Does Brownian motion visit every point uncountably many times? Let $B_t$ be a one-dimensional standard Brownian motion.


  Is it true that, almost surely, for every $x \in \mathbb{R}$ the set $\{t : B_t = x\}$ is uncountable?


Let $A_x$ be the event that $\{t : B_t = x\}$ is uncountable.  It is well known that $\mathbb{P}(A_0) = 1$.  By recurrence and the strong Markov property, we also have $\mathbb{P}(A_x) = 1$ for every $x$.  I am asking about the probability of $A = \bigcap_{x \in \mathbb{R}} A_x$.  Because of the uncountable intersection, it is not even obvious that this set is measurable.

George Lowther in this comment says the answer is yes, but I don't see how to prove it.
Actually there is an almost sure lower bound on the Hausdorff dimension of the level sets of Brownian motion that follows from the Ray-Knight theorem. 
$$ a.s \quad \forall a\in \mathbb{R}, \quad \text{dim}\{t \geq 0 | B(t) =a\} \geq \frac{1}{2}.  $$

This is a theorem (6.48, p. 170) in the book, "Brownian Motion" by Peres and Morters.
Consider $C_x$ be the event that the set ${t: B_t = x}$ is countable.

For any $x$, the measure of $C_x$ is zero. Countable zero-measure sets union is also a zero-measure set. $A = \bigcap A_x = -\bigcup -A_x = -\bigcup C_x$ and hence has a measure of $1$.