The Variation Norm on Complex Measures

From Total Variation to a Norm

For complex measures, the total variation records the largest amount of mass that can be detected by partitioning a measurable set into finitely many pieces. Its value on the whole space gives a natural size for a measure. The key point is that this size does more than satisfy the triangle inequality: it makes the space of finite complex measures complete.

Throughout, $(X,\mathcal{M})$ is a measurable space. A complex measure $\nu$ is a countably additive, complex-valued set function on $\mathcal{M}$ with $\nu(\varnothing)=0$. Its total variation is defined, for $E\in\mathcal{M}$, by

|\nu|(E)=\sup\left\{\sum_{j=1}^{k}|\nu(E_j)|:\ E_1,\ldots,E_k\in\mathcal{M}\text{ form a finite partition of }E\right\}.

A finite partition consists of finitely many pairwise disjoint measurable sets whose union is $E$. We call $\nu$ a finite complex measure when $|\nu|(X)<\infty$. The earlier material on total variation establishes that $|\nu|$ is a positive measure; in particular, it is countably additive. No finiteness or sigma-finiteness assumption on a separate reference measure is needed here.

Definition: The variation norm of a finite complex measure $\nu$ is $\|\nu\|_{\mathrm{var}}=|\nu|(X)$. The collection of all finite complex measures on $(X,\mathcal{M})$, with pointwise addition and scalar multiplication, is denoted here by $\mathcal{M}_{\mathrm{c}}(X)$.

The partition in the definition must be finite. In particular, when $X$ is countably infinite, its partition into all singleton sets is not an admissible partition. Computations on countable spaces need either an upper bound valid for every finite partition or a sequence of finite partitions.

The Norm Properties

Theorem: The finite complex measures $\mathcal{M}_{\mathrm{c}}(X)$, with the variation norm, form a normed complex vector space. In particular, for finite complex measures $\nu,\lambda$ and $c\in\mathbb{C}$, $$ \|\nu\|_{\mathrm{var}}\ge0,\qquad \|\nu\|_{\mathrm{var}}=0\iff \nu=0,\qquad \|c\nu\|_{\mathrm{var}}=|c|\,\|\nu\|_{\mathrm{var}},\qquad \|\nu+\lambda\|_{\mathrm{var}}\le\|\nu\|_{\mathrm{var}}+\|\lambda\|_{\mathrm{var}}. $$

Proof. Addition and scalar multiplication preserve countable additivity, so they give complex measures. For every finite measurable partition $E_1,\ldots,E_k$ of $X$, the complex triangle inequality gives $\sum_{j=1}^k|(\nu+\lambda)(E_j)|\le\sum_{j=1}^k|\nu(E_j)|+\sum_{j=1}^k|\lambda(E_j)|$. The right side is at most $|\nu|(X)+|\lambda|(X)$. Taking the supremum over the partitions proves $\|\nu+\lambda\|_{\mathrm{var}}\le\|\nu\|_{\mathrm{var}}+\|\lambda\|_{\mathrm{var}}$; in particular $\nu+\lambda$ has finite variation.

If $c\ne0$, then for every finite partition, $\sum_j|(c\nu)(E_j)|=|c|\sum_j|\nu(E_j)|$, so taking suprema gives $\|c\nu\|_{\mathrm{var}}=|c|\|\nu\|_{\mathrm{var}}$. If $c=0$, both sides equal zero, so homogeneity also holds in that case. Nonnegativity follows because every partition sum is nonnegative. Finally, for any $E\in\mathcal{M}$, the two-set partition $E,E^c$ of $X$ gives $|\nu(E)|\le|\nu(E)|+|\nu(E^c)|\le\|\nu\|_{\mathrm{var}}$. Thus if $\|\nu\|_{\mathrm{var}}=0$, then $\nu(E)=0$ for every measurable $E$, so $\nu=0$. Conversely, the zero measure has norm zero. This proves all the norm properties. $\square$

Worked Example: The Norm and Triangle Inequality on a Finite Space

Let $X=\{a,b\}$ with $\mathcal{M}$ the set of all subsets. Define complex measures by their masses: $\nu(\{a\})=1,\ \nu(\{b\})=i,\ \lambda(\{a\})=i,\ \lambda(\{b\})=1$. For a measure on this finite space, the singleton partition gives the sum of the absolute masses. Any other partition groups some of those masses, and the triangle inequality shows its sum cannot be larger. Consequently, $\|\nu\|_{\mathrm{var}}=|1|+|i|=1+1=2$, and $\|\lambda\|_{\mathrm{var}}=|i|+|1|=1+1=2$.

The sum measure has mass $1+i$ at $a$ and $1+i$ at $b$. Since $|1+i|=\sqrt{1^2+1^2}=\sqrt{2}$, its norm is $\|\nu+\lambda\|_{\mathrm{var}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$. Thus $2\sqrt{2}\le 2+2=4$, as the triangle inequality requires. Check: squaring the two nonnegative sides gives $8\le16$. The norm computation agrees with the finite partition into $\{a\}$ and $\{b\}$.

Computing Variation Without Infinite Partitions

For a positive finite measure, every finite measurable partition $E_1,\ldots,E_k$ of $X$ satisfies $\sum_{j=1}^k|\nu(E_j)|=\sum_{j=1}^k\nu(E_j)=\nu(X)$. Hence its variation norm is exactly its total mass. For complex atomic measures, a useful upper bound comes from the triangle inequality on each partition piece; finite partitions can then capture arbitrarily much of the sum of the absolute masses.

Worked Example: A Positive Measure on a Countable Space

On $X=\mathbb{N}=\{1,2,\ldots\}$ with all subsets measurable, define the positive measure $\nu(E)=\sum_{k\in E}3/4^k$. The geometric series gives $\nu(X)=\sum_{k=1}^{\infty}3/4^k=(3/4)/(1-1/4)=1$. For every finite measurable partition $E_1,\ldots,E_r$ of $X$, countable additivity and positivity give $\sum_{j=1}^r|\nu(E_j)|=\sum_{j=1}^r\nu(E_j)=\nu(X)=1$. Therefore $\|\nu\|_{\mathrm{var}}=1$. This argument uses a finite partition and applies to every such partition; it does not treat the countably infinite singleton partition as admissible.

Check: the finite partition consisting just of $X$ has partition sum $|\nu(X)|=1$, and every finite partition has sum $1$, so the supremum is exactly $1$. For an explicit partition with a tail piece, take $\{1\},\ldots,\{n\},\{n+1,n+2,\ldots\}$. Its sum is $\sum_{k=1}^n3/4^k+\sum_{k=n+1}^{\infty}3/4^k=(1-4^{-n})+4^{-n}=1$.

For a complex measure given by a density, an earlier result on variation states that if $f$ is complex-valued and integrable on a measure space $(X,\mathcal{M},\mu)$, then the measure $\nu(E)=\int_E f\,d\mu$ has $|\nu|(E)=\int_E|f|\,d\mu$. Here integrable means $\int|f|\,d\mu<\infty$. A direct partition check can also confirm the norm in simple examples.

Worked Example: A Piecewise Constant Complex Density

Take $X=[0,1]$, its Lebesgue sigma-algebra $\mathcal{L}$, and Lebesgue measure $m$. Set $f(x)=2i$ on $[0,1/3)$ and $f(x)=-3$ on $[1/3,1]$, and define $\nu(E)=\int_E f\,dm$. The density is integrable because $\int_0^1|f|\,dm=|2i|(1/3)+|-3|(2/3)=2/3+2=8/3$. The density formula for variation therefore gives $\|\nu\|_{\mathrm{var}}=8/3$.

For a direct check, the partition $[0,1/3),[1/3,1]$ has measure values $2i/3$ and $-2$. Its partition sum is $|2i/3|+|-2|=2/3+2=8/3$. No partition can do better, since for every measurable piece $E_j$, $|\int_{E_j}f\,dm|\le\int_{E_j}|f|\,dm$, and summing over the partition gives at most $\int_0^1|f|\,dm=8/3$.

Check: the total complex mass is $\nu(X)=2i/3-2$, whose absolute value is $\sqrt{(-2)^2+(2/3)^2}=\sqrt{40}/3$. This is less than $8/3$, since $40<64$; using only $|\nu(X)|$ would miss the larger variation detected by the two-piece partition.

Completeness in Variation Norm

Theorem: The normed space $\mathcal{M}_{\mathrm{c}}(X)$ is complete. That is, if $(\nu_n)$ is Cauchy in variation norm, there is a finite complex measure $\nu$ on $(X,\mathcal{M})$ such that $\|\nu_n-\nu\|_{\mathrm{var}}\to0$.

Proof. Let $(\nu_n)$ be Cauchy in variation norm. For any $E\in\mathcal{M}$, the partition $E,E^c$ gives $|\nu_n(E)-\nu_m(E)|\le\|\nu_n-\nu_m\|_{\mathrm{var}}$. Thus $(\nu_n(E))$ is Cauchy in $\mathbb{C}$. Define $\nu(E)=\lim_{n\to\infty}\nu_n(E)$. In particular, $\nu(\varnothing)=0$. Given $\varepsilon>0$, choose $N$ so that $\|\nu_n-\nu_m\|_{\mathrm{var}}<\varepsilon$ for $n,m\ge N$. Fix $n\ge N$, let $m\to\infty$ in the preceding setwise bound, and obtain the uniform estimate $|\nu_n(E)-\nu(E)|\le\varepsilon$ for every $E\in\mathcal{M}$.

For disjoint $A,B\in\mathcal{M}$, countable additivity of each $\nu_n$ gives $\nu_n(A\cup B)=\nu_n(A)+\nu_n(B)$; passing to the limit proves finite additivity of $\nu$. We next show that its variation, initially defined by the finite-partition supremum, is bounded. Since a Cauchy sequence in a normed space is bounded, there is $C<\infty$ with $\|\nu_n\|_{\mathrm{var}}\le C$ for every $n$. For any finite partition $E_1,\ldots,E_k$ of $X$, the sum has finitely many terms, so setwise convergence yields $\sum_{j=1}^k|\nu(E_j)|=\lim_{n\to\infty}\sum_{j=1}^k|\nu_n(E_j)|\le C$. Taking the supremum over these partitions shows that the partition supremum for $\nu$ is at most $C$.

It remains to prove countable additivity. Let $A_1,A_2,\ldots$ be pairwise disjoint measurable sets, let $A=\bigcup_{k=1}^{\infty}A_k$, and put $R_q=\bigcup_{k=q+1}^{\infty}A_k$. Fix $\varepsilon>0$. The uniform estimate above, applied with $\varepsilon/2$, gives an index $N$ such that $|\nu(E)-\nu_N(E)|\le\varepsilon/2$ for every measurable $E$. The measure $|\nu_N|$ is finite. Since $R_q$ decreases to the empty set, continuity from above for the finite measure $|\nu_N|$ implies $|\nu_N|(R_q)\to0$. Choose $q$ large enough that $|\nu_N|(R_q)<\varepsilon/2$. Then $|\nu(R_q)|\le|\nu(R_q)-\nu_N(R_q)|+|\nu_N(R_q)|\le\varepsilon/2+|\nu_N|(R_q)<\varepsilon$. Finite additivity gives $\nu(A)=\sum_{k=1}^q\nu(A_k)+\nu(R_q)$. As $\varepsilon$ was arbitrary, the remainder tends to zero, proving $\nu(A)=\sum_{k=1}^{\infty}\nu(A_k)$. Thus $\nu$ is countably additive. Its bounded partition supremum means it has finite total variation, so $\nu\in\mathcal{M}_{\mathrm{c}}(X)$.

Finally, let $\varepsilon>0$ and choose $N$ so that $\|\nu_n-\nu_m\|_{\mathrm{var}}<\varepsilon$ whenever $n,m\ge N$. Fix $n\ge N$ and any finite measurable partition $E_1,\ldots,E_k$ of $X$. Sending $m\to\infty$ gives $\sum_{j=1}^k|(\nu_n-\nu)(E_j)|=\lim_{m\to\infty}\sum_{j=1}^k|(\nu_n-\nu_m)(E_j)|\le\varepsilon$. Taking the supremum over all finite partitions yields $\|\nu_n-\nu\|_{\mathrm{var}}\le\varepsilon$. Since this holds for every $\varepsilon>0$ with a suitable $N$, the norms tend to zero. This proves completeness. $\square$

Worked Example: A Cauchy Sequence of Truncated Atomic Measures

On $\mathbb{N}$ with all subsets measurable, define $\nu_n(E)=\sum_{\substack{k\in E\\1\le k\le n}}2^{-k}$. The limit candidate is $\nu(E)=\sum_{k\in E}2^{-k}$, a positive finite measure with total mass $\sum_{k=1}^{\infty}2^{-k}=(1/2)/(1-1/2)=1$. As in the positive-measure example, every finite measurable partition has partition sum equal to the measure of the whole space. Hence the variation norm of a positive measure is its total mass, including for these measures on $\mathbb{N}$.

The difference $\nu-\nu_n$ is positive and has mass $\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}$. Therefore $\|\nu-\nu_n\|_{\mathrm{var}}=2^{-n}\to0$. More generally, for $m>n$, $\|\nu_m-\nu_n\|_{\mathrm{var}}=\sum_{k=n+1}^{m}2^{-k}=2^{-n}-2^{-m}\le2^{-n}$, so the sequence is Cauchy. Check: at $n=3$, the omitted tail is $\sum_{k=4}^{\infty}2^{-k}=(1/16)/(1-1/2)=1/8=2^{-3}$, matching the norm formula.

A Partition Pitfall and Why Completeness Matters

Common mistake: for the positive measure on $\mathbb{N}$ with masses $3/4^k$, one might say, “the singleton partition attains the sum of the masses.” The calculation points to the right value, but the partition $\{\{1\},\{2\},\ldots\}$ is countably infinite and is not allowed in the definition of variation. The corrected argument is that every finite measurable partition has partition sum $\nu(\mathbb{N})=1$, or, equivalently, that the finite partition $\{1\},\ldots,\{n\},\{n+1,n+2,\ldots\}$ has sum exactly $1$. The distinction matters whenever a supremum is defined using only finite partitions.

The norm axioms let us measure distance between complex measures, while completeness guarantees that a Cauchy sequence does not converge merely set by set to an object outside the space. In the proof, uniform control over all measurable sets comes from the variation norm; finiteness of the approximating measure’s variation controls the tails needed for countable additivity. These are the central reasons the variation norm is useful.

Check Your Understanding

Use finite partitions in each variation calculation, and identify where finiteness is used in the completeness proof.

For a positive finite measure $\rho$, prove directly from the definition that $\|\rho\|_{\mathrm{var}}=\rho(X)$.
On a two-point measurable space, let the masses of $\nu$ be $2$ and $-i$. Compute its variation norm and explain which finite partition attains it.
Why does $|\nu_n(E)-\nu_m(E)|\le\|\nu_n-\nu_m\|_{\mathrm{var}}$ hold for every measurable $E$?
In the completeness proof, explain why finiteness of $|\nu_N|(X)$ is needed to conclude that $|\nu_N|(R_q)\to0$.
For $\nu_n(E)=\sum_{k\in E,\,1\le k\le n}2^{-k}$ on $\mathbb{N}$, compute $\|\nu_5-\nu_3\|_{\mathrm{var}}$.