# Is every Hilbert space an $L^2$ space?

Let $H$ be any Hilbert space. Must there exist a measure space $(X,\scr{M},\mu)$ such that we have a Hilbert space isomorphism: $$H\cong L^2(\mu)$$

Thank you

Yes. If $\kappa>0$ is the (Hilbert) dimension of $H$, then you can take for $\mu$ the counting measure on $\kappa$ and then it's easy to see that $H\cong L^2(\mu)=\ell^2(\kappa)$.
If $\kappa=0$, then take any $\mu$ such that $\mu(X)=\infty$ for any nonempty set $X$. Then $L^2(\mu)=\{0\}$.
• $H\cong L^2(\mu)\cong \ell^2(k)$? – user51514 Jul 27 '15 at 15:33
• It is true for separable Hilbert space. Is it true that every Hilbert space $H$ is the same as $L^2(\mu)$? – user51514 Jul 27 '15 at 16:02