# Proof Check: For a completed filtration, $\mathcal{F}_{t}^{B}$ is right continuous where $B$ is a standard Brownian motion

Let $$B$$ be a standard Brownian motion on $$(\Omega, \mathcal{F}, \mathbb P)$$ and further let $$(\mathcal{F}_{t}^{B})_{t \geq 0}$$ be the natural filtration associated with $$B$$ such that $$\mathcal{F}_{t}^{B}$$ for $$t \geq 0$$ contains all null sets. Show that the filtration is right-continuous.

My approach:

Trivially, we have $$\mathcal{F}_{t}^{B}\subseteq \mathcal{F}_{t+}^{B}$$.

Now for the "$$\mathcal{F}_{t+}^{B}\subseteq \mathcal{F}_{t}^{B}$$", we assume that this does not hold:

we choose $$A \in \mathcal{F}_{t+}^{B}\setminus \mathcal{F}_{t}^{B}$$ and let $$N$$ be the null set such that $$B$$ is continuous on $$\overline{\Omega}:=\Omega\setminus N$$

Then we can construct a sequence $$(\varepsilon_{n})_{n \in \mathbb N}\subseteq(0,\infty)$$ with $$\varepsilon_{n}\downarrow 0$$ as $$n \to \infty$$ such that $$A$$ is $$B_{t+\varepsilon_{n}}-$$ measurable for any $$n \in \mathbb N$$.

Furthermore $$B$$ is continuous on $$A\setminus N_{A}$$ where $$N_{A}$$ is some null set and thus since $$A\setminus N_{A}$$ is $$B_{t+\varepsilon_{n}}-$$ measurable for any $$n \in \mathbb N$$, we have on $$A\setminus N_{A}$$ that $$B_{t+\varepsilon_{n}}\xrightarrow{n \to \infty} B_{t}$$ and thus $$A \setminus N_{A}$$ must be $$B_{t}$$ measurable. Hence $$A = (A \setminus N_{A} )\cup N_{A}$$ is $$B_{t}$$-measurable which implies $$A \in \mathcal{F}_{t}^{B}$$ which contradicts the initial assumption.

Is my proof correct? Any improvements?

• I don't know if it's obvious that $B_{t + \epsilon_n} \rightarrow B_t$ on $A \setminus N_A$ implies $A \setminus N_A$ is $B_t$ measurable. Also, the statement that $A$ is $B_{t + \epsilon_n}$-measurable is not necessarily true: $A$ is $\mathcal F_{t+\epsilon_n}^B$ measurable, but doesn't have to be $B_{t + \epsilon_n}$-measurable. This may also affect the first part of the proof I commented on. Dec 1, 2020 at 22:54

(I'm going to abbreviate $$\mathcal F^B_t$$ to $$\mathcal F_t$$, etc.)
You need to show that $$E[G\mid\mathcal F_{t+}] = E[G\mid\mathcal F_{t}]\qquad\qquad(\dagger)$$ for each bounded $$\mathcal F$$-measurable $$G$$. Once this is done, consider $$A\in\mathcal F_{t+}$$ and take $$G=1_A$$. Then ($$\dagger$$) implies that $$1_A=E[1_A\mid\mathcal F_{t+}] =E[1_A\mid\mathcal F_t]$$ a.s. Because $$\mathcal F_t$$ contains all the null sets, this shows that $$A$$ is $$\mathcal F_t$$-measurable. Therefore $$\mathcal F_{t+}\subset\mathcal F_t$$.
The identity ($$\dagger$$) is a consequence of two things: (i) the (right) continuity of the paths of Brownian motion, and (ii) the stationary independent increments of the Brownian motion.
Fix $$t>0$$. By the monotone class theorem it's enough to show ($$\dagger$$) for $$G$$ of the form $$H\cdot K_t$$, where $$H$$ is bounded and $$\mathcal F_{t}$$-measurable, and $$K_u:=\prod_{i=1}^m f_i(B_{u+s_i}-B_u),\qquad u\ge 0,$$ where $$m$$ is a positive integer, the $$s_i$$ are strictly positive numbers and the $$f_i$$ are bounded and continuous. Notice that $$u\mapsto K_u$$ is (a.s.) continuous, and $$u\mapsto E[K_u]$$ is constant. Also, $$K_u$$ is independent of $$\mathcal F_u$$ because of the independent increments mentioned before.
Now fix an event $$C\in\mathcal F_{t+}$$. Let $$\{t_n\}$$ be a strictly decreasing sequence of reals with limit $$t$$. Then \eqalign{ E[1_C\cdot G] &=E[1_CHK_t]=\lim_{n\to\infty}E[1_CHK_{t_n}]\cr &=\lim_{n\to\infty}E[1_CH]\cdot E[K_{t_n}]\cr &=E[1_CH]\cdot E[K_{t}]\cr &=E[1_CH]\cdot E[K_0]\cr &=E\left[1_CH\cdot E[K_0]\right]. } (The third equality follows because $$C\in \mathcal F_{t_n}$$, and $$K_{t_n}$$ is independent of $$\mathcal F_{t_n}$$.) This calculation shows that $$E[G\mid\mathcal F_{t+}]=H\cdot E[K_0]$$, which is $$\mathcal F_t$$-measurable. Thus ($$\dagger$$) follows.
• I am not familiar with the monotone class theorem, would you mind pointing me to some reference and explaining why it implies that it's enough to show (†) for $G$ of the form $H \cdot K_t$? Thank you. Mar 25, 2023 at 15:33
• The MCT is a sort of induction principle. A fine discussion appears at almostsuremath.com/2019/10/27/… In the notation of Theorem 1 there, take $\mathcal K$ to be all functions of the form $H\cdot K_t$ as indicated in my discussion, and $\mathcal H$ to be all the bounded $\mathcal F$ measurable $G$ for which ($\dagger$) holds. The MCT then implies that $\mathcal H$ comprises all bounded $\mathcal F$ measurable functions. Mar 26, 2023 at 16:26