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I would like to prove the following:

Let $\mu$ be a signed measure on $X$. I want to prove that $|\mu|(X) \neq 0$ iff there exists a measurable $A\subseteq X$ such that $\mu(A)\neq 0$.

I think the following proof works:

$\Rightarrow$: Let's take the Hahn decomposition $(P, N)$. We have $|\mu|(X) = \mu(P) - \mu(N) \neq 0$, meaning that at least one of $\mu(P)$ and $\mu(N)$ is non-zero.

$\Leftarrow$: Let $A$ be such that $\mu(A)\neq 0$. Then, $0 < |\mu(A)| \le |\mu|(A) \le |\mu|(X)$.

I have had little experience with signed measures, so I've been wondering if the above proof is correct.

Paweł Czyż
  • 3,238
  • Yes, it is correct. – geetha290krm Jan 22 '24 at 23:07
  • How is $|\mu|(X)$ defined in your book? The usual definition is $|\mu|(X) = \sup \sum_{i=1}^\infty |\mu(A_i)|$, the sup taken over all countable partitions of $X$ into measurable sets $A_1, A_2, \dots$. This then makes your statement trivial, as the right side is nonzero if and only if $X$ has a measurable subset of nonzero measure. – Nate Eldredge Jan 22 '24 at 23:14
  • Thank you for both answers! @NateEldredge this is a beautiful argument. Would you mind posting it as an answer, preferably linking this post? It explains why Jordan decomposition definition of $|\mu|$ (which I implicitly had in mind) and the supremum over partitions are equivalent. – Paweł Czyż Jan 23 '24 at 09:27

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