Total Variation from Measurable Partitions

Why Partition Sums Measure Sign Changes

A signed measure can assign positive values to some measurable sets and negative values to others. Consequently, $|\nu(E)|$ records the net value on $E$, but can conceal cancellation between positive and negative contributions inside $E$. To measure the size of those contributions without allowing cancellation, we examine finite measurable partitions of $E$ and add the absolute values of the signed measure on their pieces.

Throughout, $(X,\mathcal{M})$ is a measurable space and $\nu$ is a finite signed measure: $\nu$ is real-valued and countably additive on $\mathcal{M}$, and its Jordan parts $\nu^+$ and $\nu^-$ are finite positive measures satisfying $\nu=\nu^+-\nu^-$. In particular, each $\nu(E)$ is a finite real number. We use the Jordan decomposition established earlier in the course; no completeness or sigma-finiteness assumption is needed here.

Definition: If $E\in\mathcal{M}$, a finite measurable partition of $E$ is a finite family $E_1,\ldots,E_n\in\mathcal{M}$ of pairwise disjoint sets whose union is $E$. Define the total variation of $\nu$ on $E$ by $$ |\nu|(E)=\sup\left\{\sum_{i=1}^{n}|\nu(E_i)|:\ E_1,\ldots,E_n\text{ form a finite measurable partition of }E\right\}. $$ For $E=\varnothing$, the empty partition has sum $0$, so $|\nu|(\varnothing)=0$. A partition may include empty pieces without changing its sum.

The supremum ranges over all finite measurable partitions, not merely over a partition chosen in advance. Each candidate sum is finite. Moreover, it is bounded above by $\nu^+(E)+\nu^-(E)$: for every partition, $|\nu(E_i)|=|\nu^+(E_i)-\nu^-(E_i)|\le\nu^+(E_i)+\nu^-(E_i)$, and finite additivity of the positive measures gives the bound after summing. Thus $|\nu|(E)$ is finite under our assumptions.

What Refinement Does

A partition $\mathcal{P}$ refines a partition $\mathcal{Q}$ if every cell of $\mathcal{P}$ is contained in a cell of $\mathcal{Q}$. Refinement separates some of the contributions that may have cancelled in a coarser cell. It cannot lower the associated sum.

Proposition: For any finite signed measure $\nu$ and any $E\in\mathcal{M}$, the partition sum of a finite measurable partition of $E$ does not decrease when the partition is refined.

Proof. Let $\mathcal{Q}=\{Q_1,\ldots,Q_r\}$ be a partition of $E$, and let $\mathcal{P}$ be a finite refinement. For each $j$, let $P_{j,1},\ldots,P_{j,k_j}$ be the cells of $\mathcal{P}$ contained in $Q_j$. They partition $Q_j$, so finite additivity gives $\nu(Q_j)=\sum_{\ell=1}^{k_j}\nu(P_{j,\ell})$. The triangle inequality therefore yields $$ |\nu(Q_j)|\le\sum_{\ell=1}^{k_j}|\nu(P_{j,\ell})|. $$ Summing this inequality over $j=1,\ldots,r$, with each cell of $\mathcal{P}$ counted once, shows that the sum for $\mathcal{Q}$ is at most the sum for $\mathcal{P}$. This also covers zero-valued cells and empty pieces. $\square$

This proposition explains the supremum: each partition gives a lower bound for $|\nu|(E)$, and splitting cells can reveal more of the positive and negative contributions. In general, the definition does not say that one particular partition attains the supremum. For a finite measurable space whose measurable sets are unions of finitely many atoms, however, a partition into those atoms attains it.

Theorem: Suppose $A_1,\ldots,A_k$ are pairwise disjoint, nonempty measurable sets with union $X$, and every set in $\mathcal{M}$ is a union of some of these sets. Let $c_j=\nu(A_j)$. For every measurable $E=\bigcup_{j\in I}A_j$, where $I\subseteq\{1,\ldots,k\}$, $$ |\nu|(E)=\sum_{j\in I}|c_j|. $$

Proof. If $I$ is empty, then $E=\varnothing$, and both sides are zero. Suppose $I$ is nonempty. Consider any finite measurable partition $E_1,\ldots,E_n$ of $E$. Each $E_i$ is a union of some of the sets $A_j$. Since the partition pieces are disjoint and cover $E$, each $A_j$ for $j\in I$ is included in exactly one piece. For each $i$, write $J_i=\{j\in I:A_j\subseteq E_i\}$. Finite additivity gives $\nu(E_i)=\sum_{j\in J_i}c_j$, so the triangle inequality gives $|\nu(E_i)|\le\sum_{j\in J_i}|c_j|$. Summing over $i$ counts each $j\in I$ exactly once. Thus every partition sum is at most $\sum_{j\in I}|c_j|$, and the supremum is at most that value. Conversely, the sets $A_j$, $j\in I$, form a finite measurable partition of $E$, whose sum is exactly $\sum_{j\in I}|\nu(A_j)|=\sum_{j\in I}|c_j|$. Hence the supremum equals this value. If some $c_j=0$, its contribution is zero and the argument is unchanged. $\square$

Worked Finite-Partition Computations

Worked Example: Four Points with Positive and Negative Weights

Let $X=\{a,b,c,d\}$, let $\mathcal{M}$ be the collection of all subsets of $X$, and define $\nu(\{a\})=5$, $\nu(\{b\})=-2$, $\nu(\{c\})=0$, and $\nu(\{d\})=-4$. For any $E\subseteq X$, define $\nu(E)$ as the sum of the weights of its points. Compute $|\nu|(X)$, and compare two partitions.

The four singleton sets are the atoms of this finite measurable space. By the theorem, the total variation on $X$ is the sum of the absolute weights: $$ |\nu|(X)=|5|+|-2|+|0|+|-4|=5+2+0+4=11. $$ The singleton partition attains this value.

Now take the coarser partition $\{\{a,b\},\{c,d\}\}$. The first cell has signed measure $5+(-2)=3$, and the second has signed measure $0+(-4)=-4$. Its partition sum is $|3|+|-4|=3+4=7$. For the partition $\{\{a,c\},\{b,d\}\}$, the cell values are $5+0=5$ and $-2+(-4)=-6$, so its sum is $|5|+|-6|=5+6=11$. The latter grouping does not mix the negative weights with the positive one in a way that reduces the sum. Check: every partition sum is at most $11$ by the finite-atom theorem, and the singleton partition gives $11$, so the supremum is exactly $11$.

Worked Example: Total Variation on a Subset

Use the same four-point space and signed measure. Compute $|\nu|(\{b,c,d\})$, and compare it with the absolute signed measure of that set.

The subset $\{b,c,d\}$ consists of the atoms with weights $-2,0,-4$. The finite-atom formula gives $$ |\nu|(\{b,c,d\})=|-2|+|0|+|-4|=2+0+4=6. $$ On the other hand, its signed measure is $\nu(\{b,c,d\})=-2+0-4=-6$, so $|\nu(\{b,c,d\})|=|-6|=6$. Here there is no cancellation among nonzero weights, and the two quantities agree. Check: partitioning into $\{b\},\{c,d\}$ gives $|-2|+|0-4|=2+4=6$, again attaining the computed supremum.

Worked Example: Signed Point Masses

On $X=\{0,1,2\}$ with all subsets measurable, define $\nu=3\delta_0-2\delta_1+4\delta_2$, where $\delta_x(E)=1$ if $x\in E$ and $\delta_x(E)=0$ otherwise. Find $|\nu|(X)$ and $|\nu|(\{0,2\})$.

The singleton weights are $3,-2,4$. Applying the finite-atom formula to $X$ gives $$ |\nu|(X)=|3|+|-2|+|4|=3+2+4=9. $$ For $E=\{0,2\}$, only the weights $3$ and $4$ contribute, so $$ |\nu|(\{0,2\})=|3|+|4|=3+4=7. $$ As an independent check on the first value, the singleton partition has sum $3+2+4=9$. Every other partition groups some of the three weights, and the triangle inequality bounds its sum by $9$. For the second value, the partition $\{\{0\},\{2\}\}$ gives $3+4=7$, while grouping the points gives $|3+4|=7$ as well.

A Common Mistake: Taking One Absolute Value

A frequent error is to calculate $|\nu|(E)$ as $|\nu(E)|$. The latter is only the absolute value of the net signed measure on the whole set; the definition of total variation instead takes the supremum over all finite measurable partitions.

For a concrete correction, return to $X=\{a,b,c,d\}$ and take $E=X$. Its signed measure is $\nu(X)=5-2+0-4=-1$, hence $|\nu(X)|=|-1|=1$. But the singleton partition gives the partition sum $5+2+0+4=11$, and the finite-atom theorem shows that no partition sum exceeds $11$. Therefore $|\nu|(X)=11$, not $1$. The discrepancy comes from cancellation: the positive contribution $5$ and negative contributions $-2$ and $-4$ nearly cancel in $\nu(X)$, while their magnitudes add in the singleton partition.

Partition sums are the defining quantities here. The definition and the finite-atom calculation do not by themselves establish that $E\mapsto|\nu|(E)$ is countably additive; that property requires a separate proof. For present computations, the practical procedure is to list the measurable atoms contained in the set, add the absolute values of their signed weights, and use the theorem only when every measurable set is a union of those atoms.

Check Your Understanding

In each question, specify the measurable set and the partition sum being used.

For weights $4,-3,2$ on a three-point space with all subsets measurable, compute the total variation of the whole space.
For weights $4,-3,2$, compute the partition sum of the single cell consisting of all three points, and compare it with total variation.
Prove that if a finite partition is refined, its partition sum cannot decrease, using finite additivity and the triangle inequality.
If $E$ is the union of atoms with weights $c_1,\ldots,c_r$, explain why the singleton-atom partition attains $\sum_{j=1}^r|c_j|$.
Give an example in which $|\nu(E)|$ is strictly less than $|\nu|(E)$, and identify the cancellation responsible.

Total Variation from Measurable Partitions

What You'll Learn