Let $(\Omega,\mathcal{F},\mathbb{P})$ be a complete probability space, $X\in L_{\mathbb{P}}^2(\mathcal{F})$, $\mathcal{F}^X$ be the $\sigma$-algebra gneerated by $X$, and let $m$ be the law of $X$ $\mathbb{R}$.
Since both $L_{\mathbb{P}}^2(\mathcal{F}^X)$ and $L_m^2(\mathscr{B}(\mathbb{R}))$ are separable infinite-dimensional Hilbert spaces, they must be isometrically isomorphic. My question is, does the map $$ f\mapsto f(X), $$ give an isometric isomorphism from $L_m^2(\mathbb{R})$ to $L_{\mathbb{P}}^2(\mathcal{F}^X)$? If so,
The answer is YES. This is easy from the definition of $m$ in terms of $P$ and $X$ and the fact that any random variable measurable w.r.t. $\mathcal F^{X}$ is of the form $f(X)$ for some measurable function $f: \mathbb R \to \mathbb R$.
[Note that $Ef(X)^{2}=\int f^{2} dm$].