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?
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.