Separate positive and negative mass

Try

A signed ledger records masses +2, −4 and +3 on three separate accounts. The net total is one, while the total amount moved is nine. A signed measure records the net; its total variation records the amount before cancellation. Identify the positive and negative accounts first, then compare these totals.

Predict

Predict whether choosing a set with a negative net total makes every subset of it negative. It does not: combining a large negative account with a small positive account still gives a negative total. A negative set in the Hahn theorem means every measurable subset has nonpositive signed measure. That stronger property is what makes the decomposition work.

Experiment

For the fixed three-account example, construct the decomposition.

Account	Signed mass	Positive part	Negative part	Variation
a	2	2	0	2
b	−4	0	4	4
c	3	3	0	3

The positive and negative parts are both positive measures; the minus sign occurs when subtracting them. Check that their supports are disjoint and that variation is not the absolute value of the net total.

Understand

Here a finite signed measure means ν=α−β for finite positive measures α and β. This convention provides finite values, bounded range and continuity under increasing or decreasing measurable sets. It covers the finite signed measures needed below. The equivalence with a general finite-valued countably additive definition is not needed in this construction.

Let b be the finite infimum of ν(E) over measurable E. Choose A_k with ν(A_k)≤b+2^-k. For two sets, inclusion-exclusion and ν(A union B)≥b imply ν(A intersection B)≤ν(A)+ν(B)−b. Induction bounds the signed measure of the finite intersection from k=n to m by b+Σ_(k=n)^m2^-k. By continuity above, B_n=intersection over k≥n of A_k has measure at most b+Σ_(k≥n)2^-k. These B_n increase. Their union N therefore has signed measure b by continuity below and the vanishing geometric tail.

Any measurable subset of N with positive signed measure could be removed to lower ν(N), contradicting minimality. Thus N is negative. Any negative-measure subset of its complement P could be added to lower the minimum, so P is positive. This proves the Hahn decomposition.

Define ν⁺(E)=ν(E intersection P) and ν⁻(E)=−ν(E intersection N). They are positive finite measures, mutually singular because they live on disjoint measurable sets, and ν=ν⁺−ν⁻. Comparing two such disjoint-support decompositions shows their positive and negative measures agree: on a piece simultaneously positive for one and negative for the other the measure must vanish. The Hahn sets themselves need not be unique, because null pieces can move between them.

Define the total variation abs (ν)=ν⁺+ν⁻. Every finite partition sum Σabs (ν(E_j)) is bounded by abs (ν)(E). Splitting E along the Hahn partition attains the bound, giving a representation independent of the chosen partition.

Apply

For a density h relative to a positive measure μ, identify the Hahn sets where h≥0 and h<0, and show the positive and negative variations have densities h⁺ and h⁻. This converts an abstract decomposition into familiar positive and negative parts of an integrable function.

Continue

Signed measures require control of cancellation before applying positive-measure theorems. The next lesson asks when a positive measure can itself be represented by a density relative to another measure. The Hahn–Jordan construction then extends that density theorem to signed measures with integrable variation.

Continue to Recover a density from absolute continuity.