Hey, I know this is commonly a homework question, but it came up in my own studies; so this isn't a homework question for me. I hope it's alright that I put it here.(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to show that if [itex] f dx = d\lambda[/itex] for some [itex] f \in L^1(\mathbb{R}^d) [/itex] and complex Borel measure [itex]\lambda[/itex] then [itex] |f| dx = d|\lambda| [/itex] (i.e., [itex] f \to fdx [/itex] is an isometry).

What I've done so far is constructed a [itex] \psi \in L^1(\lambda) [/itex] so that [itex] \psi d\lambda = d|\lambda| [/itex] with [itex]|\psi| = 1 [/itex]. So we then have that [itex] f \psi dx = \psi d\lambda = d|\lambda| [/itex], so it looks like what I need to show is that [itex] f \psi = |f| [/itex]. I'm not really sure how to do that though.

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# Embedding L1 in the Banach space of complex Borel measures

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