Does a random sequence of vectors span a Hilbert space? Let $\mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $\mathcal{H}$ such that $P(v \perp h) < 1$ for all $h \in \mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, \ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, \ldots$ is all of $\mathcal{H}?$ 
 A: (This may turn out to be a simplified version of J. E. Pascoe's answer).
The support of (the distribution of) $v$, that we denote by $\operatorname{supp} v$, is the set of vectors $h \in \mathcal{H}$ such that $P(v \in B(h, \varepsilon)) > 0$ for every $\varepsilon > 0$. We list some properties of this set.


*

*The set $\operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v \in B) = 0$. Thus, the support is a closed set.

*Since $\mathcal{H}$ is a separable metric space, it has a countable topological base $\mathcal{B}$, and $\operatorname{supp} v$ is the complement of the union of all $B \in \mathcal{B}$ such that $P(v \in B) = 0$. By countable additivity, it follows that $P(v \in \operatorname{supp} v) = 1$ (the support is a set of full measure).

*With probability one, the closure of the random set $V = \{v_1, v_2, \ldots\}$ contains $\operatorname{supp} v$. Indeed, let $\{h_1, h_2, \ldots\}$ be a countable, dense subset of $\operatorname{supp} v$. For every $i, n = 1, 2, \ldots$ we have $P(v \in B(h_i, \tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V \cap B(h_i, \tfrac{1}{n}) = \varnothing) = 0$. It follows that $h_i \in \overline{V}$ for every $i = 1, 2, \ldots$, and consequently $\operatorname{supp} v \subseteq \overline{V}$.

*For every $h \in \mathcal{H}$, we have $P(h \perp v) < 1$, and therefore $h$ is not orthogonal to $\operatorname{supp} v$. It follows that the closed span of $\operatorname{supp} v$ is $\mathcal{H}$.
It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $\operatorname{supp} v$, which we have shown to be equal to $\mathcal{H}$.
(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)
A: Another Try
We say a $\mathcal{H}$-valued random variable $h$ is a random vector if $P(h \perp g)<1$ for all $g\in \mathcal{H}.$
If $h_1, h_2, \ldots$ is a sequence independent identically distributed of random vectors,
            then, almost surely, the closed span of the $h_i$ is equal to $\mathcal{H}.$
First we will need a lemma.
Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $\mathcal{H}$ such that the closed span of the elements of $A$ is equal to $\mathcal{H}$
and for every point $a\in A,$ $P(h\in U)>0$ for any neighborhood $U$ of $a.$
Proof
            For any subset $A$ such that for every point $a\in A,$ $P(h\in U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $\mathcal{H},$ 
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $\mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)
Choose $g$ such that $g \perp a$ for all $a\in A.$ Now, $P(h \perp g)<1.$ So there must be a point $b$ such that
$P(h\in U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED
Suppose $h_1, h_2, \ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_i\in B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:\mathbb{N}\rightarrow \mathbb{N}^2$ is surjective with infinite multiplicity.)
