is an expression $\nu=\alpha-\beta$ on every measurable set, where $\alpha,\beta$ are finite positive measures and $\alpha\perp\beta$.

The earlier Jordan Decomposition Theorem gives existence under its hypotheses. Here the question is uniqueness: if another pair $\gamma,\delta$ also consists of finite positive measures, satisfies $\gamma\perp\delta$, and has $\nu=\gamma-\delta$, must $\alpha=\gamma$ and $\beta=\delta$? The proof will not assume that the concentration sets from the two decompositions are the same.

The Four-Intersection Argument

We first record the comparison on the four pieces formed by the two concentration partitions. The set-by-set argument matters: equality of the signed measures alone does not permit us to equate their positive terms without using positivity and mutual singularity.

Lemma: Let $\nu=\alpha-\beta=\gamma-\delta$ on $(X,\mathcal{M})$, where $\alpha,\beta,\gamma,\delta$ are finite positive measures, $\alpha\perp\beta$, and $\gamma\perp\delta$. Choose $A,C\in\mathcal{M}$ with $\alpha(A^c)=\beta(A)=0$ and $\gamma(C^c)=\delta(C)=0$. Write $B=A^c$ and $D=C^c$. For every measurable $E$ contained in the indicated intersection:

on $A\cap C$, $\alpha(E)=\gamma(E)$;
on $A\cap D$, $\alpha(E)=\delta(E)=0$;
on $B\cap C$, $\beta(E)=\gamma(E)=0$;
on $B\cap D$, $\beta(E)=\delta(E)$.

Proof. Suppose first that $E\subseteq A\cap C$. Since $\beta(A)=0$ and $\delta(C)=0$, positivity gives $\beta(E)=\delta(E)=0$. Evaluating both expressions for $\nu(E)$ gives $\alpha(E)=\nu(E)=\gamma(E)$. If $E\subseteq A\cap D$, then $\beta(E)=\gamma(E)=0$, so $\nu(E)=\alpha(E)=-\delta(E)$. Both $\alpha(E)$ and $\delta(E)$ are nonnegative; the equality of one with the negative of the other forces both to be zero. If $E\subseteq B\cap C$, then $\alpha(E)=\delta(E)=0$, so $-\beta(E)=\nu(E)=\gamma(E)$. Nonnegativity forces $\beta(E)=\gamma(E)=0$. Finally, if $E\subseteq B\cap D$, then $\alpha(E)=\gamma(E)=0$, and $-\beta(E)=\nu(E)=-\delta(E)$, which gives $\beta(E)=\delta(E)$. These four arguments establish the claims. $\square$

Theorem (Uniqueness of the Jordan Decomposition): Let $(X,\mathcal{M})$ be a measurable space, and let $\nu$ be a finite signed measure on it. If $$ \nu=\alpha-\beta=\gamma-\delta $$ where $\alpha,\beta,\gamma,\delta$ are finite positive measures, $\alpha\perp\beta$, and $\gamma\perp\delta$, then $\alpha=\gamma$ and $\beta=\delta$ on every set in $\mathcal{M}$.

Proof. By mutual singularity, choose $A,C\in\mathcal{M}$ so that $\alpha(A^c)=\beta(A)=0$ and $\gamma(C^c)=\delta(C)=0$. Put $B=A^c$ and $D=C^c$. The four measurable sets $A\cap C$, $A\cap D$, $B\cap C$, and $B\cap D$ are disjoint and have union $X$.

Fix any $E\in\mathcal{M}$, and split it into its intersections with these four sets. Since $\alpha(B)=0$, the two pieces of $E$ inside $B$ have $\alpha$-measure zero. The lemma also gives $\alpha(E\cap A\cap D)=0$. Therefore finite additivity gives $$ \alpha(E)=\alpha(E\cap A\cap C). $$ Likewise, $\gamma(C^c)=\gamma(D)=0$, and the lemma gives $\gamma(E\cap B\cap C)=0$. Hence $$ \gamma(E)=\gamma(E\cap A\cap C). $$ The lemma says that $\alpha$ and $\gamma$ agree on every measurable subset of $A\cap C$. In particular, they agree on $E\cap A\cap C$. Combining the displayed equalities yields $\alpha(E)=\gamma(E)$.

For the other two measures, $\beta(A)=0$, and the lemma gives $\beta(E\cap B\cap C)=0$, so $\beta(E)=\beta(E\cap B\cap D)$. Also $\delta(C)=0$, and the lemma gives $\delta(E\cap A\cap D)=0$, so $\delta(E)=\delta(E\cap B\cap D)$. On $B\cap D$, the lemma states that $\beta$ and $\delta$ agree on every measurable subset. Thus $\beta(E)=\delta(E)$. Since $E$ was arbitrary, both pairs of measures are equal on all of $\mathcal{M}$. $\square$

Worked Examples

Worked Example: A Signed Measure on Four Points

Let $X=\{a,b,c,d\}$, let $\mathcal{M}$ be its power set, and let $\mu$ be counting measure. Define $\nu(\{a\})=4$, $\nu(\{b\})=-3$, $\nu(\{c\})=0$, and $\nu(\{d\})=2$. Find its Jordan decomposition and check the values.

For any $E\subseteq X$, add the singleton values that belong to $E$. The positive values occur at $a,d$, and the negative value occurs at $b$. Thus define positive measures by $$ \alpha(E)=4\chi_E(a)+2\chi_E(d),\qquad \beta(E)=3\chi_E(b). $$ Here $\chi_E(x)=1$ if $x\in E$ and $0$ otherwise. The measures are concentrated on the disjoint sets $\{a,d\}$ and $\{b\}$, respectively, so $\alpha\perp\beta$. For every $E\subseteq X$, $\alpha(E)-\beta(E)$ adds $4$ if $a\in E$, adds $-3$ if $b\in E$, adds $2$ if $d\in E$, and adds $0$ for $c$. This is exactly $\nu(E)$.

Check: For $E=\{a,b,d\}$, $\nu(E)=4-3+2=3$, while $\alpha(E)-\beta(E)=(4+2)-3=3$. For $E=\{b,c\}$, $\nu(E)=-3+0=-3$, while $\alpha(E)-\beta(E)=0-3=-3$. Any other mutually singular positive pair representing $\nu$ must equal this pair by the theorem.

Worked Example: A Signed Measure Given by a Density

On $(\mathbb{R},\mathcal{L},m)$, where $\mathcal{L}$ is the Lebesgue sigma-algebra and $m$ is Lebesgue measure, define $f=2$ on $[0,1)$, $f=-3$ on $[1,2)$, and $f=0$ elsewhere. Set $\nu(E)=\int_E f\,dm$. Compute a Jordan decomposition and verify it on a test set.

The function $f$ is real-valued and measurable. It vanishes outside a set of finite measure, and its absolute integral is $2m([0,1))+3m([1,2))=2\cdot1+3\cdot1=5$; consequently $\nu$ is a finite signed measure. Define $$ \alpha(E)=2m(E\cap[0,1)),\qquad \beta(E)=3m(E\cap[1,2)). $$ Both are finite positive measures. The sets $[0,1)$ and $[1,2)$ are disjoint, and $\alpha$ and $\beta$ are concentrated on these respective sets; hence $\alpha\perp\beta$. Splitting the integral over these intervals gives $\nu(E)=\alpha(E)-\beta(E)$ for every $E\in\mathcal{L}$. Their total masses are $\alpha(\mathbb{R})=2\cdot1=2$ and $\beta(\mathbb{R})=3\cdot1=3$.

For a numerical check, take $E=[0,3/2)$. Its intersections with the two intervals have measures $m([0,1))=1$ and $m([1,3/2))=1/2$. Therefore $\alpha(E)=2\cdot1=2$, $\beta(E)=3\cdot(1/2)=3/2$, and $\alpha(E)-\beta(E)=2-3/2=1/2$. Direct integration gives $\nu(E)=2\cdot1-3\cdot(1/2)=1/2$, as required. The uniqueness theorem rules out any different mutually singular pair representing this same $\nu$.

Worked Example: Why Mutual Singularity Is Essential

Let $X=\{x\}$ with its power-set sigma-algebra, and let $\delta_x$ be the Dirac measure, defined by $\delta_x(E)=1$ when $x\in E$ and $0$ otherwise. Consider the zero signed measure $\nu(E)=0$ for every $E\subseteq X$.

One representation is $\nu=0-0$; its two zero measures are mutually singular, since the set $A=\varnothing$ satisfies $0(A^c)=0$ and $0(A)=0$. Another representation is $\nu=\delta_x-\delta_x$, because for $E=\varnothing$ both sides are $0$, and for $E=\{x\}$ the difference is $1-1=0$. But $\delta_x$ and $\delta_x$ are not mutually singular: a set $A\subseteq X$ with $\delta_x(A^c)=0$ must contain $x$, while $\delta_x(A)=0$ requires $x\notin A$. Both cannot hold. Thus positive measures whose difference is $\nu$ need not be unique if the singularity hypothesis is dropped.

A Common Mistake and the Role of the Result

Common mistake: From $\alpha-\beta=\gamma-\delta$, one might rearrange to $\alpha-\gamma=\beta-\delta$ and conclude that $\alpha=\gamma$ and $\beta=\delta$. That conclusion does not follow from equality of differences alone. In the preceding example, $\delta_x-\delta_x=0-0$, although $\delta_x\ne0$. The correction is to use mutual singularity: it provides concentration sets, and their four intersections let positivity force the cross terms to vanish. The theorem then proves equality on every measurable set, rather than merely suggesting equality from a formal rearrangement.

Uniqueness means that the two measures produced by the Jordan Decomposition Theorem depend only on $\nu$, not on the Hahn decomposition or other choices used to construct them. It also explains why mutual singularity is part of the definition: without it, cancellation between two positive measures can hide many different pairs behind the same signed measure. The four-intersection proof establishes exactly the uniqueness needed, without requiring a formula for the parts or any properties of total variation.

Check Your Understanding

For each question, keep track of the concentration sets and whether the equality being asked for is on all measurable sets.

State the definition of mutual singularity for two positive measures $\alpha$ and $\beta$.
In the uniqueness proof, why does $E\subseteq A\cap D$ imply both $\alpha(E)=0$ and $\delta(E)=0$?
For the four-point example, compute $\nu(\{a,c,d\})$ and verify it using $\alpha-\beta$.
For the density example, compute $\alpha([1/2,3/2))$, $\beta([1/2,3/2))$, and their difference.
Which hypothesis fails for the representation $0=\delta_x-\delta_x$ on a one-point space?

Uniqueness of the Jordan Decomposition

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