Does a random sequence of vectors span a Hilbert space?

(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.

1. 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.

2. 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).

3. 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 $\operatorname{supp} v \subseteq \overline{V}$.

4. 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.)