Let $(X,\mathcal{A},\mu)$ be a measure space and f integrable (on it). The signed measure $v$ is defined by $v(A)=\int 1_A f d \mu,$ $\text{ }$$A \in \mathcal{A}$. Determine a unique decomposition $v=v^+-v^-$ in two to each other singular measures $v^+$ and $v^-$.

Two measures $\mu_1$ and $\mu_2$ are singular together, if the sets $A,B \in \mathcal{A}$ exists mit $A \cup B=X,A \cap B=\emptyset,\mu_1(A)=0$ and $\mu_2(B)=0$.

My suggestion (basically i haven't any idea because very unknown topic):

$v(A\cup B)=\int 1_{A\cup B}fd\mu=\int_{A\cup B}f^+d\mu-\int_{A\cup B}f^-d\mu$

If i now could assume that $\mu_2(A)=0=\mu_1(B)$ is a precondition, then i would say: $=\int_A f^+d\mu_2-\int_B f^-d\mu_1=v^+(A)-v^-(B)$

What is the right solution?


1 Answer 1


$\nu^{+}(E)=\int 1_E f^{+}d\mu$ and $\nu^{-}(E)=\int 1_E f^{-}d\mu$ defines two measures with $\nu^{+}-\nu^{-}=\nu$. Let $A=\{x: f(x) \geq 0\}$ and $B=\{x: f (x) <0\}$. Then $A\cup B=X$ and $A\cap B$ is empty. Also $\nu^{+}(E)=\nu (A \cap E)$, $\nu^{-}(E)=\nu (B \cap E)$, $\nu^{+}(B)=0$ and $\nu^{-}(A)=0$. Hence $\nu^{+}\perp \nu^{-}$

  • $\begingroup$ Thanks for the solution. So A and B are preconditions. I thought that there is a more non trivial general derivation (because of the conceptual formulation) , but it isn't that difficult. $\endgroup$
    – GE94
    Jul 2, 2020 at 9:31

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