TITLE: Is $L^2(\mathbb R)$ isometrically isomorphic with $\ell^2(\mathbb Z)?$
QUESTION [5 upvotes]: Is $L^2(\mathbb R)$ isometrically isomorphic with $\ell^2(\mathbb Z)?$
My thoughts:
We can define an operator $\mathcal L:L^2(\mathbb R)\rightarrow \ell^2(\mathbb Z)$ : $\mathcal Lf=\{\hat f(ξ)\}_{ξ\in \mathbb Z}$ 
(obviously $\mathcal L$ is linear & $1-1$ by uniqueness) and  by the Parseval identity we have that $\lVert f\rVert_{L^2(\mathbb R)}^2=\lVert \hat f\rVert_{\ell^2(\mathbb Z)}^2$ Hence we have an isometry.
is that enough?
Also, can we claim that $:L^2(\mathbb R)≅ \ell^2(\mathbb Z)?$ 
Thanks you.
EDIT: how about $L^2([a,b])?$

REPLY [3 votes]: Another explicit example of an orthonormal basis for $L^2(\mathbb R)$ is $(e^{2\pi i m x} I_{n \le x < n+1})_{(m,n) \in \mathbb Z\times\mathbb Z}$.  And since $\mathbb Z \times \mathbb Z$ is in one to one correspondence with $\mathbb Z$, this gives another isometry from $L^2(\mathbb R)$ to $\ell^2(\mathbb Z)$.