Computing Variation for Atomic and Density Measures

Variation Counts Mass Before Cancellation

A complex measure can assign values in different directions in the complex plane. When those values are added, they may cancel; total variation instead records the largest total magnitude visible across measurable pieces. For atomic measures, the pieces can isolate the atoms. For a measure given by a complex density, sufficiently fine partitions can separate regions where the density points in different directions.

Throughout, \((X,\mathcal{M},\mu)\) is a measure space when a reference measure is needed. A complex measure \(\nu\) here is countably additive on \(\mathcal{M}\) and has finite total variation. For \(E\in\mathcal{M}\), a finite measurable partition of \(E\) is a finite collection of pairwise disjoint measurable sets whose union is \(E\); empty pieces may be omitted.

The formulas in this tutorial compute this supremum for two useful forms of finite complex measures. In each case, the key is to identify partitions that either isolate contributions or group together contributions with nearly the same complex direction.

Atomic Measures: Isolate Each Point

Proof. Absolute summability ensures that the series defining \(\nu(E)\) converges absolutely for every \(E\), with \(|\nu(E)|\leq\sum_k|a_k|<\infty\). If \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets, each point \(x_k\) belongs to at most one of them. Thus the sum over atoms in \(\bigcup_jE_j\) is the sum of the corresponding sums over the \(E_j\). Absolute convergence permits grouping these terms, proving countable additivity.

Let \(E=\bigcup_{j=1}^r E_j\) be any finite measurable partition. The triangle inequality gives $$ \sum_{j=1}^r|\nu(E_j)| \leq\sum_{j=1}^r\sum_{\{k:\,x_k\in E_j\}}|a_k| =\sum_{\{k:\,x_k\in E\}}|a_k|. $$ Taking the supremum over partitions proves the upper bound for \(|\nu|(E)\). For the reverse bound, fix \(N\) and partition \(E\) into the singletons \(\{x_k\}\) for \(k\leq N\) with \(x_k\in E\), together with the measurable remainder after removing those finitely many singletons. The partition sum is at least \(\sum_{k\leq N,\,x_k\in E}|a_k|\), since the remainder contributes a nonnegative magnitude. Letting \(N\to\infty\) gives the lower bound. The two bounds establish the formula. The same argument with all atoms isolated proves the finite case. \(\square\)

If several listed contributions are located at the same point, they must first be combined: a measurable partition cannot separate contributions at one point. The resulting coefficient at that point, not the sum of the separate absolute values, is what enters the atomic formula.

Complex Densities: Partition by Direction

Proof. For any finite measurable partition \(E=\bigcup_{j=1}^r E_j\), the triangle inequality for the Lebesgue integral gives $$ \sum_{j=1}^r|\nu(E_j)| =\sum_{j=1}^r\left|\int_{E_j}h\,d\mu\right| \leq\sum_{j=1}^r\int_{E_j}|h|\,d\mu =\int_E|h|\,d\mu. $$ Thus the supremum defining \(|\nu|(E)\) is at most the proposed answer.

For the other bound, choose \(0<\delta<\pi/2\) and divide the unit circle into finitely many measurable arcs, each of angular width at most \(2\delta\). For each arc choose its central unit complex number \(u_j\). Partition \(E\cap\{h\ne0\}\) according to the arc containing \(h(x)/|h(x)|\); these are measurable sets because \(h\) is measurable. Include \(E\cap\{h=0\}\) as one more piece. On the piece \(E_j\) corresponding to \(u_j\), the angle between \(h(x)\) and \(u_j\) is at most \(\delta\). Therefore $$ \operatorname{Re}(\overline{u_j}h(x))\geq \cos(\delta)|h(x)|. $$ Integrating and using \(|z|\geq\operatorname{Re}(\overline{u_j}z)\) for every complex number \(z\), we obtain $$ \left|\int_{E_j}h\,d\mu\right| \geq\operatorname{Re}\left(\overline{u_j}\int_{E_j}h\,d\mu\right) =\int_{E_j}\operatorname{Re}(\overline{u_j}h)\,d\mu \geq\cos(\delta)\int_{E_j}|h|\,d\mu. $$ The zero-density piece contributes zero to both sides. Summing the inequalities shows that this finite partition has sum of magnitudes at least \(\cos(\delta)\int_E|h|\,d\mu\). Consequently \(|\nu|(E)\geq\cos(\delta)\int_E|h|\,d\mu\) for every such \(\delta\). Letting \(\delta\downarrow0\), so that \(\cos(\delta)\to1\), gives the lower bound. Together the bounds prove the formula. \(\square\)

A Common Mistake: Taking the Magnitude Too Soon

A tempting but incorrect shortcut is to compute the total mass first and declare its absolute value to be the variation. For the three-atom example, that calculation would report \(|\nu|(X)=|\nu(X)|=5\). It is wrong: the partition into singletons has sum \(3+4+\sqrt{2}=7+\sqrt{2}\), which is greater than \(5\); the atomic formula proves that \(7+\sqrt{2}\) is the actual variation. The corresponding density mistake would report zero variation for the piecewise constant example because its total mass is zero, even though its two-piece partition has sum \(2\).

The underlying distinction is whether cancellation is allowed before or after taking magnitudes. The expression \(|\nu(E)|\) adds all contributions in \(E\) first. The variation \(|\nu|(E)\) allows a finite measurable partition to reduce cancellation, then takes the supremum. The formulas proved above show how the best possible result is attained, or approached, by separating atoms or sorting density values by direction.