In "The Free Markoff Field" by E. Nelson, it is claimed that if $H$ is an infinite-dimensional real Hilbert space, then there is a non-atomic measure space $(K,m)$ such that $L^{2}(K,m)$ is isomorphic to $H$. How does this work? I know that if $\{u_{\alpha}\}_{\alpha \in \mathcal{A}}$ is a complete orthonormal set, then the Plancherel formula $\langle x, y \rangle = \sum_{\alpha \in \mathcal{A}} \langle x, u_{\alpha}\rangle \langle u_{\alpha},y\rangle$ exhibits a unitary isomorphism between $H$ and $L^{2}(\mathcal{A},c)$, where $c$ is the counting measure. However, $(\mathcal{A},c)$ is certainly atomic. Apparently this is untrue if $H$ is finite-dimensional, but it's not obvious to me how to get something better than $(\mathcal{A},c)$ when $\mathcal{A}$ is infinite.
[Math] Infinite dimensional Hilbert space is isomorphic to $L^{2}$
functional-analysishilbert-spacesmeasure-theory
Best Answer
For $H$ infinite-dimensional and separable, you have $H\simeq L^2[0,1]$ (or, instead of $[0,1]$, you can use any Borel subset of $\mathbb R^n$ of $\mathbb C^n$ with nonempty interior; other choices are possible, too).
When $H$ has dimension higher than countable, by decomposing its basis in a union of countable orthonomal sets we may write $H=\bigoplus_j H_j$, where each $H_j$ is separable. This will be obviusly isomorphic to $\bigoplus _j L^2[0,1]$, and now you can identify this latter space with $L^2(T)$, where $T$ is the disjoint union of copies of $[0,1]$, and the measure is the product measure.