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?
 A: (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.
