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The reflection principle of Brownian motion states that Brownian motion reflected at some stopping time $\tau$ is still a Brownian motion.

The proof found in Mörters & Peres (as well as in several other books), rests on the fact that the concatenation mapping, which takes a continuous path $\{g(t) : t \geq 0\}$ and glues it to the end point of a finite continuous path$\{f(t) : 0 \leq t \leq \tau\}$ to form a new continuous path, is measurable.

I don't see how to prove that the concatenation mapping is measurable.


Formally, the reflection principle states (I follow Schilling & Partzsch):

Let $(B(t), \mathcal{F}_t)_{t \geq 0}$ be a standard, $d$-dimensional Brownian motion and let $\tau$ be an a.s. finite stopping time. Consider the process $$ W(t,\omega) := \begin{cases} B(t,\omega), & 0 \leq t < \tau(\omega) \leq +\infty \\ 2B(\tau(\omega), \omega) - B(t,\omega), & \tau(\omega) \leq t < \infty \end{cases} $$ Then $(W(t))_{t \geq 0}$ is again a standard Brownian motion.

To prove this theorem we may assume, without loss of generality, that $\tau(\omega) < \infty$ for all $\omega \in \Omega$, and we define the mapping $\Phi : \mathbf{C}_{(0)}\times[0,\infty)\times\mathbf{C}_{(0)} \rightarrow \mathbf{C}_{(0)}$, where $\mathbf{C}_{(0)} := \{f \in \mathbf{C}[0,\infty) : f(0) = 0\}$, thus $$ \Phi(f,t,g) := \begin{cases} f(s), & 0 \leq s < t \\ f(t) + g(s - t), & t \leq s < \infty \end{cases} $$

It is claimed that $\Phi$ is $\mathcal{B}(\mathbf{C}_{(0)})\otimes\mathcal{B}[0,\infty)\otimes\mathcal{B}(\mathbf{C}_{(0)})/\mathcal{B}(\mathbf{C}_{(0)})$ measurable.

Schilling and Partzsch offer the following hint for proving this fact: equip $\mathbf{C}_{(0)}\times[0,\infty)\times\mathbf{C}_{(0)}$ with the topology of locally uniform convergence (in $\mathbf{C}_{(0)}$) and pointwise convergence (in $[0,\infty)$).

I'd appreciate help in proving why $\Phi$ is measurable, either by following Schilling & Partzsch's cue, or by another method. Additionally, if it makes life easier, it can be assumed that $d = 1$.


Works cited

  • Mörters, Peter and Peres, Yuval. Brownian Motion. The version freely available from Peres's home page. Accessed 2014-02-18. (Theorem 2.19, pp. 44-45)
  • Schilling, René L. and Partzsch, Lothar. Brownian Motion - An introduction to stochastic processes. De Gruyter. 2012. (Theorem 6.12, pp. 71-72)
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1 Answer 1

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THE MAIN RESULT

Theorem Employing the notation and terminology introduced in the original post, we have: $\Phi$ is $\mathcal{B}(\mathbf{C}_{(0)})\otimes\mathcal{B}[0,\infty)\otimes\mathcal{B}(\mathbf{C}_{(0)})/\mathcal{B}(\mathbf{C}_{(0)})$ measurable.

Proof (For a listing of pertinent results, notations and definitions consult the next section, "Auxiliary facts and definitions". References cited below - e.g. [K], [M] - are listed in full at the bottom of this post in the section "Works cited".)

I will prove only the case $d = 1$.

As suggested in [S] p. 72, it suffices to show that $\Phi$ is continuous w.r.t. the product topology $\mathcal{T}$ on $\mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ where $\mathbf{C}_{(0)}$ is equipped with the $\mathcal{O}_\rho$ topology, and $[0,\infty)$ is equipped with the standard, Euclidean topology, since if $\Phi$ is $\mathcal{T}/\mathcal{O}_\rho$-continuous, then by lemma 16 ("continuity and measurability"), $\Phi$ is $\sigma(\mathcal{T})/\sigma(\mathcal{O}_\rho)$-measurable, but by lemma 15 ("product and Borel $\sigma$-fields) and in light of proposition 8 ("$\mathbf{C}_{(0)}$ is a Polish space") we have $\sigma(\mathcal{T}) = \sigma(\mathcal{O}_\rho)\otimes\mathcal{B}[0,\infty)\otimes\sigma(\mathcal{O}_\rho)$ and by definition 7, $\sigma(\mathcal{O}_p) = \mathcal{B}(\mathbf{C}_{(0)})$.

Now, to prove that $\Phi$ is $\mathcal{T}/\mathcal{O}_\rho$-continuous, let $(f^*, t^*, g^*) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ and let $\varepsilon \in (0,\infty)$. We shall define some $\delta \in (0,\infty)$, such that for all $(f, t, g) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ with $\rho(f, f^*), |t - t^*|, \rho(g, g^*) < \delta$, we will have $$ \rho(\Phi(f,t,g), \Phi(f^*, t^*, g^*)) < \varepsilon $$

Choose $k, n \in \mathbb{N}_1$ such that

a) $t^* + 1/8 \leq k$,

b) $(k + 1)2^{-k} \leq \varepsilon$

c) $k \leq n$

d) Whenever $s, s' \in [0, k]$ are such that $|s - s'| < 2^{- (n + 2)}$, $|g^*(s) - g^*(s')| < 2^{- (k + 2)}$

And define $\delta := 2^{- (n + 2)}$.

Let $(f, t, g) \in \mathbf{C}_{(0)} \times [0,\infty) \times \mathbf{C}_{(0)}$ be such that $\rho(f, f^*), |t - t^*|, \rho(g, g^*) < \delta$, and set $$ \begin{aligned} \phi & := \Phi(f,t,g) \\ \phi^* & := \Phi(f^*, t^*, g^*) \\ m & := \min(t, t^*) \\ M & := \max(t, t^*) \end{aligned} $$ Note that $M \leq k$.

We have $$ \sup_{0 \leq s \leq k}|\phi(s) - \phi^*(s)| = \\ \max(\sup_{0 \leq s \leq m}|\phi(s) - \phi^*(s)|, \sup_{m \leq s \leq M}|\phi(s) - \phi^*(s)|, \sup_{M \leq s \leq k)} |\phi(s) - \phi^*(s)|) $$

Now, by lemma 6a ("Equivalence of $\rho$ and $\sup$"), $$ \sup_{0 \leq s \leq m}|\phi(s) - \phi^*(s)| = \sup_{0 \leq s \leq m}|f(s) - f^*(s)| < 2^{- (n + 2)} < 2^{-k} $$

Secondly, we have (by lemma 6a and by the assumptions on $k$ and on $n$), $$ \sup_{m \leq s \leq M} |\phi(s) - \phi^*(s)| \leq \sup_{m \leq s \leq M} |f(s) - f^*(s)| + \sup_{0 \leq s \leq 2^{- (n + 2)}} |g^*(s)| + \sup_{0 \leq s \leq 2^{- (n + 2)}} |g(s) - g^*(s)| \\ < \frac{2^{-n}}{4} + \frac{2^{-k}}{4} + \frac{2^{-n}}{4} < 2^{-k} $$

Thirdly, we have (again, by lemma 6a and by the assumptions on $k$ and on $n$), $$ \sup_{M \leq s \leq k} |\phi(s) - \phi^*(s)| \leq \sup_{M \leq s \leq k} |f(s) - f^*(s)| + \sup_{0 \leq s \leq k - M} |g(s) - g^*(s)| \\ + \sup_{0 \leq s \leq k - m}|g^*(s + 2^{- (n + 2)}) - g^*(s)| < \frac{2^{-n}}{4} + \frac{2^{-n}}{4} + \frac{2^{-k}}{4} < 2^{-k} $$

In conclusion, $$ \sup_{0 \leq s \leq k} |\phi(s) - \phi^*(s)| < 2^{-k} $$

Hence, by lemma 6b, $\rho(\phi, \phi^*) < (k + 1)2^{-k} \leq \varepsilon$.

Q.E.D.


AUXILIARY FACTS AND DEFINITIONS

Definition 1 ($\pi_t$, the canonical projection onto the $t$th coordinate). ([S], section 4.1, p. 40) Setting $I := [0,\infty)$, denote by $\pi_t : (\mathbb{R}^d)\rightarrow \mathbb{R}^d$, $\omega \mapsto \omega(t)$ the canonical projection onto the $t$th coordinate.

Definition 2 (The product $\sigma$-algebra $\mathcal{B}^I(\mathbb{R}^d)$). ([S], section 4.1, p. 40) The natural $\sigma$-algebra on the infinite product $(\mathbb{R}^d)^I$ is the product $\sigma$-algebra $$ \mathcal{B}^I(\mathbb{R}^d) := \sigma \{\pi_t^{-1}(B) : B \in \mathcal{B}(\mathbb{R}^d), t \in I\} = \sigma \{\pi_t : t \in I\} $$

Definition & Proposition 3 (The trace $\sigma$-algebra). (a: [S2] Example 3.3 (vi), p. 16; b: [H] Theorem E, p. 25) Let $\Omega$ be some non-empty set, let $\mathcal{F}$ be a $\sigma$-algebra over $\Omega$ and let $S \subseteq \Omega$.

a) $S \cap \mathcal{F} := \{S \cap T : T \in \mathcal{F}\}$ (called the trace $\sigma$-algebra of $S$ w.r.t. $\mathcal{F}$) is a $\sigma$-algebra over $S$.

b) If $\mathcal{E} \subseteq \mathcal{F}$ is a generator of $\mathcal{F}$ (i.e. if $\mathcal{F} = \sigma(\mathcal{E})$), then $S \cap \mathcal{F} = \sigma(S \cap \mathcal{E})$.

Corollary 4 ($\mathbf{C}_{(0)} \cap \mathcal{B}^I(\mathbb{R}^d) = \sigma(\pi_t\mid_{\mathbb{C}_{(0)}} : t \in I)$). ([S] section 4.1, p. 41) $$ \mathbf{C}_{(0)} \cap \mathcal{B}^I(\mathbb{R}^d) = \sigma(\pi_t\mid_{\mathbb{C}_{(0)}} : t \in I) $$

Definition & Proposition 5 ($\rho$, the locally uniform convergence metric). (Based on, but not identical to, [S] section 4.1, p. 41) The function $\rho: \mathbf{C}_{(0)} \rightarrow [0,\infty)$, $$ \rho(w,v) := \sum_{n = 1}^\infty \left(2^{-n} \wedge \sup_{0 \leq t \leq n}|w(t) - v(t)|\right) $$ is a metric on $\mathbf{C}_{(0)}$.

Lemma 6 (Equivalence of $\rho$ and $\sup$). Let $n \in \mathbb{N}_1$ and let $w,v \in \mathbf{C}_{(0)}$.

a. If $\rho(w,v) < 2^{-n}$, $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$.

b. If $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$, $\rho(w,v) < (n + 1) 2^{-n}$.

Proof of lemma 6

a. Assume that $\rho(w,v) < 2^{-n}$. Suppose, by way of contradiction, that $\sup_{0 \leq t \leq n}|w(t) - v(t)| \geq 2^{-n}$. Then $$ \rho(w,v) \geq 2^{-n} \wedge \sup_{0 \leq t \leq n} |w(t) - v(t)| = 2^{-n} $$ contrary to the assumption.

b. Assume that $\sup_{0 \leq t \leq n}|w(t) - v(t)| < 2^{-n}$. Then $$ \rho(w, v) \leq \sum_{m = 1}^n 2^{-n} + \sum_{m = n+1}^\infty 2^{-m} = n2^{-n} + 2^{-n} $$

Q.E.D.

Definition 7 ($\mathcal{O}_\rho$, $\mathcal{B}(\mathbf{C}_{(0)})$). ([S] section 4.1, p. 41) Denote by $\mathcal{O}_\rho$ the topology induced by $\rho$ and consider the Borel $\sigma$-algebra $\mathcal{B}(\mathbf{C}_{(0)}) := \sigma(\mathcal{O}_\rho)$ on $\mathbf{C}_{(0)}$.

Proposition 8 ($\mathbf{C}_{(0)}$ is a Polish space). ([S] section 4.1, p. 41) Equipped with the metric $\rho$ defined above, $\mathbf{C}_{(0)}$ becomes a Polish space, i.e. a complete, separable metric space.

Lemma 9 ($\mathbf{C}_{(0)}\cap\mathcal{B}^I(\mathbb{R}^d) = \mathcal{B}(\mathbf{C}_{(0)})$). ([S] Lemma 4.1, p. 41) We have $$ \mathbf{C}_{(0)}\cap\mathcal{B}^I(\mathbb{R}^d) = \mathcal{B}(\mathbf{C}_{(0)}) $$

Proposition 10 (A Brownian motion $\omega \mapsto (t \mapsto B(\omega, t))$ is $\mathcal{F}/\mathbf{C}_{(0)}$-measurable). ([S] top paragraph on p. 42) A standard, $d$-dimensional Brownian motion $(B(t))_{t \geq 0}$ over the probability space $(\Omega, \mathcal{F}, P)$, when considered as a function $\in \Omega \mapsto \mathbf{C}_{(0)}$, is $\mathcal{F}/\mathcal{B}(\mathbf{C}_{(0)})$-measurable.

Definition 11 (Box topology). ([M] p. 114) Let $\{X_\alpha\_{\alpha \in J}$ be an indexed family of topological spaces. Let us take as a bases for a topology on the product space $\prod_{\alpha \in J}X_\alpha$ the collection of all sets of the form $\prod_{\alpha \in J}U_\alpha$ where $U_\alpha$ is open in $X_\alpha$, for each $\alpha \in J$. The topology generated by this basis is called the box topology.

Definition 12 (Product topology, product space). ([M] p. 114) Let $\mathcal{S}_\beta$ denote the collection $$ \mathcal{S}_\beta := \{\pi_\beta^{-1}(U_\beta) \mid U_\beta \textrm{ open in } X_\beta \} $$ and let $\mathcal{S}$ denote the union of these collections, $\mathcal{S} := \bigcup_{\beta \in J}\mathcal{S}_\beta$. The topology generated by the sub basis $\mathcal{S}$ is called the product topology. In this topology, $\prod_{\alpha \in J} X_\alpha$ is called a product space.

Theorem 13 (Comparison of the box and product topologies). ([M] Theorem 19.1, p. 115) The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.

Corollary 14 (Equality of the box and product topologies for finite products). ([M] p. 115) For finite products $\prod_{\alpha = 1}^n X_\alpha$, the two topologies are precisely the same.

Lemma 15 (Product and Borel $\sigma$-fields). ([K] Lemma 1.2, p. 3) If $S_1, S_2, \dots$ are separable metric spaces with induced topologies $\mathcal{T}_1, \mathcal{T}_2, \dots$, respectively, and if $\mathcal{T}$ is the product topology on $S_1 \times S_2 \times \cdots$, then $$ \sigma(\mathcal{T}) = \sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \otimes \cdots $$

Lemma 16 (Continuity and measurability). ([K] Lemma 1.5, p. 4) Let $f$ be a continuous mapping between two topological spaces $S$ and $T$ with Borel $\sigma$-fields $\mathcal{S}$ and $\mathcal{T}$. Then $f$ is $\mathcal{S}/\mathcal{T}$-measurable.


WORKS CITED

[H] Halmos, Paul R. Measure Theory. Springer-Verlag, 1974

[K] Kallenberg, Olav. Foundations of Modern Probability. 2nd edition. Springer, 2002

[M] Munkres, James R. Topology. 2nd edition. Prentice Hall, 2000

[S] Schilling, René L. and Partzsch, Lothar. Brownian motion - an introduction to stochastic processes. De Gruyter, 2012

[S2] Schilling, René L. Measures, integrals and martingales. Cambridge University Press, 2011

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