Total Variation and the Hahn Decomposition

How a Hahn Decomposition Reveals Total Variation

A signed measure can have a small value on a set because its positive and negative contributions cancel. Total variation records the size of both contributions before cancellation. The connection with a Hahn decomposition is exact: on each measurable set, total variation is the signed measure restricted to the positive set plus the negative of the signed measure restricted to the negative set.

Let \((X,\mathcal{M})\) be a measurable space and let \(\nu\) be a finite signed measure on it. Here, finite means that \(\nu\) is the difference of two finite positive measures, so its total variation is finite. The space need not have an additional positive measure, and no completeness or sigma-finiteness assumption is needed. A Hahn decomposition is a pair of disjoint sets \(P,N\in\mathcal{M}\) with \(P\cup N=X\), such that \(\nu(A)\ge 0\) for every measurable \(A\subseteq P\), and \(\nu(A)\le 0\) for every measurable \(A\subseteq N\).

A finite measurable partition of \(E\in\mathcal{M}\) is a finite family of pairwise disjoint measurable sets whose union is \(E\). The total variation of \(\nu\) on \(E\) is the supremum of the sums of the absolute values of \(\nu\) over all such partitions.

This definition does not say that \(|\nu|(E)=|\nu(E)|\). The latter quantity takes the absolute value only after the contributions on \(E\) have been added. The partition supremum can keep positive and negative contributions separate. We use the standard result that total variation, as defined above, is a positive measure; its measure property was established earlier. The result proved here identifies its values using a Hahn decomposition.

The Total-Variation Identity

Proof. Fix \(E\in\mathcal{M}\). If \(E=\varnothing\), both sides of the first identity are zero. Suppose \(E\ne\varnothing\), and take any finite measurable partition \(E_1,\ldots,E_k\) of \(E\). For each \(j\), the Hahn properties give \(\nu(E_j\cap P)\ge 0\) and \(\nu(E_j\cap N)\le 0\), because both intersections are measurable subsets of the corresponding Hahn sets. Additivity across the disjoint sets \(E_j\cap P\) and \(E_j\cap N\) gives \(\nu(E_j)=\nu(E_j\cap P)+\nu(E_j\cap N)\). If \(a\ge0\) and \(b\le0\), then \(|a+b|\le a-b\): when \(a+b\ge0\), \(a+b\le a-b\) follows from \(b\le0\); when \(a+b<0\), \(-(a+b)\le a-b\) follows from \(a\ge0\). Applying this inequality with \(a=\nu(E_j\cap P)\) and \(b=\nu(E_j\cap N)\), then summing, yields \(\displaystyle \sum_{j=1}^k|\nu(E_j)|\le\sum_{j=1}^k\bigl(\nu(E_j\cap P)-\nu(E_j\cap N)\bigr)=\nu(E\cap P)-\nu(E\cap N).\) The last equality follows from finite additivity and the fact that the \(E_j\) partition \(E\). Taking the supremum over all partitions proves \(|\nu|(E)\le\nu(E\cap P)-\nu(E\cap N)\). For the reverse inequality, partition \(E\) into its intersections with the Hahn sets, omitting an empty intersection if necessary. Each intersection is measurable, and the partition sum is \(|\nu(E\cap P)|+|\nu(E\cap N)|=\nu(E\cap P)-\nu(E\cap N)\), because the first term is nonnegative and the second signed measure value is nonpositive. This partition is among those in the defining supremum, so \(|\nu|(E)\ge\nu(E\cap P)-\nu(E\cap N)\). The two inequalities give the identity for every \(E\). Finally, \(\nu^+\) and \(\nu^-\) are positive measures: their values are nonnegative by the Hahn properties, and countable additivity follows by restricting the countably additive measure \(\nu\) to \(P\) and \(N\), with a minus sign on \(N\). Thus the identity also gives \(|\nu|=\nu^++\nu^-\). \(\square\)

The proof shows why the Hahn partition is decisive. Every other partition can split positive and negative contributions across the same cell and lose some of their magnitude when it takes an absolute value. The partition made from \(E\cap P\) and \(E\cap N\) separates the two signs, and therefore attains the supremum.

The Jordan Parts and a Common Pitfall

The restrictions in the theorem are the Jordan parts: \(\nu^+(E)=\nu(E\cap P)\) and \(\nu^-(E)=-\nu(E\cap N)\). The signed measure is their difference, since \(E\cap P\) and \(E\cap N\) partition \(E\): \(\nu(E)=\nu(E\cap P)+\nu(E\cap N)=\nu^+(E)-\nu^-(E)\). Together with the theorem, this also shows that the parts can be recovered from \(\nu\) and its total variation alone.

Indeed, adding \(|\nu|(E)=\nu^+(E)+\nu^-(E)\) and \(\nu(E)=\nu^+(E)-\nu^-(E)\) gives \(2\nu^+(E)\); subtracting the second equality from the first gives \(2\nu^-(E)\). These formulas show that the positive and negative measures are determined by \(\nu\), even if a zero-measure subset is assigned differently in another Hahn decomposition.

Common mistake: replacing total variation by \(|\nu(E)|\). In the first example, \(\nu(X)=0\), so \(|\nu(X)|=0\). It would be incorrect to conclude that \(|\nu|(X)=0\). The correction is to use the partition definition, or the theorem: the Hahn-set calculation gives \(|\nu|(X)=3-(-3)=6\). In general, the one-cell partition only proves \(|\nu(E)|\le|\nu|(E)\); it need not attain the supremum.

This identity is useful whenever a signed measure is already understood through its sign sets. Instead of optimizing over every finite measurable partition separately for each \(E\), one fixed Hahn decomposition computes total variation on all measurable sets. It also makes the distinction between the signed value and the total variation concrete: the signed value records net mass, while total variation records the mass of both signs separately.