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Jordan Decomposition Theorem 📂Measure Theory

Jordan Decomposition Theorem

Theorem

Let’s assume a measurable space (X,E)(X,\mathcal{E}) and a signed measure ν\nu defined on it. Then, there uniquely exist two positive measures ν+\nu^{+}, ν−\nu^{-} that satisfy the following conditions, and ν=ν+−ν−\nu=\nu^{+}-\nu^{-} is called the Jordan decomposition of ν\nu.

ν=ν+−ν− \nu=\nu^{+}-\nu^{-}

ν+⊥ν− \nu^{+} \perp \nu^{-}

If X=P∪NX=P \cup N is called a partition, then ν+,ν−\nu^{+}, \nu^{-} is as follows.

ν+(E)=ν(E∩P)ν−(E)=−ν(E∩N) \begin{align*} \nu^{+} (E) &= \nu ( E \cap P) \\ \nu^{-}(E) &= -\nu (E \cap N) \end{align*}

Proof

  • Part 1. Existence

    Let’s assume a measurable space (X,E)(X,\mathcal{E}) and a signed measure ν\nu defined on it. Then, according to the partition theorem, there exist the positive set PP and the negative set NN that satisfy X=P∪NX=P \cup N, P∩N=∅P\cap N=\varnothing for ν\nu. Let’s define the positive measures ν+\nu^{+}, ν−\nu^{-} as follows.

    ν+(E):=ν(E∩P)ν−(E):=−ν(E∩N)∀ E∈E \begin{align*} \nu^{+} (E) &:= \nu (E \cap P) \\ \nu^{-}(E) &:= -\nu (E\cap N) \end{align*} \quad \forall\ E\in\mathcal{E}

    Then, since P∪N=XP\cup N=X, the following is trivially true.

    ν(E)=ν(E∩P)+ν(E∩N)=ν+(E)−ν−(E) \nu (E)=\nu (E\cap P)+\nu (E \cap N)=\nu^{+} (E) -\nu^{-} (E)

    Moreover, if E1⊂PE_{1} \subset P, E2⊂NE_2 \subset N, then every E1E_{1} is a null set for ν−\nu^{-}, and every E2E_2 is a null set for ν+\nu^{+}. Therefore, PP is ν−−null\nu^{-} -\mathrm{null}, and NN is ν+−null\nu^{+} -\mathrm{null}. Hence, ν+⊥ν−\nu^{+} \perp \nu^{-}.

  • Part 2. Uniqueness

    Let μ+\mu^+, μ−\mu^- be other two positive measures different from ν+\nu^{+}, ν−\nu^{-} that satisfy the above.

    ν=μ+−μ− \nu=\mu^+ - \mu^-

    And let’s assume two sets that satisfy E, F∈EE,\ F \in \mathcal{E}, μ+⊥μ−\mu^+\perp \mu^-.

    μ(E∩N)=μ−(E)=0=μ+(F)=μ(F∩P) \mu (E\cap N)=\mu^-(E)=0=\mu^+(F)=\mu (F\cap P)

    E∪F=XandE∩F=∅ E \cup F=X \quad \text{and} \quad E\cap F = \varnothing Due to the above two conditions, it can be seen that X=E∪FX=E\cup F is another partition. Here, EE is the positive set, and NN is the negative set. Hence, by the partition theorem, (P−E)∪(E−P) (P-E)\cup (E-P) is a null set for ν\nu. Therefore, for any A∈EA \in \mathcal{E}, the following holds.

    μ+(A)=μ+(A∩E)=ν(A∩E)=ν(A∩P)=ν+(A) \mu^+(A)=\mu^+ (A\cap E)=\nu (A\cap E)=\nu (A \cap P)=\nu^{+}(A)

    Similarly, the following formula is valid.

    μ−(A)=μ−(A∩F)=ν(A∩F)=ν(A∩N)=ν−(A) \mu^-(A)=\mu^- (A\cap F)=\nu (A\cap F)=\nu (A \cap N)=\nu^{-}(A)

    Thus, the two positive measures satisfying the Jordan decomposition theorem are unique.

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