Total Variation of Atomic Signed Measures

Atomic Weights Determine Total Variation

A signed measure supported on countably many points is controlled by the weights assigned to those points. The signed measure adds the weights with their signs; its total variation adds their magnitudes. The distinction matters: positive and negative weights can cancel in the value of the signed measure, but they do not cancel in its total variation.

Throughout, let \((X,\mathcal{M})\) be a measurable space in which each singleton under discussion is measurable. There is no ambient positive measure required. We consider a finite real signed measure \(\nu\), meaning a countably additive signed measure with finite total variation. The formulas below apply whether the set of atoms is finite, countably infinite, or empty.

Definition: A finite or countable atomic representation is an expression \(\nu=\sum_{j\in J}a_j\delta_{x_j}\), where \(J\) is finite or countable, the points \(x_j\) are distinct, the weights \(a_j\) are nonzero real numbers, and \(\sum_{j\in J}|a_j|<\infty\). Here \(\delta_{x_j}(E)=1\) if \(x_j\in E\) and \(0\) otherwise. The empty index set represents the zero measure. In set form, \(\nu(E)=\sum_{j:\,x_j\in E}a_j\); this series is absolutely convergent.

Distinctness is part of the representation: if several Dirac masses are located at the same point, their weights must first be combined. The positive and negative weights will be separated into \(P=\{x_j:a_j>0\}\) and \(N=\{x_j:a_j<0\}\). Both are measurable because they are countable unions of measurable singletons.

The Atomic Series Defines a Finite Signed Measure

Proposition: Given distinct measurable points \(x_j\) and real weights \(a_j\) with \(\sum_{j\in J}|a_j|<\infty\), the set function \(\nu(E)=\sum_{j:\,x_j\in E}a_j\) is a finite signed measure on \((X,\mathcal{M})\).

Proof. Define \(\nu^+(E)=\sum_{j:\,x_j\in E}a_j^+\) and \(\nu^-(E)=\sum_{j:\,x_j\in E}a_j^-\), where \(a_j^+=\max(a_j,0)\) and \(a_j^-=\max(-a_j,0)\). Each is nonnegative and bounded above on \(X\) by \(\sum_j|a_j|<\infty\). To check countable additivity, take pairwise disjoint \(E_k\in\mathcal{M}\). For either set of nonnegative weights \(c_j=a_j^+\) or \(c_j=a_j^-\), the index sets \(I_k=\{j:x_j\in E_k\}\) are disjoint. Thus \(\sum_k\sum_{j\in I_k}c_j=\sum_{j\in\cup_k I_k}c_j\): both sides are the supremum of sums over finite subsets of the same indices, since all terms are nonnegative. This equals \(\sum_{j:x_j\in\cup_k E_k}c_j\). So \(\nu^+\) and \(\nu^-\) are finite positive measures, and \(\nu=\nu^+-\nu^-\) is a finite signed measure. \(\square\)

This construction also separates the Jordan parts. The measures \(\nu^+\) and \(\nu^-\) are concentrated on the disjoint measurable sets \(P\) and \(N\), respectively. They are therefore mutually singular, and \(\nu=\nu^+-\nu^-\) is the Jordan decomposition.

Total Variation and Jordan Parts

For a finite signed measure \(\nu\) on \((X,\mathcal{M})\), its total variation on \(E\in\mathcal{M}\) is defined by \(|\nu|(E)=\sup\sum_{i=1}^r|\nu(E_i)|\), where the supremum is over all finite measurable partitions \(E=\bigcup_{i=1}^r E_i\) into pairwise disjoint sets. For \(E=\varnothing\), set \(|\nu|(E)=0\). This is the partition definition of total variation.

Theorem: For the finite or countable atomic signed measure \(\nu=\sum_{j\in J}a_j\delta_{x_j}\) on \((X,\mathcal{M})\), \[ |\nu|(E)=\sum_{j:\,x_j\in E}|a_j|,\qquad \nu^+(E)=\sum_{j:\,x_j\in E}a_j^+,\qquad \nu^-(E)=\sum_{j:\,x_j\in E}a_j^-. \] In particular, \(|\nu|=\nu^++\nu^-\), and \(|\nu|(X)=\sum_{j\in J}|a_j|\).

Proof. Fix \(E\in\mathcal{M}\). For any finite measurable partition \(E_1,\ldots,E_r\) of \(E\), absolute convergence and the triangle inequality give \[ \sum_{i=1}^r|\nu(E_i)| \le \sum_{i=1}^r\sum_{j:\,x_j\in E_i}|a_j| =\sum_{j:\,x_j\in E}|a_j|. \] The equality holds because each atom in \(E\) belongs to exactly one member of the partition. Taking the supremum over partitions proves \(|\nu|(E)\le\sum_{j:\,x_j\in E}|a_j|\).

For the reverse inequality, the case \(E=\varnothing\) has both sides zero. If \(E\ne\varnothing\), use the measurable partition consisting of \(E\cap P\), \(E\cap N\), and \(E\setminus(P\cup N)\), omitting any empty pieces. These sets cover all of \(E\), including points outside the atomic support. On the first piece, \(\nu(E\cap P)=\sum_{x_j\in E\cap P}a_j\ge0\); on the second, \(\nu(E\cap N)=\sum_{x_j\in E\cap N}a_j\le0\); and on the third, \(\nu(E\setminus(P\cup N))=0\). Consequently, the sum of the absolute values of the measures of these partition pieces is \[ \sum_{x_j\in E\cap P}a_j+\sum_{x_j\in E\cap N}|a_j| =\sum_{j:\,x_j\in E}|a_j|. \] This is one of the sums used to define \(|\nu|(E)\), so it proves the reverse inequality. The displayed formulas for \(\nu^+\) and \(\nu^-\) follow from their construction in the proposition, and \(|a_j|=a_j^++a_j^-\) gives \(|\nu|=\nu^++\nu^-\). \(\square\)

Worked Calculations

Worked Example: A Finite Measure and a Subset

Let \(X=\{a,b,c,d\}\), \(\mathcal{M}=\mathcal{P}(X)\), and \(\nu=4\delta_a-3\delta_b+2\delta_d\). Compute the Jordan parts, \(|\nu|(X)\), and \(|\nu|(\{a,b,c\})\).

The positive points are \(a,d\), with weights \(4,2\), and the negative point is \(b\), with weight \(-3\). Therefore \(\nu^+=4\delta_a+2\delta_d\) and \(\nu^-=3\delta_b\). The whole-space total variation and signed mass are \[ |\nu|(X)=|4|+|-3|+|2|=4+3+2=9,\qquad \nu(X)=4-3+2=3. \] For \(A=\{a,b,c\}\), the atom at \(d\) is not included, so \[ |\nu|(A)=|4|+|-3|=7,\qquad \nu(A)=4-3=1. \] As a direct partition check on \(A\), the pieces \(\{a\},\{b\},\{c\}\) give \(4+3+0=7\) in the variation definition. The whole-space Jordan masses also check: \(\nu^+(X)+\nu^-(X)=6+3=9\).

Worked Example: Infinitely Many Alternating Atoms

On \(X=\mathbb{N}=\{1,2,\ldots\}\) with \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), let \(\nu=\sum_{n=1}^{\infty}(-1)^{n+1}2^{-n}\delta_n\). Find its Jordan masses, total variation, and signed mass.

The absolute weights sum to \(\sum_{n=1}^{\infty}2^{-n}=\frac{1/2}{1-1/2}=1\), so the series defines a finite signed measure. The positive weights occur at odd \(n=2k-1\), and the negative weights at even \(n=2k\): \[ \nu^+(X)=\sum_{k=1}^{\infty}2^{-(2k-1)} =\frac{1/2}{1-1/4}=\frac23,\qquad \nu^-(X)=\sum_{k=1}^{\infty}2^{-2k} =\frac{1/4}{1-1/4}=\frac13. \] Hence \(|\nu|(X)=\frac23+\frac13=1\), while \(\nu(X)=\frac23-\frac13=\frac13\). On the even integers \(A=\{2,4,6,\ldots\}\), every weight is negative, so \(\nu(A)=-\frac13\) and \(|\nu|(A)=\frac13\). As checks, the first two signed weights are \(1/2\) and \(-1/4\), and the first two absolute weights sum to \(3/4\), consistent with the full sums \(1/3\) and \(1\) after the remaining tails are included.

Worked Example: Combining Weights at the Same Point

On \(\mathbb{R}\) with its Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\), consider the signed measure written as \(4\delta_0-3\delta_0+2\delta_1\). Compute its Jordan parts and total variation.

The first two terms are at the same point, so their combined weight at \(0\) is \(4-3=1\). The distinct-point representation is therefore \(\nu=\delta_0+2\delta_1\). Both weights are positive, giving \(\nu^+=\delta_0+2\delta_1\), \(\nu^-=0\), and \(|\nu|(\mathbb{R})=|1|+|2|=3\). Directly, \(\nu(\{0,1\})=3\), and the partition into the singleton atoms gives \(1+2=3\) for total variation. By contrast, adding the magnitudes of the uncombined terms gives the incorrect value \(4+3+2=9\); that calculation treats two descriptions of the same atom as separate support points.

Common Pitfalls and Scope

Do not omit the part of a partition outside the signed support. In the first example, the whole space includes \(c\), which is outside \(P\cup N\). A partition containing only \(X\cap P\) and \(X\cap N\) would not cover \(X\). The correct partition includes \(X\setminus(P\cup N)\) as a third piece when it is nonempty. Its signed measure is zero, so adding it does not change the sum of absolute values, but it is necessary for a valid partition.

Do not confuse signed mass with total variation. For the alternating measure, \(\nu(X)=1/3\), whereas \(|\nu|(X)=1\). The signed mass records cancellation between positive and negative weights; total variation records both contributions without cancellation. The Jordan parts retain this distinction: \(\nu=\nu^+-\nu^-\), while \(|\nu|=\nu^++\nu^-\).

Absolute summability is the finiteness condition. The formula for a finite atomic signed measure requires \(\sum_j|a_j|<\infty\). In this case the theorem shows that this sum is exactly \(|\nu|(X)\). The signed series might appear to converge through cancellation even when the absolute weights do not sum finitely, but that does not give a finite signed measure with finite total variation as considered here.

Check Your Understanding

Use the partition definition and the atomic formulas to check each answer.

For \(\nu=5\delta_a-2\delta_b\) on a space containing \(a,b,c\), find \(\nu^+\), \(\nu^-\), and \(|\nu|(\{a,c\})\).
For \(\nu=\sum_{n=1}^{\infty}(-1)^n3^{-n}\delta_n\) on \(\mathbb{N}\), compute \(|\nu|(\mathbb{N})\) and \(\nu(\mathbb{N})\).
Explain why the partition of \(E\) used in the theorem’s lower bound must include \(E\setminus(P\cup N)\) when this set is nonempty.
Combine the terms \(7\delta_x-4\delta_x-\delta_y\) into a distinct-point representation and compute the total variation.
State the condition on the weights that makes the countable atomic signed measure finite, and identify the resulting value of \(|\nu|(X)\).