Why Signed-Measure Errors Are Easy to Make
A signed measure records a net contribution: positive and negative mass can cancel when they are evaluated on the same set. Total variation records the size of both contributions, without cancellation. Confusing these two quantities can lead to incorrect claims about zero-mass sets, Jordan decompositions, or the size of a signed measure.
Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a finite signed measure, meaning that \(\nu\) is countably additive and has finite total variation on \(X\). A Hahn decomposition is a pair of disjoint measurable sets \(P,N\) with \(P\cup N=X\), such that \(\nu(A)\ge0\) for every measurable \(A\subseteq P\), and \(\nu(A)\le0\) for every measurable \(A\subseteq N\). The Jordan parts associated with this decomposition are the positive measures \(\nu^+(E)=\nu(E\cap P)\) and \(\nu^-(E)=-\nu(E\cap N)\), for \(E\in\mathcal{M}\). They satisfy \(\nu=\nu^+-\nu^-\), and are mutually singular: one is concentrated on \(P\), the other on the disjoint set \(N\). The Hahn and Jordan decomposition theorems, including uniqueness up to \(\nu\)-null sets, are standard results established earlier in the course.
Zero Signed Mass Does Not Mean No Mass
A set \(E\) with \(\nu(E)=0\) has equal positive and negative Jordan mass, not necessarily zero positive and negative mass. Nor does \(\nu(E)=0\) imply that \(\nu\) vanishes on every measurable subset of \(E\). Here, “\(\nu\) vanishes on \(E\)” means \(\nu(A)=0\) for every measurable \(A\subseteq E\), a stronger assertion.
Proof. The first identity is the defining difference of the Jordan parts. The second follows from the partition characterization, proved in the next section; alternatively, it is the standard total-variation identity for Jordan parts. If \(\nu(E)=0\), the first identity gives \(\nu^+(E)=\nu^-(E)\), and if those quantities are equal then the same identity gives \(\nu(E)=0\). Since \(\nu^+(E)\) and \(\nu^-(E)\) are nonnegative, their sum is zero exactly when each is zero. The second identity therefore gives the stated equivalence for \(|\nu|(E)=0\). \(\square\)
Worked Example: Equal Positive and Negative Mass
Let \(X=\{a,b\}\) with \(\mathcal{M}\) the power set of \(X\), and define \(\nu(\{a\})=4\), \(\nu(\{b\})=-4\). Compute the signed mass and variation of \(E=X\).
A Hahn decomposition is \(P=\{a\}\), \(N=\{b\}\). Thus \(\nu^+(X)=4\) and \(\nu^-(X)=4\). The signed mass is \(\nu(X)=4-4=0\), while the variation is \(|\nu|(X)=4+4=8\). The partition \(\{\{a\},\{b\}\}\) gives the variation sum \(|4|+|-4|=8\); the one-set partition \(\{X\}\) gives only \(|\nu(X)|=|0|=0\). Check: on the subset \(\{a\}\), the signed measure is \(4\), so it does not vanish on \(X\), despite \(\nu(X)=0\).
The Partition Formula: Do Not Replace a Supremum by One Partition
The signed mass of the whole set is one particular partition sum, using the partition with just one member. Its absolute value can be much smaller than the supremum over all finite partitions. The following argument proves the exact formula and explains why a Hahn decomposition supplies a partition that captures both signs.
Proof. Fix \(E\in\mathcal{M}\) and a finite measurable partition \(E=\bigcup_{j=1}^{k}E_j\). Put \(a_j=\nu^+(E_j)\) and \(b_j=\nu^-(E_j)\); both are nonnegative and \(\nu(E_j)=a_j-b_j\). The inequality \(|a_j-b_j|\le a_j+b_j\) gives $$ \sum_{j=1}^{k}|\nu(E_j)|\le\sum_{j=1}^{k}(a_j+b_j)=\nu^+(E)+\nu^-(E). $$ This bounds every partition sum above by \(\nu^+(E)+\nu^-(E)\). For the reverse inequality, use the partition \(E\cap P,E\cap N\), omitting an empty member if necessary. On \(E\cap P\), \(\nu\) is nonnegative and \(\nu(E\cap P)=\nu^+(E)\). On \(E\cap N\), \(\nu(E\cap N)=-\nu^-(E)\). The sum of the absolute values on these members is therefore \(\nu^+(E)+\nu^-(E)\). If either member is empty, its corresponding Jordan mass is zero and the same equality holds using the nonempty member; if \(E\) is empty, both sides are zero. This partition attains the upper bound, proving the formula. \(\square\)
Worked Example: The Whole-Space Mass Misses the Variation
Let \(X=\{1,2,3\}\) with its power-set sigma-algebra, and assign the singleton masses \(\nu(\{1\})=5\), \(\nu(\{2\})=-4\), and \(\nu(\{3\})=1\). Compare \(|\nu(X)|\) with \(|\nu|(X)\).
Adding the singleton masses gives \(\nu(X)=5-4+1=2\), so \(|\nu(X)|=|2|=2\). The singleton partition has sum \( |5|+|-4|+|1|=5+4+1=10\). The Hahn sets are \(P=\{1,3\}\) and \(N=\{2\}\), so the partition \(\{P,N\}\) gives \(|\nu(P)|+|\nu(N)|=|5+1|+|-4|=6+4=10\). The partition formula proves no partition can exceed \(\nu^+(X)+\nu^-(X)=6+4=10\). Hence \(|\nu|(X)=10\), not \(2\). Check: both the singleton partition and the Hahn partition give \(10\), and the formula bounds every other partition by \(10\).
Jordan Parts Must Not Overlap
It is not enough to write a signed measure as the difference of two positive measures. Such a difference need not be the Jordan decomposition. The Jordan parts must also be mutually singular: there must be disjoint measurable sets on which the positive and negative measures, respectively, are concentrated. This condition rules out a common positive contribution being added to both parts.
Worked Example: A Difference of Positive Measures That Is Not a Jordan Decomposition
Let \(X=\{a,b,c\}\) with its power-set sigma-algebra, and let \(\nu(\{a\})=3\), \(\nu(\{b\})=-2\), and \(\nu(\{c\})=0\). Find its Jordan parts, then test a proposed alternative.
The Hahn sets can be taken as \(P=\{a,c\}\) and \(N=\{b\}\). The Jordan parts are \(\nu^+=3\delta_a\) and \(\nu^-=2\delta_b\), where \(\delta_x\) is the unit mass at \(x\). Their difference assigns masses \(3,-2,0\) to \(a,b,c\), as required. They are mutually singular: \(\nu^+\) is concentrated on \(\{a,c\}\), and \(\nu^-\) on the disjoint set \(\{b\}\).
Now set \(\alpha=3\delta_a+\delta_c\) and \(\beta=2\delta_b+\delta_c\). These are positive measures, and \(\alpha-\beta\) still assigns masses \(3,-2,0\): at \(c\) the difference is \(1-1=0\). But both \(\alpha\) and \(\beta\) assign mass \(1\) to \(\{c\}\), so they cannot be concentrated on disjoint sets and are not mutually singular. Moreover, \((\alpha+\beta)(X)=4+3=7\), while the actual variation is \((\nu^++\nu^-)(X)=3+2=5\). Check: the singleton partition gives \(3+2+0=5\), confirming that the proposed overlapping parts do not give total variation.
A Reliable Checklist
When checking a signed-measure calculation on a measurable space \((X,\mathcal{M})\), keep the following distinctions in view:
| Claim to check | What is valid | Common error |
|---|---|---|
| \(\nu(E)=0\) | \(\nu^+(E)=\nu^-(E)\) | Concluding both are zero |
| \(\nu=\alpha-\beta\) | This gives a representation if \(\alpha,\beta\) are positive measures | Calling the pair Jordan parts without checking mutual singularity |
| \(|\nu|(E)\) | Take the supremum over finite measurable partitions of \(E\) | Replacing the supremum by \(|\nu(E)|\) |
| \(|\nu|(E)=0\) | Both Jordan parts assign zero mass to \(E\) | Confusing zero variation with zero signed mass |
A useful correction to the most common wrong calculation is to replace \(|\nu|(E)=|\nu(E)|\) by the partition formula. The one-set partition always shows \(|\nu(E)|\le|\nu|(E)\), but equality holds only when no finer measurable partition produces a larger sum. The Hahn partition shows exactly how separating positive and negative regions can prevent cancellation.
These checks matter whenever a signed measure is restricted to a set, added to another signed measure, or compared with its total variation. Always identify which quantity is being computed: net signed mass, positive and negative Jordan mass, or the supremum of partition sums. They are related, but they are not interchangeable.
Check Your Understanding
For each question, distinguish the signed mass from the positive, negative, or total-variation quantities being asked about.
- On \(X=\{u,v\}\), suppose \(\nu(\{u\})=6\) and \(\nu(\{v\})=-6\). Find \(\nu(X)\), \(\nu^+(X)\), \(\nu^-(X)\), and \(|\nu|(X)\).
- Explain why \(\nu(E)=0\) implies \(\nu^+(E)=\nu^-(E)\), but does not imply \(|\nu|(E)=0\).
- For singleton masses \(2,-5,4\), compute the whole-space signed mass and the total variation. Give a partition that attains the variation.
- Why do two positive measures \(\alpha,\beta\) with \(\nu=\alpha-\beta\) not necessarily form the Jordan decomposition?
- Prove directly from the partition definition that \(|\nu(E)|\le|\nu|(E)\) for every measurable \(E\).