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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.

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@ Nate Eldredge : Hi here is a naive question, by taking a dense subset $B$ of or $\mathbb{R}$ and using the continuity of Brownian motion, is there a problem to show that $A^B=\bigcap_{x \in B} A_x$ ? Best regards –  TheBridge Jul 3 '12 at 6:50
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@TheBridge: I thought about that a little, but it can't be quite that simple. For instance, one can construct a continuous function $\gamma$ such that $\{t : \gamma(t)=x\}$ is uncountable precisely when $x$ is rational. (Enumerate the rationals as $q_0, q_1, \dots$, and let $\gamma(t) = q_n$ for $t \in [2n, 2n+1]$ and piecewise linear in between.) So something else is needed. –  Nate Eldredge Jul 3 '12 at 7:06
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@ Nate Eldredge: I don't know if this helps but still naively, I just can't help myself thinking about support of Local Times, and remembering about them that Local Times seen as indexed by time and space are jointly continuous (or at least for a good version). Best Regards –  TheBridge Jul 3 '12 at 14:02
    
I believe the fact that the local time process is continuous in space follows from the Ray Knight Theorem, which says that the space distribution of the local time is a squared two-dimensional Bessel Process. –  Jeremy Voltz Jul 4 '12 at 15:10

2 Answers 2

up vote 8 down vote accepted

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.

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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 $1$.

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... but $\bigcup_x C_x$ is an uncountable union ... –  martini Jul 3 '12 at 6:59
    
Sure. I misread the question, thinking you're asking about simple random walk. –  penartur Jul 3 '12 at 7:42

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