Let $n \in \mathbb{N}_1 := \{1, 2, \dots\}$ and let $B:\Omega \times [0, \infty) \rightarrow \mathbb{R}^n$ be a standard, $n$-dimensional Brownian motion over the probability space $(\Omega, \mathcal{F}, P)$. Let $t \in (0, \infty)$ and consider the following set:

$$ S := \{(\omega, x) \in \Omega\times\mathbb{R}^n \mid: \exists s \in [0,t], x = B(\omega, s)\} $$

Is $S$ $\mathcal{F} \otimes \mathcal{B}_n$-measurable? ($\mathcal{B}_n$ denotes the Borel field over $\mathbb{R}^n$ generated by the Euclidean topology.)


I'm trying to understand why equation (2) in saz's answer makes sense; specifically, why the integrand $1_{B([0,1])}(x,y)$ is measurable.


For simplicity (of notation), we assume $n=2$, i.e. that $(B_s)_{s \geq 0}$ is a $2$-dimensional Brownian motion, and $t=1$. For $k,j \in \mathbb{Z}$ and $m \in \mathbb{N}$ set

$$A_{k,j}^m := \left[ \frac{k}{2^m}, \frac{k+1}{2^m} \right) \times \left[ \frac{j}{2^m}, \frac{j+1}{2^m} \right)$$


$$B_{k,j}^m := \left[ \frac{k-1}{2^m}, \frac{k+2}{2^m} \right] \times \left[ \frac{j-1}{2^m}, \frac{j+2}{2^m} \right].$$

We define random variables

$$X_m(x,\omega) := \sum_{\ell=0}^{2^{8m}} \sum_{k,j} 1_{B_{k,j}^m}(x) \cdot 1_{A_{k,j}^m}(B(\ell/2^{8m},\omega)).$$

(Note that for fixed $(x,\omega)$ and $\ell \in \{0,\ldots,2^{8m}\}$ exactly one of the terms in the series does not equal zero; in particular, the random variable is well-defined.)

$\hspace{80pt}$enter image description here

We claim that $$X := \limsup_{m \to \infty} X_m \in [0,\infty]$$ (which is a measurable) satisfies $\{X \neq 0\}=S$. Indeed: Since $B(\cdot,\omega)$ is continuous, we know that $B([0,1],\omega)$ is compact. Hence, $$(x,\omega) \in S \Leftrightarrow x \in B([0,1],\omega) \Leftrightarrow d:=d(x,B([0,1],\omega))=0.$$

  • Pick $x \in \mathbb{R}^2$, $\omega \in \Omega$ and suppose that $d>0$. Then it follows straight from the definition that $1_{B_{k,j}^m}(x) \cdot 1_{A_{k,j}^m}(B(\ell/2^{8m},\omega))=0$ for any $\ell \in \{0,\ldots,2^{8m}\}$ and $m \geq m_0=m_0(x,\omega)$ sufficiently large. Hence, $X(x,\omega)=0$.
  • Now suppose that $d=0$. Then there exists $t \in [0,1]$ such that $x=B(t,\omega)$. Since the Brownian motion has almost surely Hölder continuous sample paths of order $<1/2$ (see e.g. René Schilling & Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 9), there exists a constant $C=C(\omega)$ such that $$|B_t(\omega)-B_s(\omega)| \leq C(\omega) |t-s|^{1/4}.$$ In particular, $$|B_t(\omega)-B_{t_m}(\omega)| \leq C(\omega) 2^{-2m} = (C(\omega) 2^{-m}) \cdot 2^{-m}$$ where $t_m := \lfloor t 2^{8m} \rfloor/2^{8m}$. This shows that $$x=B_t(\omega) \in B(B(t_m,\omega),2^{-m})$$ for $m$ sufficiently large. Hence, $x \in B_{k,j}^m$ for $k,j$ such that $B(t_m,\omega) \in A_{k,j}^m$. In particular, we see that $X(x,\omega) \geq 1$.

Remark Since $(x,\omega) \mapsto 1_{B([0,1],\omega)}(x)$ is jointly measurable, this implies, by Fubini's theorem, that $$\omega \mapsto \lambda(B([0,1],\omega))$$ is measurable.

  • $\begingroup$ Very nice. Thank you. And you kinda put to waste all my hard work of proving that $L$ is measurable with your little remark there. Chapeau... $\endgroup$ – Evan Aad Apr 12 '14 at 10:21
  • $\begingroup$ @EvanAad Yeah, I'm sorry about that... (But I wouldn't call it a waste; it is good to know how to prove such things on your own.) $\endgroup$ – saz Apr 12 '14 at 11:31

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