Augmentation generated by Brownian motion are right continuous? I understand that the augmentation generated by Brownian motions are right continuous. However, I do not know how to come up with the answer using the proof of stochastic differential equation and diffusion process by Ikeda and Watanabe (Lemma 6.1. of chapter II).
The author does it as follow:
He defines a strongly continuous semigroup of operators $\{H_t\}$ on $C_{0}(R^{d})$, where $(H_t f)(x)=\int p(t,x-y)f(y)dy$, Then he compute what $E[f_{1}(X_{t_1})\cdots f_{n}(X_{t_n})|F_{t}^{X}]$ if $t_{k-1}\leq t<t_{k}$, then conclude that
$$E[f_{1}(X_{t_1})\cdots f_{n}(X_{t_n})|F_{t+0}^{X}]
=\lim_{h\downarrow 0} E[f_{1}(X_{t_1})\cdots f_{n}(X_{t_n})|F_{t+h}^{X}]=E[f_{1}(X_{t_1})\cdots f_{n}(X_{t_n})|F_{t}^{X}]$$, which finally proves that $F_{t+0}^{X}=F_{t}^{X}$. Indeed, I do not understand why the equalities hold. (I guess the first equality use Levy downward theorem, the second equality make use the continuous semigroup; I hope someone can answer my question) What's more, I do not understand quite well how one can make the conclusion if the above equalities holds. Even I show the above equalities, how do I conclude that $F_{t+0}^{X}=F_{t}^{X}$?
 A: The first equality is indeed a direct consequence of Lévy's backward convergence theorem. For the second one the reasoning goes as follows: Choose $h>0$ sufficiently small such that $t_{k-1} \leq t < t+h < t_{k}$. Since $f_j(X_{t_j})$ is $\mathcal{F}_{t+h}$ measurable for each $j \leq k-1$, we have
$$\mathbb{E} \left( \prod_{j=1}^n f_j(X_{t_j}) \mid \mathcal{F}_{t+h}^X \right) = \prod_{j=1}^{k-1} f_j(X_{t_j}) \mathbb{E} \left( \prod_{j=k}^n f_j(X_{t_j}) \mid \mathcal{F}_{t+h}^X \right). \tag{1}$$
Since $t+h \leq t_k$, it follows the Markov property that
$$\begin{align*} \mathbb{E} \left( \prod_{j=k}^n f_j(X_{t_j}) \mid \mathcal{F}_{t+h}^X \right) &= \mathbb{E}\left( \prod_{j=k}^n f_j(X_{t_j-(t+h)}+x) \right) \bigg|_{x=X_{t_{t+h}}} \\ &:= \int_{\Omega} \mathbb{E} \left( \prod_{j=k}^n f_j(X_{t_j-(t+h)}+X_{t+h}(\omega)) \right) \, d\mathbb{P}(\omega). \end{align*}$$
As each $f_j$ is bounded and continuous and $(X_t)_{t \geq 0}$ has continuous sample paths, we can let $h \to 0$ using the dominated convergence theorem to conclude that
$$\lim_{h \downarrow 0} \mathbb{E} \left( \prod_{j=k}^n f_j(X_{t_j}) \mid \mathcal{F}_{t+h} \right) = \mathbb{E} \left( \prod_{j=k}^n f_j(X_{t_j-t}+x) \right) \bigg|_{x=X_t} \tag{2} $$ almost surely. Using again the Markov property, we find that the right-hand side of $(2)$ equals
$$\mathbb{E}\left( \prod_{j=k}^n f_j(X_{t_j}) \mid \mathcal{F}_t^X \right).$$
Combining this with $(2)$, we get the second equality, i.e. we have shown that
$$\mathbb{E} \left( \prod_{j=1}^n f_j(X_{t_j}) \mid \mathcal{F}_{t+0}^X \right) = \mathbb{E} \left( \prod_{j=1}^n f_j(X_{t_j}) \mid \mathcal{F}_t^X \right) \tag{3}$$
almost surely. So far, we have assumed that each $f_j$ is bounded and continuous. The first step is to extend $(3)$ to a larger class of functions. Since the indicator function of any closed set $C_j \subseteq \mathbb{R}^d$ can be approximated by continuous bounded functions, we get
$$\mathbb{E} \left( \prod_{j=1}^n 1_{C_j}(X_{t_j}) \mid \mathcal{F}_{t+0}^X \right) = \mathbb{E} \left( \prod_{j=1}^n 1_{C_j}(X_{t_j}) \mid \mathcal{F}_t^X \right) \quad \text{a.s.}$$
for any closed sets $C_j$. Using that
$$\mathcal{D} := \{B \in \mathcal{F}_{\infty}^X; \mathbb{E}(1_B  \mid \mathcal{F}_{t+h}^X) = \mathbb{E}(1_B \mid \mathcal{F}_t^X) \, \text{a.s.} \}$$
is a Dynkin system which contains the $\cap$-stable generator
$$\mathcal{G} := \left\{\bigcap_{j=1}^n \{X_{t_j} \in C_j\}; C_j \, \, \text{closed}, t_1<\ldots<t_n \right\} \cup \mathcal{N}$$
of $\mathcal{F}_{\infty}^X$, we get $\mathcal{D} = \mathcal{F}_{\infty}^X$. In particular, for any $F \in \mathcal{F}_{t+0}^X$,
$$1_F = \mathbb{E}(1_F \mid \mathcal{F}_{t}^X) \quad \text{a.s.}$$
Since $\mathcal{F}_{t}^X$ contains all null sets, this implies that $1_F$ is $\mathcal{F}_{t}^X$-measurable. As $F \in \mathcal{F}_{t+0}^X$ is arbitrary, this means that $\mathcal{F}_{t+0}^X \subseteq \mathcal{F}_t^X$.
Remark: For an alternative proof see e.g. René Schilling & Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Theorem 6.21.
