Total Variation of a Complex Measure

Partition Sums Measure the Size of a Complex Measure

A complex measure can have substantial cancellation: the values it assigns to disjoint sets may point in different directions in the complex plane. The absolute value of the measure of their union can therefore be much smaller than the sum of their absolute values. Total variation records the largest sum of absolute values that can be obtained by partitioning a set into finitely many measurable pieces.

Let $(X,\mathcal{M})$ be a measurable space. A complex measure here is a function $\nu:\mathcal{M}\to\mathbb{C}$ with $\nu(\varnothing)=0$ that is countably additive: for pairwise disjoint $E_n\in\mathcal{M}$, the series $\sum_{n=1}^{\infty}\nu(E_n)$ converges in $\mathbb{C}$ to $\nu(\bigcup_{n=1}^{\infty}E_n)$. No finiteness or sigma-finiteness assumption is imposed on the measurable space. A finite measurable partition of $E\in\mathcal{M}$ is a finite family of pairwise disjoint measurable sets whose union is $E$. The empty set has the empty partition.

Definition: For $E\in\mathcal{M}$, the total variation of $\nu$ on $E$ is the extended nonnegative number $$ |\nu|(E)=\sup\left\{\sum_{j=1}^{k}|\nu(E_j)|:\ (E_1,\ldots,E_k)\text{ is a finite measurable partition of }E\right\}. $$ The supremum is allowed to be $+\infty$. When $|\nu|(X)<\infty$, $\nu$ is called a finite complex measure.

One partition gives one candidate sum, and hence a lower bound for $|\nu|(E)$. It need not give the supremum: splitting its pieces can reveal cancellation that was hidden in the original partition. The next result makes this precise.

Refining Partitions

Proposition: Let $(X,\mathcal{M})$ be a measurable space, let $\nu$ be a complex measure, and let $\mathcal{Q}$ refine a finite measurable partition $\mathcal{P}$ of $E$. Then $$ \sum_{Q\in\mathcal{Q}}|\nu(Q)|\ \ge\ \sum_{P\in\mathcal{P}}|\nu(P)|. $$ A common refinement of two finite measurable partitions is obtained by taking all their intersections.

Proof. Each $P\in\mathcal{P}$ is the disjoint union of the members $Q\in\mathcal{Q}$ contained in it. Finite additivity, which follows from countable additivity by appending empty sets, gives $\nu(P)=\sum_{Q\subseteq P}\nu(Q)$. The triangle inequality therefore gives $|\nu(P)|\le\sum_{Q\subseteq P}|\nu(Q)|$. Summing this inequality over $P$ counts every member of $\mathcal{Q}$ exactly once, proving the claim. If $\mathcal{P}$ and $\mathcal{R}$ are two partitions, the nonempty intersections $P\cap R$ are pairwise disjoint and cover $E$; each is contained in one member of each partition. They form a common refinement. $\square$

This proposition is useful even when the supremum over all partitions is hard to evaluate. A partition gives a lower bound; replacing it by a refinement can only increase that bound. For any two proposed partitions, their common refinement lets us compare their sums by refining both.

Worked Example: Refining a Partition Reveals Cancellation

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be its power set, and define $\nu(E)=\sum_{x\in E}w_x$, where $w_a=1+i$, $w_b=-1+i$, and $w_c=2$. Compare the partition $\{\{a,b\},\{c\}\}$ with the singleton partition of $X$.

For the first piece, $\nu(\{a,b\})=(1+i)+(-1+i)=2i$, so $|\nu(\{a,b\})|=|2i|=2$. For the second, $\nu(\{c\})=2$, so its absolute value is $2$. The partition sum is therefore $2+2=4$.

The singleton partition has sum $|1+i|+|-1+i|+|2|=\sqrt{1^2+1^2}+\sqrt{(-1)^2+1^2}+2=\sqrt{2}+\sqrt{2}+2=2+2\sqrt{2}$. Since $2+2\sqrt{2}>4$, refining this partition increases the sum. The increase occurs because $1+i$ and $-1+i$ partially cancel when combined: their sum is $2i$, whose absolute value is $2$, less than $2\sqrt{2}$.

Check: $\nu(X)=(1+i)+(-1+i)+2=2+2i$, so the one-piece partition has sum $|2+2i|=\sqrt{2^2+2^2}=2\sqrt{2}$. This is below the two-piece sum $4$, which is below the singleton sum $2+2\sqrt{2}$, as the refinement inequality requires.

Total Variation Is a Positive Measure

Theorem: Let $\nu$ be a complex measure on $(X,\mathcal{M})$. The set function $E\mapsto|\nu|(E)$, defined by finite partition suprema, is a measure from $\mathcal{M}$ to $[0,\infty]$. In particular, it is countably additive, and $|\nu|(\varnothing)=0$.

Proof. The only partition of $\varnothing$ is the empty partition, whose sum is $0$, so $|\nu|(\varnothing)=0$. Let $E,F\in\mathcal{M}$ be disjoint. We first prove finite additivity. If $\{C_1,\ldots,C_k\}$ partitions $E\cup F$, then $\{C_j\cap E\}_j$ and $\{C_j\cap F\}_j$, after empty sets are omitted, partition $E$ and $F$, respectively. Finite additivity of $\nu$ and the triangle inequality give $$ \sum_{j=1}^k|\nu(C_j)| \le \sum_{j=1}^k\bigl(|\nu(C_j\cap E)|+|\nu(C_j\cap F)|\bigr) \le |\nu|(E)+|\nu|(F). $$ Taking the supremum over partitions of $E\cup F$ proves $|\nu|(E\cup F)\le|\nu|(E)+|\nu|(F)$. Conversely, combine any finite partition of $E$ with any finite partition of $F$; the result is a finite partition of $E\cup F$. Its sum is the sum of the two original partition sums. Taking suprema gives $|\nu|(E\cup F)\ge|\nu|(E)+|\nu|(F)$, with the same conclusion if either supremum is infinite. Thus $|\nu|$ is finitely additive on disjoint sets.

Now let $E_n\in\mathcal{M}$ be pairwise disjoint and put $E=\bigcup_{n=1}^{\infty}E_n$. For each $N$, finite additivity and nonnegativity give $|\nu|(E)\ge\sum_{n=1}^{N}|\nu|(E_n)$. Taking the supremum over $N$ yields $|\nu|(E)\ge\sum_{n=1}^{\infty}|\nu|(E_n)$. For the reverse inequality, take any finite measurable partition $A_1,\ldots,A_k$ of $E$. Countable additivity of $\nu$ gives $\nu(A_j)=\sum_{n=1}^{\infty}\nu(A_j\cap E_n)$. The triangle inequality for a convergent complex series yields $$ \sum_{j=1}^{k}|\nu(A_j)| \le \sum_{j=1}^{k}\sum_{n=1}^{\infty}|\nu(A_j\cap E_n)| = \sum_{n=1}^{\infty}\sum_{j=1}^{k}|\nu(A_j\cap E_n)| \le \sum_{n=1}^{\infty}|\nu|(E_n). $$ The equality reorders nonnegative terms; it follows by taking increasing finite partial sums on both indices. For each fixed $n$, the sets $A_j\cap E_n$, with empty ones omitted, partition $E_n$, which proves the last inequality. Taking the supremum over partitions of $E$ proves the reverse bound. Hence $|\nu|(E)=\sum_n|\nu|(E_n)$, as required. $\square$

The proof also explains why refinement matters: the countable-additivity argument compares any partition of a union with its intersections with the disjoint pieces of that union. The definition of variation ensures that each resulting inner sum is bounded by the variation of the corresponding piece.

Basic Bounds and Further Partition Checks

Corollary: For every $E\in\mathcal{M}$, $|\nu(E)|\le|\nu|(E)$. For disjoint $E,F\in\mathcal{M}$, $|\nu|(E\cup F)=|\nu|(E)+|\nu|(F)$. More generally, $|\nu|$ is monotone: if $E\subseteq F$, then $|\nu|(E)\le|\nu|(F)$.

Proof. The one-piece partition $\{E\}$ has sum $|\nu(E)|$, so this sum is at most the defining supremum $|\nu|(E)$. The disjoint-set equality is finite additivity proved above. If $E\subseteq F$, then $F=E\cup(F\setminus E)$, a disjoint union; additivity and nonnegativity give $|\nu|(F)=|\nu|(E)+|\nu|(F\setminus E)\ge|\nu|(E)$. $\square$

Worked Example: Two Different Partitions and Their Common Refinement

Let $X=\{1,2,3,4\}$, take $\mathcal{M}$ to be the power set, and define $\nu(E)=\sum_{x\in E}w_x$ with $w_1=1$, $w_2=i$, $w_3=-1$, and $w_4=-i$. Compare $\mathcal{P}=\{\{1,2\},\{3,4\}\}$ and $\mathcal{R}=\{\{1,3\},\{2,4\}\}$.

For $\mathcal{P}$, the two measure values are $1+i$ and $-1-i$. Each has absolute value $\sqrt{1^2+1^2}=\sqrt{2}$, so the partition sum is $2\sqrt{2}$. For $\mathcal{R}$, both piece values are $1+(-1)=0$ and $i+(-i)=0$, so the partition sum is $0$. This difference is possible because the partitions group the values differently.

The common refinement consists of the four singleton sets. Its sum is $|1|+|i|+|-1|+|-i|=1+1+1+1=4$. It refines both original partitions, so the proposition predicts a sum at least $2\sqrt{2}$ and at least $0$; indeed $4\ge2\sqrt{2}\ge0$. In this finite space every partition is refined by the singleton partition, so no partition can have a sum greater than $4$. Thus $|\nu|(X)=4$.

Check: $\nu(X)=1+i-1-i=0$, and the one-piece partition has sum $|0|=0$. The value $0$ is a valid lower bound, not the total variation: finer partitions recover the four separate absolute values.

Worked Example: A Partition Sum Is Only a Lower Bound

On $X=\{u,v\}$ with its power set, define $\nu(E)=\sum_{x\in E}w_x$, where $w_u=3+4i$ and $w_v=-3+4i$. The one-piece partition of $X$ gives $\nu(X)=(3+4i)+(-3+4i)=8i$, and its sum is $|8i|=8$.

The partition into singletons gives $|3+4i|+|-3+4i|=\sqrt{3^2+4^2}+\sqrt{(-3)^2+4^2}=5+5=10$. Every partition of this two-point space is either the one-piece partition or the singleton partition, up to empty parts. Consequently, the supremum is $10$, not $8$.

Check: the triangle inequality gives $|\nu(X)|=|(3+4i)+(-3+4i)|\le|3+4i|+|-3+4i|=10$; the computed union value is $8$, consistent with this bound.

A Common Mistake: Using Only the Coarsest Partition

A frequent error is to identify $|\nu|(E)$ with $|\nu(E)|$. The latter is only the sum from the one-piece partition; it can miss cancellation among the values on finer pieces. In the preceding example, replacing the supremum by the one-piece sum would give $8$, while the valid singleton partition gives $10$. The correction is to keep the supremum in the definition, or to prove that the partition being used is sufficiently fine to control all other partitions.

Conversely, an increase in a partition sum does not by itself show that the supremum has been reached. The refinement inequality compares partition sums, but it does not claim that every sequence of refinements attains a maximum. In general, the definition is a supremum, not a maximum. The measure $|\nu|$ is the positive measure that records these suprema on all measurable sets; when it is finite, its total mass $|\nu|(X)$ is finite by definition of a finite complex measure.

Check Your Understanding

Use the definition and the refinement proposition to answer the following questions.

Why does the one-piece partition imply $|\nu(E)|\le|\nu|(E)$ for every measurable set $E$?
Explain why the intersections of two finite partitions form a common refinement after empty intersections are omitted.
If a partition sum is $7$ and a refinement sum is $9$, what can you conclude about $|\nu|(E)$, and what can you not conclude?
In the two-point example with weights $3+4i$ and $-3+4i$, calculate the one-piece and singleton partition sums and explain why they differ.
In the countable-additivity proof for $|\nu|$, why is the sum over $j$ and $n$ safely reordered?