Lebesgue measure of closure of a set Let $\{X_t \mid t\in\mathbb{N}\}$ be an iid sequence of normal random variables on some probability space $(\Omega,\mathcal{F},P)$. Define $A(\omega) = \{X_t(\omega) \mid t\in\mathbb{N}\}$ for all $\omega\in\Omega$ and let $\overline{A(\omega)}$ be its closure. Let $\ell$ be Lebesgue measure. I am interested in the number
$$
P(\{\omega \mid \ell(\overline{A(\omega)}) = 0\}).
$$
Since $A(\omega)$ is a countable set for any $\omega$ we have $\ell(A(\omega)) = 0$. However, for some $\omega$ we could have for example $A(\omega)$ is $\mathbb{Q}$, so that $\ell(\overline{A(\omega)}) = 1$. In fact, I believe there are uncountably many $\omega$ for which  $\ell(\overline{A(\omega)}) > 0$.
Does anyone have an idea on how to tackle this problem? Any references to texts that might help would be great too. Thank you in advance!
 A: Let $q\in\mathbb{Q}$ and $n\in\mathbb{N}$ be arbitrary. Then, for
each $k\in\mathbb{N}$, the event $A_{q,n}^{\left(k\right)}:=\left\{ \left|X_{k}-q\right|<\frac{1}{n}\right\} $
has the probability $\mathbb{P}\left(A_{q,n}^{\left(k\right)}\right)=\mathbb{P}\left(A_{q,n}^{\left(1\right)}\right)>0$
which is independent of $k\in\mathbb{N}$. Since the $\left(A_{q,n}^{\left(k\right)}\right)_{k\in\mathbb{N}}$
form a family of independent events, the Borel Cantelli lemma implies
$$
\mathbb{P}\left(A_{q,n}\right)=1,\text{ i.e. }\mathbb{P}\left(A_{q,n}^{c}\right)=0
$$
for the event
$$
A_{q,n}:=\left\{ A_{q,n}^{\left(k\right)}\text{ infinitely often}\right\} .
$$
Thus, the event
$$
A:=\bigcup_{\substack{q\in\mathbb{Q}\\
n\in\mathbb{N}
}
}A_{q,n}^{c}
$$
has probability zero.
We conclude that almost surely, the event $A^{c}=\bigcap_{q\in\mathbb{Q},n\in\mathbb{N}}A_{q,n}$
occurs, which means that for fixed $q\in\mathbb{Q}$, for any $n\in\mathbb{N}$,
we have $\left|X_{k}-q\right|<\frac{1}{n}$ for infinitely many $k\in\mathbb{N}$.
This implies that $q$ is an accomulation point of the sequence $\left(X_{k}\right)_{k}$.
Since this holds for arbitrary $q\in\mathbb{Q}$ and since the set
of accumulations points of a sequence is closed, we get
$$
\overline{A\left(\omega\right)}=\mathbb{R}\text{ almost surely}.
$$
In particular,
$$
\mathbb{P}\left(\left\{ \omega\,\mid\,\ell\left(\overline{A\left(\omega\right)}\right) = 0\right\} \right)=0.
$$
Finally, note that we did not really use that each $X_k$ is Gaussian. All we needed was that each interval $(a,b) \subset \Bbb{R}$ has positive probability $\mathbb{P}(X_k \in (a,b)) >0$.
