Algebra of the Positive and Negative Parts

From the Jordan Decomposition to Algebra

For a signed measure, the positive and negative parts are measures, not pointwise positive and negative values of a number. Their algebra is nevertheless as direct as the corresponding real-number identities. Once a finite signed measure is decomposed into its Jordan parts, adding and subtracting those parts recovers both the signed measure and its total variation. Solving these two equations gives formulas for each Jordan part separately.

Throughout, let $(X,\mathcal M)$ be a measurable space and let $\mu$ be a finite signed measure on it. By the Jordan decomposition theorem, there are finite positive measures $\mu^+$ and $\mu^-$ on $(X,\mathcal M)$ such that $\mu=\mu^+-\mu^-$, and they are mutually singular: there are disjoint measurable sets supporting the two measures. The uniqueness theorem for the Jordan decomposition says that this pair is unique.

Definition: The total variation measure of $\mu$, denoted by $|\mu|$, is the positive measure $\mu^++\mu^-$. Thus for every $E\in\mathcal M$, $|\mu|(E)=\mu^+(E)+\mu^-(E)$. In this tutorial all measure identities are setwise: they hold for every measurable set $E$, not merely almost everywhere. The measures $\mu^+$, $\mu^-$, and $|\mu|$ are finite.

The definition of total variation used here is the sum of the Jordan parts. In particular, $|\mu|(E)$ is not generally the absolute value $|\mu(E)|$: the former records positive and negative mass within $E$ before cancellation, while the latter is the absolute value after cancellation.

The Algebraic Identities

Theorem: Let $\mu$ be a finite signed measure on $(X,\mathcal M)$, with Jordan parts $\mu^+$ and $\mu^-$, and define $|\mu|=\mu^++\mu^-$. Then, as equalities of finite signed measures, $$ \mu=\mu^+-\mu^-,\qquad |\mu|=\mu^++\mu^-, $$ and $$ \mu^+=\frac{|\mu|+\mu}{2},\qquad \mu^-=\frac{|\mu|-\mu}{2}. $$ Equivalently, for every $E\in\mathcal M$, $$ \mu^+(E)=\frac{|\mu|(E)+\mu(E)}{2},\qquad \mu^-(E)=\frac{|\mu|(E)-\mu(E)}{2}. $$

Proof. The Jordan decomposition gives $\mu(E)=\mu^+(E)-\mu^-(E)$ for every $E\in\mathcal M$, and the definition gives $|\mu|(E)=\mu^+(E)+\mu^-(E)$. Adding these two real-number equalities yields $|\mu|(E)+\mu(E)=2\mu^+(E)$, so $\mu^+(E)=(|\mu|(E)+\mu(E))/2$. Subtracting the first equality from the second yields $|\mu|(E)-\mu(E)=2\mu^-(E)$, so $\mu^-(E)=(|\mu|(E)-\mu(E))/2$. These hold for every measurable $E$, proving the measure identities. In particular, the right-hand expressions define the already-existing positive measures $\mu^+$ and $\mu^-$; their positivity is guaranteed by the Jordan decomposition, not by treating an arbitrary difference of positive measures as positive. $\square$

These identities can be remembered as the solution of two simultaneous equations. The signed measure is the difference of the unknown positive parts, and total variation is their sum. Adding and subtracting isolates each part. The factor $1/2$ is essential: each addition or subtraction produces twice the corresponding measure.

Corollary: For every $E\in\mathcal M$, $$ |\mu(E)|\leq|\mu|(E). $$ Also, $$ \mu^+(X)=\frac{|\mu|(X)+\mu(X)}{2},\qquad \mu^-(X)=\frac{|\mu|(X)-\mu(X)}{2}. $$

Proof. For a fixed $E$, write $a=\mu^+(E)$ and $b=\mu^-(E)$. Both are nonnegative real numbers. Then $\mu(E)=a-b$ and $|\mu|(E)=a+b$. The inequalities $-(a+b)\leq a-b\leq a+b$ follow respectively from $-2a\leq0$ and $-2b\leq0$. Hence $|\mu(E)|\leq|\mu|(E)$. Taking $E=X$ in the theorem’s setwise formulas gives the two total-mass identities. $\square$

Worked Applications

Worked Example: A Signed Measure on Three Atoms

Let $X=\{a,b,c\}$, let $\mathcal M$ be its power set, and let $\mu$ be the finite signed measure determined by $\mu(\{a\})=3$, $\mu(\{b\})=-2$, and $\mu(\{c\})=0$. Find the Jordan parts and check the formulas on a set containing both positive and negative mass.

The positive and negative singleton masses give $$ \mu^+=3\delta_a,\qquad \mu^-=2\delta_b,\qquad |\mu|=3\delta_a+2\delta_b, $$ where $\delta_x$ is the measure assigning mass $1$ to any set containing $x$ and mass $0$ otherwise. These are the Jordan parts because they are positive, mutually singular, and their difference has the specified values on every singleton. Since every subset of $X$ is a disjoint union of singletons, those singleton values determine $\mu$ on all of $\mathcal M$.

Take $E=\{a,b\}$. Directly, $\mu(E)=3-2=1$ and $|\mu|(E)=3+2=5$. The algebraic formulas give $$ \frac{| \mu|(E)+\mu(E)}2=\frac{5+1}{2}=3=\mu^+(E),\qquad \frac{| \mu|(E)-\mu(E)}2=\frac{5-1}{2}=2=\mu^-(E). $$ As a further check on the whole space, $\mu(X)=3-2+0=1$ and $|\mu|(X)=3+2+0=5$, so the total-mass formulas give $\mu^+(X)=(5+1)/2=3$ and $\mu^-(X)=(5-1)/2=2$, as required.

Worked Example: A Signed Measure Defined by a Density

Let $X=[0,2]$ with its Lebesgue sigma-algebra and let $m$ be Lebesgue measure. Define $f(x)=1$ on $[0,1]$ and $f(x)=-2$ on $(1,2]$, and define $\mu(E)=\int_E f\,dm$ for measurable $E\subseteq X$. The density is integrable, so this gives a finite signed measure. Determine its Jordan parts and use the algebraic formulas on a set with mass of both signs.

The density is nonnegative on $[0,1]$ and negative on $(1,2]$. Thus the positive and negative measures are $$ \mu^+(E)=m(E\cap[0,1]),\qquad \mu^-(E)=2m(E\cap(1,2]). $$ Their supports are disjoint, and their difference is $\mu$, so they are the Jordan parts. Consequently $|\mu|(E)=m(E\cap[0,1])+2m(E\cap(1,2])$.

Choose $E=[0,\tfrac12]\cup[\tfrac32,2]$. The first interval has length $1/2$, and the second has length $2-3/2=1/2$. Therefore $\mu^+(E)=1\cdot(1/2)=1/2$, $\mu^-(E)=2\cdot(1/2)=1$, $\mu(E)=1/2-1=-1/2$, and $|\mu|(E)=1/2+1=3/2$. Substitution gives $$ \frac{|\mu|(E)+\mu(E)}2=\frac{\frac32-\frac12}{2}=\frac12,\qquad \frac{|\mu|(E)-\mu(E)}2=\frac{\frac32+\frac12}{2}=1. $$ These agree with the directly computed Jordan masses. On all of $X$, $\mu(X)=1-2=-1$ and $|\mu|(X)=1+2=3$, so the total formulas give $\mu^+(X)=(3-1)/2=1$ and $\mu^-(X)=(3+1)/2=2$. These also match the interval lengths and density values.

Worked Example: Cancellation in a Difference of Positive Measures

Let $X=\{a,b\}$ with the power-set sigma-algebra. Let $\alpha,\beta$ be positive measures with $\alpha(\{a\})=4$, $\alpha(\{b\})=1$, $\beta(\{a\})=1$, and $\beta(\{b\})=3$, and set $\mu=\alpha-\beta$. Find the Jordan parts. This example shows why $\alpha$ and $\beta$ need not themselves be the Jordan parts.

Subtract the singleton values: $\mu(\{a\})=4-1=3$ and $\mu(\{b\})=1-3=-2$. Thus the Jordan parts are $\mu^+=3\delta_a$ and $\mu^-=2\delta_b$; their supports are disjoint and their difference has the computed singleton values. Hence $|\mu|=3\delta_a+2\delta_b$.

For $E=X$, the signed mass is $\mu(X)=3+(-2)=1$, while the total variation is $|\mu|(X)=3+2=5$. The formulas recover $$ \mu^+(X)=\frac{5+1}{2}=3,\qquad \mu^-(X)=\frac{5-1}{2}=2. $$ The original measures have $\alpha(X)=4+1=5$ and $\beta(X)=1+3=4$, so neither equals the corresponding Jordan part. At $a$, the common mass $1$ cancels in $\alpha-\beta$; at $b$, the common mass $1$ also cancels. The Jordan parts record only the net positive and negative masses, not the particular representation as a difference.

Common Sign Errors and What the Identities Say

A frequent mistake is to reverse the sign in the formula for $\mu^+$. For the three-atom example, the incorrect expression $(\mu-|\mu|)/2$ evaluated on $\{a\}$ gives $(3-3)/2=0$, not $\mu^+(\{a\})=3$. On $\{b\}$, it gives $(-2-2)/2=-2$, which cannot be the value of a positive measure. The correction is $$ \mu^+=\frac{|\mu|+\mu}{2},\qquad \mu^-=\frac{|\mu|-\mu}{2}. $$ For example, on the negative atom $\{b\}$, the second formula gives $(2-(-2))/2=2$, the correct nonnegative mass.

A second error is to replace $|\mu|(E)$ by $|\mu(E)|$. In the density example, for $E=X$ the signed measure is $\mu(X)=-1$, so $|\mu(X)|=1$, whereas $|\mu|(X)=3$. The positive and negative contributions cancel in $\mu(X)=1-2$, but they add in $|\mu|(X)=1+2$. In general, the corollary gives only $|\mu(E)|\leq|\mu|(E)$, not equality.

The identities are useful whenever both the signed mass and total variation on a set are known: they recover the positive and negative masses separately without reconstructing a Hahn decomposition. Conversely, when the Jordan parts are available, their sum and difference immediately give total variation and the original signed measure. The formulas apply to finite signed measures, and every quantity in them is a finite real number on each measurable set.

Check Your Understanding

Use the Jordan-part identities, keeping track of whether each quantity is a signed mass or a positive mass.

If $\mu(E)=-4$ and $|\mu|(E)=10$, compute $\mu^+(E)$ and $\mu^-(E)$.
For the three-atom example, compute $\mu^+(\{a,c\})$ and $\mu^-(\{a,c\})$ using the setwise formulas.
Explain why $|\mu(E)|\leq|\mu|(E)$ holds even when $E$ contains both positive and negative mass.
Give a set $E$ from the density example for which $|\mu(E)|\ne|\mu|(E)$, and calculate both values.
In the finite-space example with $\mu=\alpha-\beta$, identify the common mass that cancels at each point before the Jordan parts are formed.

Algebra of the Positive and Negative Parts

What You'll Learn