# Signed Measure Decomposition Integral

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?

$$\nu^{+}(E)=\int 1_E f^{+}d\mu$$ 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 $$\nu^{+}\perp \nu^{-}$$

• 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.
– GE94
Jul 2, 2020 at 9:31