Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking at the following version of the Riesz representation theorem:

Let $X$ be a compact metric space and let $\Lambda : C(X) \to \mathbb R$ (or $\mathbb C$) be a continuous linear functional. Then there exists a unique positive measure $|\mu|$ and a measurable function $g$ with $\|g\|_\infty =1$ such that for all $f \in C(X)$: $$ \Lambda (f) = \int_X f g d |\mu|$$

My question is, could I instead say that there exists a unique complex signed mesure $\mu$ such that $$ \Lambda (f) = \int_X f d \mu$$

If yes, why do we want to write signed measures as a product of a bounded measurable function with a positive measure? If no, why not?

Thanks for your help.

share|cite|improve this question
I wasn't sure whether to add the soft-question tag or not, the question seems sort of soft. – Rudy the Reindeer Aug 19 '12 at 6:57
No, it's probably nonsense. Sorry. :-) – user20266 Aug 19 '12 at 6:59
@Thomas I'm glad : ) – Rudy the Reindeer Aug 19 '12 at 7:00
Yes, you could say that. However, to get from your second version to the first version, there's some work to do: there's a variant of Radon-Nikodym hidden in the statement, and your first version also gives the Hahn decomposition theorem essentially for free. – t.b. Aug 19 '12 at 7:03
@t.b. Thank you. – Rudy the Reindeer Aug 19 '12 at 7:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.