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?
Equivalent measures
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functional-analysis
measure-theory
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0What do you mean precisely by $\mu \sim \sigma$? – 2012-09-02
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1I mean that $\mu <<\sigma$ and also $ \sigma <<\mu$. So $\mu(E)=0$ iff $\sigma(E)=0$ – 2012-09-02
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0$L^2(\mu,\mathbb{T})$ is a measure space. In what sense do you see it as a topological space? – 2012-09-02
1 Answers
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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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0Wouldn'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. – 2012-09-02
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1No. 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. – 2012-09-02
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1"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. – 2012-09-02
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0But what about if the space is a probability space (or has finite measure)? Can you find an example then? – 2013-04-05
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0@chango: I'm afraid I don't understand your question. – 2013-04-05
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0Your example above uses the space $\mathbb{R}$ which has infinite measure. Look at my recently posted question: http://math.stackexchange.com/questions/352431/equivalence-of-measures-and-l1-functions – 2013-04-05
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0@chango: Ok, now I understand what you mean. I don't know the answer right ahead. – 2013-04-05