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let $\mu,\sigma$ be finite measures on the complex unit circle $\mathbb{T}$. Would it be correct to say that $\mu\sim\sigma$ implies that $L^2(\mu,\mathbb{T})=L^2(\sigma,\mathbb{T})$ as topological spaces?

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What do you mean precisely by $\mu \sim \sigma$? –  Giuseppe Negro Sep 2 '12 at 15:54
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I mean that $\mu <<\sigma$ and also $ \sigma <<\mu$. So $\mu(E)=0$ iff $\sigma(E)=0$ –  user25640 Sep 2 '12 at 15:58
    
$L^2(\mu,\mathbb{T})$ is a measure space. In what sense do you see it as a topological space? –  Martin Argerami Sep 2 '12 at 18:32
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I would not say that $L^2(\mu)=L^2(\sigma)$ as topological spaces, meaning that they are homeomorphic, since any two separable Hilbert spaces are. So this would be a trivial claim.

I would rather say that there exists a natural Hilbert space isomorphism between $L^2(\sigma)$ and $L^2(\mu)$: indeed, by Radon-Nikodym's theorem, there exist measurable $f, g\ge 0$ such that $d\sigma=f d \mu$ and $d\mu=gd\sigma$, so that $fg=1$ both $\sigma$- and $\mu$-almost everywhere. Then the isometric mappings \begin{align} L^2(\sigma)\to L^2(\mu),& & L^2(\mu)\to L^2(\sigma) \\ u \mapsto u \sqrt{f},& & v \mapsto v \sqrt{g} \end{align} are inverse to each other and so they are Hilbert space isomorphisms.

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Wouldn't it be correct to say that every function $f$ is square integrable with respect to $\mu$ iff it is square integrable with respect to $\sigma$? I think $L^2(\mu)$ and $L^2(\sigma)$ are the same as sets and I wonder if the topologies are also the same. –  user25640 Sep 2 '12 at 17:35
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No. Let us make an example on the line $\mathbb{R}$ with the ordinary Lebesgue measure $d\mu$. Define $d\sigma=e^{-x^2}d\mu$. We have $\mu <<\sigma <<\mu$ but the constant function $1$ is in $L^2(\sigma)$ and not in $L^2(\mu)$. Similar examples can be made in $L^2(T)$, too. –  Giuseppe Negro Sep 2 '12 at 17:35
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"I think L2(μ) and L2(σ) are the same as sets and I wonder if the topologies are also the same." I disagree. I think that those sets are different, albeit they have basically the same Hilbert space structure. As for the topology, any two Hilbert spaces share the same topology, this is not very useful. –  Giuseppe Negro Sep 2 '12 at 17:37
    
But what about if the space is a probability space (or has finite measure)? Can you find an example then? –  chango Apr 5 '13 at 18:11
    
@chango: I'm afraid I don't understand your question. –  Giuseppe Negro Apr 5 '13 at 18:52
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