Let $(X, \scr{A})$ be a measure space. If $\mu, \nu$ are finite signed measure, then $$|\nu + \mu|(A) \leq |\mu|(A) + |\nu|(A)$$
where $|\mu| := \mu^+ +\mu^-$ the to tal variation of $\mu$, and $\mu^+, \mu^-$ are the Jordan-decomposition of $\mu.$
I think that it is quite clear that $\mu + \nu$ is a signed measure. Let $E, F$ be the Hahn Decomposition of $X$ with respect to $\mu + \nu$ where $E$ is a negative set and $F$ is a positive set.
Then $$(\mu+\nu)^+(A) = (\mu+\nu)(A \cap F) = (\mu^+ + \nu^+ – \mu^- -\nu^-)(A \cap F) \leq (\mu^+ + \nu^+)(A \cap F) \leq \mu^+(A) + \nu^+(A).$$
Similarly, $$(\mu + \nu)^-(A) = – (\mu + \nu)(A \cap E) \leq \mu^-(A) + \nu^-(A).$$
Then I have the result, But I do not use the assumption $\mu, \nu$ are finite. So I doubt that my proof is wrong somewhere. Can anyone point out why we need the finiteness of signed measures, and is my proof correct ?
Best Answer
It is because that if u and v are not finite then u+v may not be properly defined. If u takes on infinity and v takes on negative infinity, then both u and v are signed measures, but u+v cannot be properly defined.