Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space, let $X$ be a measurable map from $\Omega$ into a Polish space $E$, define $H_f(t)=\mathbb{P}(|f(X)|>t)$ for some $f:E\to\mathbb{R}$ and let $Q_f$ be the inverse function of $H_f$. If $w$ is a decreasing cadlag function on $[0,1]$ that satisfies $w\geqslant 1$ (where $w(0)=\infty$ is allowed), then we define a (semi-)norm \begin{equation} \|f\|_{2,w}=\Big(\int_0^1w(x)Q_f^2(x)\mathrm{d}x\Big)^{1/2} \end{equation} and we let $L_{2,w}$ be the normed space of all $f$ such that $\|f\|_{2,w}<\infty$.
Problem: I am interested in whether $(L_{2,w},\|\cdot\|_{2,w})$ is a Polish space (and if not, perhaps Lusin or Suslin), given that $\int_0^1w(x)\mathrm{d}x<\infty$.
To start, it is easy to verify that $\|f\|_2\leqslant\|f\|_{2,w}$ and moreover, if $w=1$, then $\|f\|_{2,w}=\|f\|_2$. In fact, if $\int_0^1w(x)^p\mathrm{d}x<\infty$ for some $p>1$, then $L_{2,w}$ can be continuously embedded into $L^q$ for some suitable $q\in[2,\infty)$.
However, I am unable to see whether any such continuous embedding (or any other property for that matter) may help in (dis)proving that indeed this normed space is a Polish space.
As I am not too familiar with the theory of functional analysis, I am asking whether anyone knows whether there are ways to verify that indeed this is the case, perhaps under some additional (mild) conditions?
Remark. As for some background information, a norm similar to that of $\|\cdot\|_{2,w}$ is used in the theory of empirical measures, where $w$ is a weight function to account for any dependencies. In case that this space is Polish (or the weaker condition that it is Suslin), then the supremum of empirical processes indexed by suitably measurable subsets of $L_{2,w}$ become measurable, at least under some suitable technical conditions. Indeed, $w=1$ when we have independence, in which case $L_{2,w}=L_2$ which is Polish, but then again, I am unsure how the situation changes when we introduce the weighted variant.
