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Let $(\Omega,\mathcal{A})$ be a measurable space.

(Hahn Decomposition Theorem) Let $\varphi\colon\mathcal{A}\to\mathbb{R}$ be a signed measure. Then there exist disjoint sets $\Omega^+\in\mathcal{A}$ and $\Omega^-\in\mathcal{A}$ with $\Omega^+\cup\Omega^-=\Omega$, so that $\varphi(E)\geq 0$ for all $E\in\mathcal{A}, E\subset\Omega^+$ and $\varphi(E)\leq 0$ for all $E\in\mathcal{A}, E\subset\Omega^-$.

(Singular measures) Two measures $\mu,\nu$ on $\mathcal{A}$ are called singular to each other (in signs: $\mu\bot\nu$) if it exists a $A\in\mathcal{A}$ with $\mu(A)=0=\nu(A^C)$.

(Jordan Decomposition Theorem) Let $\varphi\colon\mathcal{A}\to\mathbb{R}$ be a signed measure. Then there exist two finite measures $\varphi^+,\varphi^-$ on $\mathcal{A}$, so that $\varphi=\varphi^+-\varphi^-$ and $\varphi^+\bot\varphi^-$.

Proof of Jordan Decomposition Theorem: Let $\Omega^+\cup\Omega^-=\Omega$ be the Hahn decomposition relating to $\varphi$. For $A\in\mathcal{A}$ define $$ \varphi^+(A):=\varphi(A\cap\Omega^+),~~~~~\varphi^-(A):=-\varphi(A\cap\Omega^-). $$ Then $\varphi^+$ and $\varphi^-$ are the searched measures.


My question concerns this proof. I am wondering why $\varphi^+$ and $\varphi^-$ are finite measures, why $\varphi=\varphi^+-\varphi^-$ and why $\varphi^+\bot\varphi^-$.

Concerning the first point: $\varphi$ is finite by definition, it is $\sigma$-additive and $\varphi(\emptyset)=0$ by definition. So $\varphi$ is a finite measure. Then by definition in the proof above $\varphi^+$ and $\varphi^-$ are finite measures.

Now to the second point I am wondering about: Let $A\in\mathcal{A}$, then it is to show that $$ \varphi(A)=\varphi(A\cap\Omega^+)+\varphi(A\cap\Omega^-). $$ I think this simply follows from the fact that $\Omega^+\cap\Omega^-=\emptyset, \Omega=\Omega^+\cup\Omega^-$. So for $A\in\mathcal{A}$ it is $A\subset\Omega$, which means that $A\subset\Omega^+$ or (exclusive) $A\subset\Omega^-$. In the first case it is $A\cap\Omega^+=A$ and $A\cap\Omega^-=\emptyset$, leading to $\varphi(A)=\varphi(A)+\varphi(\emptyset)=\varphi(A)+0=\varphi(A)$. And analog in the second case.

The only point I cannot see by myself is why $\varphi^+\bot\varphi^-$. It is to show that it exists a $A\in\mathcal{A}: \varphi(A\cap\Omega^+)=0=\varphi(A^C\cap\Omega^-)$. I do not see which $A$ does fullfill that. Do you see it?

Greetings

Miro

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Set $A=\Omega^-$. Then $A^C = \Omega\backslash \Omega^-=\Omega^+$, so $A\cap \Omega^+=A^C\cap \Omega^-=\emptyset$.

Therefore, $$\varphi^+(A)=\varphi(A\cap\Omega^+)=0=-\varphi(A\cap\Omega^-)=\varphi^-(A)$$ which means that $\varphi^+$, $\varphi^-$ are mutually singular.

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