A planar Brownian motion has area zero

Question

I'm looking for proofs of Paul Lévy's theorem that a planar Brownian motion has Lebesgue measure $0$. I know of only two proofs: one is in Lévy's original paper (Théorème 12, p. 532) and the other is in Mörters & Peres's "Brownian Motion" (Theorem 2.24, p. 46).

Unfortunately, Mörters & Peres's proof leaves out many technical details, without which I find the proof hopelessly difficult to understand.

I have not attempted to read Lévy's original proof, partly because it's in French (though I can read French, with effort) and partly because it's in the middle of a long article that was not written for students, but rather for professional mathematicians, and hence I suspect that it would be a pain for me to read it.

I have not been able to find any other sources containing a proof of this theorem. If someone knows of any such sources, in English, French or German, ideally textbooks (but any other source will be helpful too), please let me know.

Thank you.

Answer 1

Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof.

---

Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and denote by $\lambda$ the (two-dimensional) Lebesgue measure.

First, we show that $L := \lambda(B([0,1]))$ satisfies $\mathbb{E}L<\infty$. To this end, note that

$$\begin{align*} \{L>4r^2\} &\subseteq \left\{ \max_{t \in [0,1]}(|B^1(t)|,|B^2(t)|) > r \right\} \\ &= \left\{ \max_{t \in [0,1]} B^1(t) > r \right\} \cup \left\{ \max_{t \in [0,1]} B^2(t) > r \right\} \\ &\quad \cup \left\{ \min_{t \in [0,1]} B^1(t) < -r \right\}\cup \left\{ \min_{t \in [0,1]} B^2(t) < -r \right\}.\end{align*}$$

From the reflection principle we know that $\max_{t \in [0,1]} B^j(t) \sim -\min_{t \in [0,1]} B^j(t) \sim |B^j(1)|$ for $j=1,2$. Hence,

$$\mathbb{P}(L>4r^2) \leq 4 \sqrt{\frac{2}{\pi}} \int_r^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \leq \frac{4 \sqrt{2}}{\pi} \frac{1}{r} \exp \left(- \frac{r^2}{2} \right).$$

Combining this estimate with the fact that $\mathbb{E}L = \int_0^{\infty} \mathbb{P}(L >r) \, dr$, we get $\mathbb{E}L<\infty$.

Next, we note that for the restarted Brownian motion $W_t := B_{t+1}-B_1$, we have $\lambda(W([0,1])) = \lambda(B([1,2]))$ and, since $W \sim B$, we conclude that $$\mathbb{E}(\lambda(B([0,1]))) = \mathbb{E}(\lambda(B([1,2]))).$$ Similarly, as $\tilde{W}_t := \frac{1}{\sqrt{2}} B_{2t}$ is a Brownian motion, we have

$$\lambda(\tilde{W}([0,1])) = \lambda \left( \frac{1}{\sqrt{2}} B([0,2]) \right) = \frac{1}{2} \lambda(B([0,2])$$

i.e. $\lambda(B([0,2])) \sim 2 \lambda(B([0,1]))=2L$. Therefore,

$$\begin{align*} 2\mathbb{E}L = \mathbb{E}(\lambda(B([0,2]) &= \mathbb{E}(\lambda(B([0,1])) + \mathbb{E}(\lambda(B([1,2])) - \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) \\ &= 2\mathbb{E}L + \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))). \end{align*}$$

Hence, $$ \mathbb{E}\big(\lambda(B([0,1]) \cap B([1,2]))\big)=0. \tag{1}$$

By Fubini's theorem, we have

$$\mathbb{E}L = \int \int 1_{B([0,1])}(x,y) d\lambda(x,y) \, d\mathbb{P} = \int \underbrace{\mathbb{P}((x,y) \in B([0,1]))}_{=:p(x,y)} \, d\lambda(x,y). \tag{2}$$

If we set $W_t := B_{1-t}-B_1$ and $\tilde{W}_t := B_{t+1}-B_1$, then both processes are Brownian motions and

$$W([0,1]) = B([0,1])-B_1 \qquad \qquad \tilde{W}([0,1]) = B([1,2])-B_1.$$

Using that $(W_t)_{t \geq 0}$ and $(\tilde{W}_t)_{t \geq 0}$ are independent, we get

$$\begin{align*}\mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) &= \mathbb{E}(\lambda(W([0,1]) \cap \tilde{W}([0,1]))) \\ &= \int \mathbb{P}((x,y) \in W([0,1]) \cap \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int \mathbb{P}((x,y) \in W([0,1])) \mathbb{P}((x,y) \in \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int p^2(x,y) \, d\lambda(x,y). \end{align*}$$

Now $(1)$ implies $p(x,y)=0$ $\lambda$-almost everywhere and therefore, by $(2)$, $\mathbb{E}L=0$. Hence, as $L \geq 0$, we finally conclude $L=0$ almost surely.

Remark It is not obvious that $\omega \mapsto L(\omega)$ and $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y))$ are random variables; see this question and Evan Aad's answer for the measurability of $L$ and this question for the measurability of the second mapping. (Note that measurability of $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y)$ entails the measurability of $L$, by Fubini's theorem.)

Answer 2

I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable.

EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end).

Introduction & proof outline 0

Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable function. The proof is based on ideas presented by Paul Lévy in [L] (p. 533), but mine is not exactly the same as the proof given there, simply because I didn't quite understand that proof.

I first establish some notations (Notation 1), which are used in the sequel. I then introduce three constants (Constants 2) that will be in effect throughout. A few definitions and lemmas lead to the main result (theorem 9). The functions introduced in Definition 5 and Definition 7 correspond, more or less, to objects used in Lévy's proof. Lemma 6 and Lemma 8 study the properties and inter-relationships between these objects.

The idea of the proof is to cover the Brownian path segment between the times $0$ and $t$ ($g_t(\omega)$) by two sets: $f_{m, r, t}(\omega)$ and $h_{r, t}(\omega)$. $f_{m, r, t}(\omega)$'s measure $F_{m, r, t}(\omega)$ can readily be shown to be a measurable function of $\omega$, whereas $h_{r, t}(\omega)$'s measure $H_{r, t}(\omega)$ can readily be shown to converge to $g_t(\omega)$'s measure $G_t(\omega)$, as $r$ diminishes to $0$. These two desireable properties ($F_{m, r, t}$'s measurability and $H_{r, t}$'s convergence to $G_t$) are combined (Theorem 9.1) using a result that I proved in another post (theorem 1 here).

The reason why $F_{m, r, t}$'s measurability is relatively easy to demonstrate, is that $f_{m, r, t}(\omega)$ has a simple structure: it is the union of a finite number ($m$) of balls of the same radius ($r$). Such unions are denoted below by $e_r(x_1, \dots, x_m)$ with $x_1, \dots, x_m$ marking the centers of the balls. To assist in the study of the properties of $e_r(x_1, \dots, x_m)$'s measure $E_r(x_1, \dots, x_m)$, we use bounds derived in Lemma 4.

All the functions listed in the previous two paragraphs are introduced in Definition 5 and studied in Lemma 6. The functions introduced in Definition 7, whose properties are studied in Lemma 8, are used in the main result (Theorem 9.1) to construct a sequence of covers $f_{m_k, 1 / k, t}$ converging to $g_t$ and to quantify the rate of convergence of $F_{m_k, 1/k, t}$ to $G_t$. To quantify this rate of convergence, a uniform upper bound ($u$) is calculated (Lemmas 8.2(c)) on the maximal distance ($d_{a, b}(\omega)$) between two points that lie on the path segment $g_t(\omega)$ between times $a$ and $b$.

The entire proof is carried out under the assumption (Constants 2.1) that $n \geq 2$. In the end (Remark 10) I indicate how to extend the proof to the case $n = 1$.

The published works referenced in the post are listed at the end of the post in the section "Works cited".

Notation 1

For every $n \in \mathbb{N}_1 := \{1, 2, \dots\}$,

1. Denote the Borel field on $\mathbb{R}^n$ by $\mathcal{B}_n$,
2. Denote the Lebesgue measure on $\mathcal{B}_n$ by $\lambda_n$ and the outer measure on $\mathbb{R}^n$ by $\lambda_n^*$,
3. Denote by $|\cdot|_n$ the standard Euclidean norm on $\mathbb{R}^n$,
4. For every $x \in \mathbb{R}^n$ and every $r \in (0,\infty)$, define $$ \begin{align} \mathbf{B}_x(r) & := \{y \in \mathbb{R}^n \mid: |y - x|_n < r\} \\ \overline{\mathbf{B}_x(r)} & := \{y \in \mathbb{R}^n \mid: |y - x|_n \leq r\} \end{align} $$
5. Denote the topology induced by $|\cdot|_n$ on $\mathbb{R}^n$ by $\mathcal{O}_n$ and for every $m \in \mathbb{N}_1$, denote the product topology induced by $|\cdot|_n$ on $(\mathbb{R}^n)^m$ by $\mathcal{O}_{n;m}$.
6. If $S$ is a set, we denote its power set by $2^S$.
7. Denote by $\mathbf{C}_{(0; n)}$ the set of continuous functions $f : [0, \infty) \rightarrow \mathbb{R}^n$, such that $f(0) = 0$. Denote by $\mathcal{B}(\mathbf{C}_{(0; n)})$ the Borel $\sigma$-algebra on $\mathbf{C}_{(0; n)}$ (cf. [S], p. 41).

Constants 2

Fix the following for the remaining of this post.

1. $n \in \mathbb{N}_1$, such that $n \geq 2$,
2. $S := (\Omega, \mathcal{F}, P)$ - A probability space,
3. $W : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ - a standard, $n$-dimensional Brownian motion over $S$. Denote $W$'s components thus: $W = (W_1, \dots, W_n)$.

Definition 3

Set $$ c := \frac{\pi^{n / 2}}{\Gamma(\frac{n}{2} + 1)} $$ (see the next lemma for the rationale.)

Lemma 4

1. Let $x, y \in \mathbb{R}^n$, let $\varepsilon \in (0,1]$ and let $r \in (0,\infty)$. If $|x - y|_n < \varepsilon$, $$ \lambda_n(\mathbf{B}_x(r) \Delta \mathbf{B}_y(r)) < 2c(r + 1)^n\varepsilon $$
2. Let $m \in \mathbb{N}_1$ and consider some pair of $m$-length sequences in $\mathcal{B}_n$: $(B_1, \dots, B_m)$, $(C_1, \dots, C_m)$ with $B_i, C_i \in \mathcal{B}_n$ for every $i \in \{1, \dots, m\}$. Let $\varepsilon \in (0,\infty)$. If, for each $i \in \{1, \dots, m\}$, we have $\lambda_n(B_i \Delta C_i) < \varepsilon$, then $$ \lambda_n\left(\left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right)\right) < m\varepsilon $$
3. Let $m\in \mathbb{N}_1$, let $\varepsilon \in (0,\infty)$, let $x := (x_1, \dots, x_m), y := (y_1, \dots, y_m) \in (\mathbb{R}^n)^m$ and let $r \in (0,\infty)$. If, for every $i \in \{1, \dots, m\}$, $|x_i - y_i|_n < \varepsilon$, we have $$ \lambda_n\left(\left(\bigcup_{i = 1}^m \mathbf{B}_{x_i}(r)\right) \Delta \left(\bigcup_{j = 1}^m \mathbf{B}_{y_j}(r)\right)\right) < 2mc(r + 1)^n\varepsilon $$

Proof

1. By the triangle inequality, $\mathbf{B}_y(r) \subseteq \mathbf{B}_x(r + |y - x|_n)$. Hence $\mathbf{B}_y(r) \setminus \mathbf{B}_x(r) \subseteq \mathbf{B}_x(r + |y - x|_n) \setminus \mathbf{B}_x(r)$. Hence $$ \lambda_n(\mathbf{B}_y(r) \setminus \mathbf{B}_x(r)) \leq \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r)) $$

   By the formula for the volume of an $n$-dimensional ball, $$ \begin{aligned} \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r)) & = c(r + |y - x|_n)^n - cr^n \\ & = c \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^k \\ & = c |y - x|_n \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^{k - 1} \\ & \leq c \varepsilon \sum_{k = 1}^n\binom{n}{k}r^{n - k} \\ & < c \varepsilon (r + 1)^n \end{aligned} $$

   By an analogous argument, $\lambda_n(\mathbf{B}_x(r) \setminus \mathbf{B}_y(r)) < c \varepsilon (r + 1)^n$.

2. By properties of the symmetric difference, $$ \left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right) \subseteq \bigcup_{i = 1}^m \left(B_i \Delta C_i\right) $$
3. The result follows from the first two parts.

Q.E.D.

Definition 5

For every $m \in \mathbb{N}_1$, every $t \in (0,\infty)$ and every $r \in (0,\infty)$,

1. Define $e_{m,r} : (\mathbb{R}^n)^m \rightarrow 2^{\mathbb{R}^n}$ and $E_{m,r} : (\mathbb{R}^n)^m \rightarrow [0,\infty)$ as follows $$ \begin{align} e_{m,r}(x_1, \dots, x_m) & := \bigcup_{i = 1}^m\overline{\mathbf{B}_{x_i}(r)} \\ E_{m,r}(x_1, \dots, x_m) & := \lambda_n(e_m(x_1, \dots, x_m)) \end{align} $$
2. Define $f_{m, r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $F_{m, r, t} : \Omega \rightarrow [0,\infty)$ as follows $$ \begin{align} f_{m, r, t}(\omega) & := e_{m,r}(W(\omega, 0), W(\omega, \frac{t}{m}), W(\omega, 2\frac{t}{m}), \dots, W(\omega, t)) \\ F_{m, r, t}(\omega) & := \lambda_n(f_{m, r, t}(\omega)) \end{align} $$
3. Define $g_t : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G_t : \Omega \rightarrow [0,\infty]$ as follows $$ \begin{align} g_t(\omega) & := \{W(\omega, s) :\mid s \in [0,t]\} \\ G_t(\omega) & := \lambda_n^*(g_t(\omega)) \end{align} $$
4. Define the functions $g^* : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G^* : \Omega \rightarrow [0,\infty]$ as follows $$ \begin{aligned} g^*(\omega) & := \{W(\omega, t) :\mid t \in [0,\infty)\} \\ G^*(\omega) & := \lambda_n^*(g^*(\omega)) \end{aligned} $$
5. Define $h_{r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $H_{r, t} : \Omega \rightarrow [0,\infty]$ as follows $$ \begin{align} h_{r, t}(\omega) & := \bigcup_{s \in [0, t]} \overline{\mathbf{B}_{W(\omega, s)}(r)} \\ H_{r, t}(\omega) & := \lambda_n^*(h_{r, t}(\omega)) \end{align} $$

Lemma 6

For every $m \in \mathbb{N}_1$, every $r \in (0,\infty)$, every $t \in (0,\infty)$, every sequence $r_1, r_2, \dots \in (0,\infty)$ such that $\lim_{k \rightarrow \infty} r_k = 0$ and every $\omega \in \Omega$,

1. $h_{r, t}(\omega)$ is compact (and therefore $\in \mathcal{B}_n$ and $\lambda_n^*(h_{r, t}(\omega)) = \lambda_n(h_{r, t}(\omega))$).
2. For every $r' \in [r,\infty)$, we have $$ \begin{aligned} h_{r, t}(\omega) & \subseteq h_{r', t}(\omega) \\ H_{r, t}(\omega) & \leq H_{r', t}(\omega) \end{aligned} $$
3. We have $$ g_t(\omega) = \bigcap_{k = 1}^\infty h_{r_k, t}(\omega) $$ (and therefore $g_t(\omega) \in \mathcal{B}_n$ and $\lambda_n^*(g_t(\omega)) = \lambda_n(g_t(\omega)$).
4. We have $$ G_t(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega) $$
5. We have $$ \begin{align} f_{m, r, t}(\omega) & \subseteq h_{r, t}(\omega) \\ F_{m, r, t}(\omega) & \leq H_{r, t}(\omega) \end{align} $$
6. $E_{m, r}$ is $\mathcal{O}_{n; m}/\mathcal{O}_1$-continuous (and hence $(\otimes_{i = 1}^m\mathcal{B}_n)/\mathcal{B}_1$-measurable).
7. $F_{m, r, t}$ is $\mathcal{F}/\mathcal{B}_1$-measurable.
8. $g^*(\omega) \in \mathcal{B}_n$ (and therefore $\lambda_n^*(g^*(\omega)) = \lambda_n(g^*(\omega))$.

Proof

1. Define $d : \mathbb{R}^n \rightarrow [0,\infty)$ as follows $$ d(x) := \inf \{|x - y|_n :\mid y \in g_t(\omega)\} $$ Then $h_{r, t}(\omega) = d^{-1}([0,r])$. Hence, by the fact that $d$ is $\mathcal{O}_n / \mathcal{O}_1$-continuous ([M2] p. 175), $h_{r, t}(\omega)$ is closed. Since $[0,t]$ is compact, so is $g_t(\omega)$. In particular, $g_t(\omega)$ is bounded. Let $R \in (0,\infty)$ be such that $g_t(\omega) \subseteq \overline{\mathbf{B}_0(R)}$. Then $h_{r, t}(\omega) \subseteq \overline{\mathbf{B}_0(R + r)}$. In particular, $h_{r, t}(\omega)$ is bounded. $h_{r, t}(\omega)$ is closed and bounded, hence compact.
2. Immediate.

3. It is evident that $$ g_t(\omega) \subseteq \bigcap_{k = 1}^\infty h_{r_k, t}(\omega) $$ I will prove the converse containment. Let $x \in \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$. For every $k \in \mathbb{N}_1$, let $y_k \in g_t(\omega)$ be such that $|y_k - x|_n \leq r_k$. Since $g_t(\omega)$ is compact, there is, by the Bolzano-Weierstrass theorem, a convergent subsequence $(y_{k_1}, y_{k_2}, \dots)$. Set $y := \lim_{i \rightarrow \infty} y_{k_i}$. Since $g_t(\omega)$ is closed, $y \in g_t(\omega)$. I will show that $x = y$.

   Let $\varepsilon \in (0,\infty)$ and let $i \in \mathbb{N}_1$ be such that $r_{k_i} < \varepsilon / 2$ and $|y_{k_i} - y|_n < \varepsilon / 2$. We have, by the triangle inequality, $$ |y - x|_n \leq |y - y_{k_i}|_n + |y_{k_i} - x|_n < \varepsilon $$ Since $\varepsilon$ was chosen arbitrarily in $(0,\infty)$, $|y - x|_n = 0$.

4. Choose a strictly decreasing subsequence $(r_{k_1}, r_{k_2}, \dots)$. Then by 6.2 and 6.3, $$ g_t(\omega) = \bigcap_{i = 1}^\infty h_{r_{k_i}, t}(\omega) $$ Hence, by the continuity of measures, and by the fact (6.1) that $h_{r_{k_1}, t}$ is bounded and therefore of finite Lebesgue measure, $$ G_t(\omega) = \lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega) $$ But by 6.2, $\lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega)$.
5. Immediate.
6. Fix some $x_1, \dots, x_m \in \mathbb{R}^n$. For every $y_1, \dots, y_m \in \mathbb{R}^n$, we have $$ |E_{m, r}(y_1, \dots, y_m) - E_{m, r}(x_1, \dots, x_m)| \leq \lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m)) $$ Let $\varepsilon \in (0,\infty)$. Define $$ \delta := \frac{\varepsilon}{2 m c (r + 1)^n} $$ If $|y_i - x_i|_n < \delta$ for all $i \in \{1, \dots, m\}$, then, by lemma 4.3, $$ \lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m)) < \varepsilon $$
7. Observe that, for all $\omega \in \Omega$, $$ F_{m, r, t}(\omega) = E_{m, r}(W(\omega, 0), W(\omega, \frac{t}{m}), \dots, W(\omega, t)) $$ The result now follows by 6.6.
8. The result follows from 6.3 by observing that $$ g^*(\omega) = \bigcup_{k = 1}^\infty g_k(\omega) $$

Q.E.D.

Definition 7

1. Define $u : (n - 2, \infty) \rightarrow (0,\infty)$ as follows $$ u(s) := \frac{1}{\sqrt{\pi}} \frac{s}{s - (n - 2)} \exp\left(-\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right) $$ (for the rationale - see the next lemma.)
2. For every $a, b \in [0, \infty)$ with $a < b$, define $d_{a, b}: \Omega \rightarrow [0,\infty]$ as follows $$ d_{a, b}(\omega) := \max \{\left|W(\omega, s) - W(\omega, a)\right|_n :\mid s \in [a, b]\} $$
3. For every $m \in \mathbb{N}_1$ and every $t \in (0,\infty)$, define $D_{m, t} : \Omega \rightarrow [0,\infty)$ as follows $$ D_{m, t}(\omega) := \max(d_{0, \frac{1}{m}t}(\omega), d_{\frac{1}{m}t, \frac{2}{m}t}(\omega), \dots, d_{\frac{m - 1}{m}t, t}(\omega)) $$

Lemma 8

1. $\lim_{s \rightarrow \infty} s u(s) = 0$

2. Let $a, b \in [0, \infty)$ with $a < b$.

   a) $d_{a,b}$ is $\mathcal{F}/\mathcal{B}_1$ - measurable (and, therefore, so is $D_{m, t}$ for all $m \in \mathbb{N}_1$, $t \in (0, \infty)$).

   b) $d_{a, b} \sim d_{0, b - a}$,

   c) For every $r \in (0,\infty)$, $$ \frac{r^2}{b} > n - 2 \implies P(d_{0, b} > r) \leq u\left(\frac{r^2}{b}\right) $$

3. Let $r \in (0,\infty)$ and $t \in (0,\infty)$.

   a) For every $m \in \mathbb{N}_1$, $$ \{D_{m , t} \leq r \} \subseteq \{g_t \subseteq f_{m, r, t}\} = \{G_t \leq F_{m, r, t}\} $$ b) There exists some $M \in \mathbb{N}_1$, such that, for all $m \in \{M, M + 1, \dots\}$, $$ P(D_{m, t} > r) \leq m u\left(\frac{r^2}{t} m\right) $$

Proof

1. Since, for every $s \in (0,\infty)$, $$ \frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right) = \frac{1}{2} \frac{s}{n} \left(n - (n - 2) \frac{\ln(s / n)}{s / n} + \frac{\ln(n) - n}{s / n} \right) $$

   we have

   $$ \lim_{s \rightarrow \infty} \frac{\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)}{\frac{s}{2\sqrt{\pi}}} = \sqrt{\pi} > 1 $$

   Hence $$ \limsup_{s \rightarrow \infty} \frac{\exp\left(- \frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right)}{\exp(- \frac{s}{2\sqrt{\pi}})} \leq 1 $$

   Hence $$ \limsup_{s \rightarrow \infty} \frac{s u(s)}{\frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}})} \leq 1 $$

   But $$ \lim_{s \rightarrow \infty} \frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}}) = 0 $$

   Therefore, $\limsup_{s \rightarrow \infty} s u(s) = 0$.

2. a) Let $\omega \in \Omega$. Since $W(\omega)$ is continuous and since the rational numbers are dense in $\mathbb{R}$, we have $$ d_{a, b}(\omega) = \sup \{\left|W(\omega, q) - W(\omega, a)\right|_n :\mid q \in [a, b] \cap \mathbb{Q}\} $$ The result now follows from the fact ([K2] theorem 1.92, p. 40) that the supremum of a denumerable family of random variables is a random variable.

   b) Define $\varphi: \mathbf{C}_{(0; n)} \rightarrow [0,\infty)$ as follows $$ \varphi(f) := \max f([0,t]) $$ Firstly, I will show that $\varphi$ is $\mathcal{B}(\mathbf{C}_{(0; n)})/\mathcal{B}_1$-measurable. It suffices to show that for all $M \in \mathbb{R}$, $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$. Let $M \in \mathbb{R}$. We have $$ \{\varphi \leq M\} = \bigcap_{s \in [0,t]} \{f \in \mathbf{C}_{(0; n)} \mid: f(s) \leq M\} = \bigcap_{q \in [0,t] \cap \mathbb{Q}} \{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\} $$ For every $q \in [0,t] \cap \mathbb{Q}$, $\{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$, hence $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$.

   Now, define the process $V : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ as follows: $$ V(\omega, t) := W(\omega, a + t) - W(\omega, a) $$ $V$ is a standard, $n$-dimensional Brownian motion ([S] 2.9, p. 12), hence $V \sim W$. Therefore, $$ d_{a, b} = \varphi \circ V \sim \varphi \circ W = d_{0, b - a} $$ c) For every $k \in \{1, \dots, n\}$, define $$ d_{0, b; k}(\omega) := \max \{\left|W_k(\omega, s) - W_k(\omega, a)\right|_1 :\mid s \in [0, b]\} $$ We have $$ d_{0, b} \leq \sqrt{d_{0, b; 1}^2 + \cdots + d_{0, b; n}^2} $$ Hence, $$ \begin{aligned} P(d_{0, b} > r) & = P(d_{0, b}^2 > r^2) & \\ & \leq P(\sum_{k = 1}^n d_{0, b; k}^2 > r^2) & \\ & = P(\sum_{k = 1}^n W_k^2(b) > r^2) & (1) \\ & = \chi^2(n)\left(\left(r^2/b, \infty\right)\right) & (2) \\ & \leq u\left(\frac{r^2}{b}\right) & (3) \end{aligned} $$ where (1) is by Bachelier's maximum process theorem ([K1] Proposition 13.13, p. 256) together with the fact that the components of an $n$-dimensional Brownian motion are independent; $\chi^2(n)$ in (2) denotes the (central) chi-squared distribution with $n$ degrees of freedom; and (3) is from [I2] equation (3.1), p. 341 (see also [I1] Lemma 1, p. 586; a proof can be found there in the appendix).

3. a) Let $\omega \in \{D_{m, r} \leq r\}$ and let $x \in g_t(\omega)$. Let $s \in [0,t]$ be such that $x = W(\omega, s)$. Define $$ K := \max \{k \in \{0, 2, \dots, m - 1\} \mid: s \geq \frac{k}{m}\} $$ Since $d_{K, K + 1}(\omega) \leq r$, $g_s(\omega) \in \mathbf{B}_{g_{K / m}(\omega)}(r) \subseteq f_{m, r, t}(\omega)$.

   b) Let $m \in \mathbb{N}_1$ be such that $$ \frac{r^2}{t} m > n - 2 $$ Then $$ \begin{aligned} P(D_{m,t} > r) & \leq \sum_{k = 0}^{m - 1} P(d_{k\frac{t}{m}, (k + 1) \frac{t}{m}} > r) & \\ & = m P(d_{0, \frac{t}{m}} > r) & (1)\\ & \leq m u\left(\frac{r^2}{t} m\right) & (2) \end{aligned} $$ where (1) is by 8.2 (b) and (2) is by 8.2 (c). Now set $$ M := \lceil\frac{t}{r^2} (n - 2)\rceil + 1 $$

Q.E.D.

Theorem 9

For every $t \in (0, \infty)$,

1. $G_t$ is $\mathcal{F}/\mathcal{B}_1$-measurable in the sense that there is an $\mathcal{F}/\mathcal{B}_1$-measurable function $G'_t$ and an event $B \in \mathcal{F}$, with $P(B) = 0$, such that $$ \{G'_t \neq G_t\} \subseteq B $$
2. $G^*$ is $\mathcal{F}/\mathcal{B}_1$-measurable (in the same sense as in 9.1).

Proof

1. For every $k \in \mathbb{N}_1$, use 8.1 to choose $I_k \in \mathbb{N}_1$ such that, for all $i \in \{I_k, I_k + 1, \dots\}$, $$ \frac{k^{-2}}{t} i \cdot u(\frac{k^{-2}}{t} i) < \frac{k^{-2}}{t} \cdot \frac{1}{k} $$ and use 8.3 (b) to choose $J_k \in \mathbb{N}_1$, such that, for all $j \in \{J_k, J_k + 1, \dots\}$, $$ P(D_{j, t} > \frac{1}{k}) \leq j u\left(\frac{k^{-2}}{t} j\right) $$ Set $m_k := \max(I_k, J_k)$, and define $$ A_k := \left\{D_{m_k, t} \leq \frac{1}{k}\right\} $$ Then $P(A_k) \geq 1 - \frac{1}{k}$.

   For every $k \in \mathbb{N}_1$ we have, by 6.5, $$ \begin{aligned} &F_{m_k, \frac{1}{k}, t} \leq H_{\frac{1}{k}, t} & (1) \end{aligned} $$

   and by 6.2 and 6.4 $$ \begin{aligned} &G_t = \lim_{j \downarrow \infty} H_{\frac{1}{j}, t} & (2) \end{aligned} $$

   and by 8.3 (a), $$ \begin{aligned} &\forall \omega \in A_k, G_t(\omega) \leq F_{m_k, \frac{1}{k}, t}(\omega) & (3) \end{aligned} $$

   The result now follows from (1), (2) and (3) by theorem 1 here.

2. Observe that for every $\omega \in \Omega$, $$ g^*(\omega) = \lim_{k \uparrow \infty} g_k(\omega) $$ Therefore, by continuity of measures, $G^* = \lim_{k \rightarrow \infty} G_k$. The result now follows from 9.1.

Q.E.D.

Remark 10

The only snag that prevents this proof from extending to the case $n = 1$, is the requirement that $n \geq 2$ in equation (3.1) of [I2], which is used to justify the upper bound used in 8.2 (c).

To extend to $n = 1$, the following bound on the expression that is tagged with "(1)" in the proof of 8.2 (c) can be used (cf. [M1] lemma 12.9, p. 349): $$ 1 - \Phi(r) \leq \frac{1}{r} \frac{1}{\sqrt{2\pi}} e^{- r^2 / 2} $$ where $\Phi$ is the c.d.f. of the standard normal distribution.

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WORKS CITED

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