Construction of random elements in Hilbert space which are almost surely orthogonal. Let $(\mathcal{H},\langle \cdot, \cdot \rangle)$ be an arbitrary Hilbert space. Can one construct two independent and identically distributed random elements $X,Y:(\Omega,\mathbb{F},P)\to (\mathcal{H},\langle \cdot, \cdot \rangle)$ with $\text{supp}(X)\not = \{0\}$,  such that
$$
\langle X(\omega),Y(\omega) \rangle  =0
$$
for almost all $\omega\in\Omega$, i.e.  $X$ and $Y$ are almost surely orthogonal. 
Question:
I have shown that this can not be done for separable Hilbert spaces $\mathcal{H}$, but is it possible to make such a construction in non-separable Hilbert spaces?
 A: Counterexample.  This counterexample is inspired by class notes by D.J.H. Garling from the 1980s.
Let $\kappa$ be a measurable cardinal https://en.wikipedia.org/wiki/Measurable_cardinal which is also a limit ordinal.  Thus there is a $\{0,1\}$ valued measure on $\kappa$ in which every subset is measurable.
Let $\mathcal H$ be the Hilbert space whose basis is of cardinality $\kappa$, and let $(e_{\alpha})_{\alpha \in \kappa}$ be a basis.  Let $X,Y:\kappa \to \mathcal H$ be the functions $X(\alpha) = e_{\alpha}$ and $Y(\alpha) = e_{\alpha'}$, where $\alpha'$ denotes the successor ordinal of $\alpha$.
Clearly $\langle X,Y\rangle = 0$ everywhere.  It remains to show that $X$ and $Y$ are independent.  But this follows because any two subsets of $\kappa$ are independent (since their measures can only take the values $0$ or $1$).
It does use measurable cardinals, whose existence cannot be proved.  But most likely, if their existence can be disproved, then probably the same proof will show ZF is inconsistent.
A: Let $\mathcal{A} = \mathbb{R}^{[0,1]}$ be the set of all functions from $[0,1]$ to $\mathbb R$, not necessarily continuous, equipped with the product Gaussian measure. Let $\mathcal H$ be $L^2(\mathcal{A})$. $\mathcal H$ is not separable.
Let $x$ be uniformly distributed on $[0,1]$ and let $X_x$ be the function that takes $a \in \mathcal A$ to $a(x)$. The marginal distribution of each $X$ is a Gaussian, so $X_x \in \mathcal H$. Furthermore, $X_x$ is independent of $X_y$ unless $x=y$, which happens with probability zero.
