Worked Hahn, Jordan, and Variation Problems

A Reliable Order for Signed-Measure Calculations

For a signed measure, the total mass of a set can be small because its positive and negative contributions cancel. The Jordan parts separate those contributions, and total variation records their combined size. In explicit problems, a reliable sequence is: find where the signed measure is positive and negative, form the Jordan parts, then compute total variation and check the results against the original measure.

Here a finite signed measure \(\nu\) means a real-valued, countably additive set function with finite total variation. A measurable set \(P\) is a positive set if \(\nu(E)\ge 0\) for every measurable \(E\subseteq P\); a measurable set \(N\) is a negative set if \(\nu(E)\le 0\) for every measurable \(E\subseteq N\). A Hahn decomposition is a measurable partition \(X=P\cup N\), \(P\cap N=\varnothing\), in which \(P\) is positive and \(N\) is negative. Its Jordan parts are the mutually singular positive measures \(\nu^+,\nu^-\) such that \(\nu=\nu^+-\nu^-\); the total variation is \(|\nu|=\nu^++\nu^-\). We use the partition formula \(|\nu|(A)=\sup\sum_{j=1}^k|\nu(A_j)|\), where the supremum is over finite measurable partitions \(A=\bigcup_{j=1}^k A_j\).

The Hahn and Jordan decomposition theorems and the partition formula have been established in the earlier material on signed measures. The calculations below use their explicit forms. Zero contributions may be assigned to either side of a Hahn decomposition; they contribute to neither Jordan part.

Finite Spaces: Sort the Weights by Sign

Proof. Set \(P=\{x\in X:a_x\ge0\}\) and \(N=\{x\in X:a_x<0\}\). These sets partition \(X\). For any \(E\subseteq P\), \(\nu(E)=\sum_{x\in E}a_x\ge0\); for any \(E\subseteq N\), \(\nu(E)=\sum_{x\in E}a_x\le0\). Thus \(P,N\) form a Hahn decomposition. The Jordan parts obtained from this decomposition assign each nonnegative weight to \(\nu^+\) and the absolute value of each negative weight to \(\nu^-\), giving the stated formulas and \(|\nu|=\nu^++\nu^-\). To check the total variation directly, any finite partition \(A=\bigcup_j A_j\) satisfies \(\sum_j|\nu(A_j)|=\sum_j|\sum_{x\in A_j}a_x|\le\sum_{x\in A}|a_x|\) by the triangle inequality. The partition into \(A\cap P\) and \(A\cap N\), omitting empty pieces, attains this bound: its sum is \(\sum_{x\in A\cap P}a_x+\sum_{x\in A\cap N}(-a_x)=\sum_{x\in A}|a_x|\). This proves the total-variation formula as well. \(\square\)

Countably Many Atoms: Check Summability First

On a countable space, the same sign rule applies, but a finite signed measure needs a summability condition. In the following proposition the space is \((\mathbb{N},\mathcal{P}(\mathbb{N}))\), where \(\mathbb{N}=\{1,2,\ldots\}\); counting measure is sigma-finite, not finite. Suppose real weights \((a_n)\) satisfy \(\sum_{n=1}^{\infty}|a_n|<\infty\), and define \(\nu(A)=\sum_{n\in A}a_n\). This is a finite signed measure. For a set \(A\), the series defining \(\nu(A)\) converges absolutely.

Proof. For \(E\subseteq P\), every term in the absolutely convergent series \(\nu(E)=\sum_{n\in E}a_n\) is nonnegative, so its sum is nonnegative. For \(E\subseteq N\), every term is negative, so its sum is nonpositive. This proves the Hahn sign conditions. The Jordan formulas follow by assigning each atom according to its sign. For any finite measurable partition \(A=\bigcup_{j=1}^k A_j\), absolute convergence and the triangle inequality give \(\sum_{j=1}^k|\nu(A_j)|\le\sum_{n\in A}|a_n|\). The partition \(A\cap P,A\cap N\) attains equality, because the first sum has only nonnegative terms and the second only negative terms; their absolute values add to \(\sum_{n\in A}|a_n|\). Empty pieces can be omitted. Taking the supremum over partitions proves the formula. \(\square\)

Density Measures: Split Where the Density Changes Sign

Let \((X,\mathcal{M},\mu)\) be a measure space and let \(g:X\to\mathbb{R}\) be measurable and integrable, meaning \(\int_X|g|\,d\mu<\infty\). Then \(\nu(E)=\int_Eg\,d\mu\) defines a finite signed measure. The sets where \(g\) is nonnegative or negative give a Hahn decomposition. The following formula also explains how to compute variation on any measurable set.

Proof. Put \(P=\{g\ge0\}\) and \(N=\{g<0\}\). For measurable \(E\subseteq P\), \(\nu(E)=\int_Eg\,d\mu\ge0\); for measurable \(E\subseteq N\), \(\nu(E)=\int_Eg\,d\mu\le0\). Thus these sets form a Hahn decomposition, and the Jordan formulas follow by restricting the positive and negative contributions of \(g\). To establish the variation formula from the partition definition, take any finite measurable partition \(A=\bigcup_{j=1}^k A_j\). Then \(\sum_j|\nu(A_j)|=\sum_j|\int_{A_j}g\,d\mu|\le\sum_j\int_{A_j}|g|\,d\mu=\int_A|g|\,d\mu\). The partition \(A\cap P,A\cap N\), with empty pieces omitted, attains equality: on \(A\cap P\) the integral is nonnegative and equals \(\int_{A\cap P}|g|\,d\mu\); on \(A\cap N\) it is nonpositive and its absolute value equals \(\int_{A\cap N}|g|\,d\mu\). The supremum over partitions is therefore \(\int_A|g|\,d\mu\). \(\square\)

Checks That Catch Common Errors

A frequent mistake is to compute variation by taking the absolute value of the total signed mass. In the density example, this would give \(|\nu|([0,1])\overset{\text{wrong}}{=}|\nu([0,1])|=|0|=0\). The correction is to take the absolute value before integrating: \(|\nu|([0,1])=\int_0^1|x-\frac12|\,dx=\frac14\). Positive and negative contributions cancel in \(\nu([0,1])\), but they add in total variation.

Two other checks are useful. First, a proposed Hahn set must have the required sign on every measurable subset, not merely on the whole set. Second, for a density or atomic sequence, locate the zero contributions as well as the strictly positive and strictly negative ones; zeros can be assigned to either Hahn side, but they do not contribute to either Jordan mass or total variation. In an infinite atomic example, verify absolute summability before using the finite signed-measure formulas.