# Equivalent measures

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
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

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.
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
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