Math 2.0

Measure and integration Signed measures and differentiation of measures Theorem

Hahn decomposition theorem

Builds on Signed measureMeasurable space

Statement

Let (X,Σ)(X, \Sigma) be a measurable space and let ν\nu be a signed measure on it. A set A∈ΣA \in \Sigma is positive for ν\nu if ν(E)≥0\nu(E) \ge 0 for every E∈ΣE \in \Sigma with E⊆AE \subseteq A, and negative if ν(E)≤0\nu(E) \le 0 for all such EE. Then there are disjoint sets P,N∈ΣP, N \in \Sigma with P∪N=XP \cup N = X such that PP is positive and NN is negative for ν\nu. If P′,N′P', N' is another such pair, then P△P′=N△N′P \triangle P' = N \triangle N' is null for ν\nu: every measurable subset of it has ν\nu-measure 00.

Remarks

Assume ν\nu does not take the value +∞+\infty; otherwise apply the argument to −ν-\nu. Let ss be the supremum of ν(A)\nu(A) over positive sets AA. A maximising sequence of positive sets has a positive union PP with ν(P)=s\nu(P) = s. Suppose N=X∖PN = X \setminus P contained a set of strictly positive measure. Then a lemma gives a positive subset of NN of strictly positive measure, and adding it to PP contradicts maximality.

Example: let ν(E)=∫Ef dμ\nu(E) = \int_E f \, d\mu with f∈L1(μ)f \in L^1(\mu) real. Then P={f≥0}P = \set{f \ge 0} and N={f<0}N = \set{f < 0} is a Hahn decomposition. The set {f=0}\set{f = 0} can be put into either part, which shows that the decomposition is unique only up to null sets.

Going deeper

For a student

The Hahn decomposition is the first step towards the Jordan decomposition ν=ν+−ν−\nu = \nu^+ - \nu^-, which reduces questions about signed measures to questions about measures. It is also used in the proof of the Radon–Nikodym theorem, applied to signed measures of the form ν−cμ\nu - c \mu.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Hahn_decomposition_theorem
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