# Proof of Blumenthal's 0-1 law for Brownian Motion

I am currently reading the book "Brownian Motion, Martingales, and Stochastic calculus" by Jean-François Le Gall and am stuck at understanding the proof of Blumenthal's 0-1 law for Brownian Motion.

The Setup is the following: Assume we have a one-dimensional Brownian $$(B_t)_{t \geq 0}$$ on some probability space $$(\Omega, \mathcal{F}, P)$$. For $$t \geq 0$$ let $$\mathcal{F}_t= \sigma(B_s: 0 \leq s \leq t)$$ and define $$\mathcal{F}_{0+}= \cap_{\epsilon > 0} \mathcal{F}_\epsilon$$. Then the following theorem holds.

Theorem: The sigma-algebra $$\mathcal{F}_{0+}$$ is trivial in the sense that for all $$A \in \mathcal{F}_{0+}$$, the probability of $$A$$ is either $$0$$ or $$1$$.

I will outline the proof that can be found in the book with the part I do not understand:

Proof outline: Using $$\cap$$-stable generators, it will be enough to proof that for all $$n \in \mathbb{N}$$ and $$0 < t_1 < ... < t_n$$ the sigma-algebra $$\mathcal{F}_{0+}$$ and $$\sigma(B_{t_1},...,B_{t_n})$$ are independent. Now let $$g: \mathbb{R}^n \rightarrow \mathbb{R}$$ be bounded and continuous and $$A \in \mathcal{F}_{0+}$$. By continuity and dominated convergence we can write for $$0 < \epsilon < t_1$$,

\begin{align} E [1_A g(B_{t_1},...,B_{t_n})] = \lim_{\epsilon \rightarrow 0} E [1_A g(B_{t_1} - B_\epsilon,...,B_{t_n}- B_\epsilon)]. \end{align}

Now by the simple Markov property of Brownian Motion, $$(B_{t+\epsilon}-B_\epsilon)_{t \geq 0}$$ is independent of $$\mathcal{F}_\epsilon$$ and thus we can continue to write the above to

\begin{align} \lim_{\epsilon \rightarrow 0} E [1_A g(B_{t_1} - B_\epsilon,...,B_{t_n}- B_\epsilon)] &= P(A) \lim_{\epsilon \rightarrow 0} E[g(B_{t_1} - B_\epsilon,...,B_{t_n}- B_\epsilon)] \\ &= P(A) E[g(B_{t_1},...,B_{t_n})]. \end{align}

This then should be enough to conclude independence. I do not know why this should suffice. If $$g$$ was allowed to be measurable, then it would be clear. But how does independence of $$\mathcal{F}_{0+}$$ and $$\sigma(B_{t_1},...,B_{t_n})$$ follow from $$g$$ only being bounded continuous?

Thanks a lot in advance!

## 1 Answer

Approach I (via characteristic functions): The identity $$E(1_A g(B_{t_1},\ldots,B_{t_n})) = P(A) E(g(B_{t_1},\ldots,B_{t_n})) \tag{1}$$ can be easily extended to complex-valued continuous functions (just write $$g= \text{Re} g + i \, \text{Im g}$$ and apply $$(1)$$ separately to the real and imaginary part of $$g$$). Choosing $$g(x_1,\ldots,x_n) := \exp \left( i \sum_{j=1}^n \xi_j x_j \right)$$ for some fixed $$\xi=(\xi_1,\ldots,\xi_n) \in \mathbb{R}^n$$ we find that $$E \left( 1_A \exp \left[ i \sum_{j=1}^n \xi_j B_{t_j} \right] \right) = P(A) E \exp \left( i \sum_{j=1}^n \xi_j B_{t_j} \right)$$ for all $$A \in \mathcal{F}_{0+}$$ and $$\xi \in \mathbb{R}^n$$. This implies by Kac's lemma that $$\mathcal{F}_{0+}$$ and $$\sigma(B_{t_1},\ldots,B_{t_n})$$ are independent, see the lemma here for details.

Approach II (via monotone class argument): For any open set $$U \subseteq \mathbb{R}^n$$ there exists a sequence of continuous bounded functions $$(g_k)_{k \in \mathbb{N}}$$ such that $$g_k \uparrow 1_U$$. Using $$(1)$$ and applying the monotone convergence theorem, it follows that \begin{align*} E(1_A 1_U(B_{t_1},\ldots,B_{t_n})) &= \sup_{k \geq 1} E(1_A g_k(B_{t_1},\ldots,B_{t_n})) \\ &= P(A) \sup_{k \geq 1} E(g_k(B_{t_1},\ldots,B_{t_n})) \\ &= P(A) E(1_U(B_{t_1},\ldots,B_{t_n})) \end{align*} for $$A \in \mathcal{F}_{0+}$$. Since the open sets are a generator of the Borel $$\sigma$$-algebra $$\mathcal{B}(\mathbb{R}^n)$$, an application of the monotone class theorem yields that $$E(1_A 1_F (B_{t_1},\ldots,B_{t_n})) = P(A) E(1_F(B_{t_1},\ldots,B_{t_n})), \qquad A \in \mathcal{F}_{0+}$$ for any $$F \in \mathcal{B}(\mathbb{R}^n)$$. Hence, $$\mathcal{F}_{0+}$$ and $$\sigma(B_{t_1},\ldots,B_{t_n})$$ are independent.

• Can you please tell we why it is enough to prove that $\mathcal F_{0^+}$ is independent of $\sigma (B_{t_1},...,B_{t_n})$ for all finite dimensional vectors $(B_{t_1},...,B_{t_n})$, where $0<t_1<...<t_n$ to conclude that $\mathcal F_{0^+}$ is "independent of him self" (i.e. that $\mathbb P(A\cap A)=\mathbb P(A)^2$). I don't get this point. Nov 21, 2020 at 9:22
• @Bruce The family $$\bigcup_{n \in \mathbb{N}} \bigcup_{0<t_1 < \ldots < t_n} \sigma(B_{t_1},\ldots,B_{t_n})$$ is a generator of $\mathcal{F}_{\infty} = \sigma(B_t; t>0)$, which is stable under intersections. Consequently, $\mathcal{F}_{0+}$ and $\mathcal{F}_{\infty}$ are independent. Since $\mathcal{F}_{\infty} \supseteq \mathcal{F}_{0+}$, this implies that $\mathcal{F}_{0+}$ is independent of itself; hence trivial.
– saz
Nov 21, 2020 at 12:40
• If $\mathcal B=\bigcup_{i=1}^nB_i$ and $\mathcal F=\sigma (\mathcal B)$, then if $\mathcal G$ is a $\sigma -$algebra, to prove that $\mathcal G$ and $\mathcal F$ are independent, it's sufficent to prove that $\mathcal G$ and $B_i$ are independent when $A_i$ are $\sigma -$field (according to a proposition of my lecture). (in our situation it's the case, so it's fine, but in my lecture I only have : $A_1,...,A_n$ are independent $\pi-$system implies $\sigma (A_1),...,\sigma (A_n)$ independent.) So does my first statement works if the $B_i$ are only $\pi-$system but not $\sigma$ field ? Nov 21, 2020 at 15:50
• And does it also work if the union is countable but not finite ? Nov 21, 2020 at 15:50
• @Bruce Sorry, can't follow your thoughts. What is $B_i$? The important point is that the generator $\mathcal{B}$ is stable under intersections, i.e. $F,G \in \mathcal{B}$ implies $F \cap G \in \mathcal{B}$. For such a generator $\mathcal{B}$ of a $\sigma$-algebra $\mathcal{F}$, we have that $\mathcal{G}$ and $\mathcal{F}$ are independent iff $\mathcal{G}$ and $\mathcal{B}$ are independent.
– saz
Nov 21, 2020 at 16:00