Signed measures and the Jordan decomposition
Separate positive and negative mass intrinsically and compute total variation without confusing cancellation with absence.
On this page
The question for today
What you’ll understand
- Use the finite signed-measure convention consistently.
- Distinguish a Hahn partition from the unique Jordan component measures.
- Compute total variation for a density and for a finite atomic example.
A net balance can vanish even when large positive and negative contributions remain. Signed measures make this distinction precise for distributed charges, differences of probability laws, and integration against signed densities. Throughout this lesson a signed measure means a countably additive, finite real-valued set function with finite total variation. This finite convention keeps every subtraction meaningful. The Hahn and Jordan theorems then separate where signs occur from the uniquely determined positive and negative masses.
Signed values retain countable additivity
A finite signed measure assigns a real number to each measurable set and adds over countably many disjoint sets. It can be constructed as the difference of two finite positive measures. Conversely the decomposition theorem will show every finite signed measure has such a representation. Unlike a positive measure, it need not respect inclusion: adding a region with negative mass can reduce the total. Consequently its value on the whole space does not bound the size of its individual positive and negative contributions.
If h is integrable with respect to a positive measure μ, then ν(A)=∫A h dμ defines a finite signed measure. Integrability of |h| ensures countable additivity and prevents ambiguous infinite cancellation. The set A may contain both signs of h, so ν(A)=0 does not imply h vanishes there or that every subset has zero ν mass. Distinguishing a zero net value from a region on which the entire measure vanishes is essential when discussing positive sets and negative sets.
A sign partition exists, but its boundary is not canonical
A positive set P for ν is a measurable set whose every measurable subset has nonnegative ν value. A negative set N has the corresponding nonpositive property. Hahn's theorem supplies a measurable partition X=P∪N into one positive and one negative set. It is stronger than merely asking ν(P)≥0 and ν(N)≤0: a positive net total could otherwise hide a negative subset. For a density h, the obvious partition uses h≥0 and h<0, with the zero region assignable to either side.
Define ν⁺(A)=ν(A∩P) and ν⁻(A)=−ν(A∩N). These are finite positive measures, they are mutually singular because they live on disjoint pieces, and their difference equals ν. Hahn partitions may differ on regions carrying no positive or negative mass, but the two measures ν⁺ and ν⁻ are independent of that choice. Their uniqueness is the Jordan decomposition theorem. An arbitrary difference of positive measures is not necessarily Jordan: shared mass can be present on both sides until cancellation is removed.
Variation measures the cost of cancellation
The variation of ν on A can also be defined as the supremum, over finite measurable partitions of A, of the sum of the absolute ν values of the pieces. Refining a partition exposes cancellations hidden inside larger pieces. For a density h, variation has density |h|, while the Jordan parts have densities max(h,0) and max(−h,0). This connects the abstract decomposition directly to positive and negative parts of integrable functions. It also gives the bound |ν(A)|≤|ν|(A) for every measurable A.
Integration against ν is defined through its Jordan parts, with absolute integrability measured using |ν|. For a suitable f, the magnitude of ∫f dν is at most ∫|f| d|ν|. This is useful when comparing two probability measures: their signed difference has total mass zero, but its variation quantifies disagreement on events. Numerical conventions for total variation distance may include a factor one half, so the variation measure and a normalized probability distance should be named separately rather than identified silently.
Work it through
A balanced density with nonzero variation
On [0,1], let . Find its Jordan parts, total mass, and total variation mass.
The density is negative below 1/2, zero at 1/2, and positive above it. A Hahn partition is P=[1/2,1] and N=[0,1/2).
The positive mass is the integral of 2x−1 from 1/2 to one, equal to 1/4. The negative-part mass is the integral of 1−2x from zero to 1/2, also 1/4.
Therefore ν(X)=1/4−1/4=0, but |ν|(X)=1/4+1/4=1/2. The Jordan densities are (2x−1)₊ and (1−2x)₊, supported on the respective halves.
Zero net mass hides positive and negative masses of 1/4 each. Total variation records their combined magnitude 1/2.
The living notebook
Explore the model
Interactive lab · waves
Cancellation is not absence of components
Set amplitude ratio a=1 and phase difference to 180 degrees in the two sine waves, then predict their zero sum. Move the phase away from 180 degrees to see the residual. This is a pointwise cancellation analogy; the lab does not compute a Hahn partition, signed-measure variation, or the absolute integral of a waveform.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA canceled sum can coexist with substantial components, just as a zero signed total can hide nonzero variation.
Pause & reason
Check your understanding
Your turn
Try it yourself
On three atoms a,b,c, let ν({a})=3, ν({b})=−2, and ν({c})=1. Compute ν(X), ν⁺(X), ν⁻(X), and |ν|(X), and give a Hahn partition.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Separate atoms according to the sign of their weights, then add magnitudes for variation.
Compare with the solution
Take P={a,c} and N={b}. Every subset of P has nonnegative weight and every subset of N has nonpositive weight. The positive mass is 3+1=4, the negative-part mass is 2, and the signed total is 4−2=2. Total variation is 4+2=6. Using the one-piece partition {X} would expose only absolute net mass two; refining into atoms exposes all six units of variation.
Keep the idea
Take it with you
- The finite convention avoids undefined differences of infinite signed masses.
- Hahn sign partitions can vary, but the mutually singular Jordan component measures are unique.
- Total variation records magnitude before positive and negative cancellation.
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