If $B$ is a BM and $\mathcal F_t=\sigma(B_s,s\le t)$, then $(B_{s+t}-B_t)_{s\ge 0}$ is independent of $\mathcal F_t^+:=\bigcap_{s>t}\mathcal F_s$ Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname{P})$, i.e. $B$ is a real-valued stochastic process with


*

*$B_0=0$ almost surely

*$B$ has independent and stationary increments

*$B_t\sim\mathcal{N}_{0,\;t}$

*$B$ is almost surely continuous


Let $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be the filtration generated by $B$ and $$\mathcal F_t^+:=\bigcap_{s>t}\mathcal F_s\;\;\;\text{for all }t\ge 0\;.$$

**Claim:**$\;$ Let $t>0$ and $$B':=(B_{s+t}-B_t)_{s\ge 0}\;.$$ Then $B'$ is independent of $\mathcal F^+_t$.

Proof: $\;$ Since $B$ is almost surely right-continuous $$B_{s+t_n}-B_{t_n}\stackrel{n\to\infty}{\to}B'_s\;\;\;\text{almost surely}\tag{1}$$ for all $t_n\downarrow t$ and $s\ge 0$. And further, observing that $B'$ is a Brownian motion independent of $\mathcal F_t$, we see that $$\left(B_{s_1}',\ldots,B_{s_m}'\right)\stackrel{(1)}{=}\lim_{n\to\infty}\underbrace{\left(B_{s_1+t_n}-B_{t_n},\ldots,B_{s_m+t_n}-B_{t_n}\right)}_{=:X_n}\;\;\;\text{almost surely}\tag{2}$$  for all $s_1,\ldots,s_m\ge 0$. Again, it's easy to verify, that $$X_n\text{ is independent of }\mathcal F_{t_n}\supseteq\mathcal F_t\;\;\;\text{for all }n\in\mathbb{N}\;.$$ So, we're really close, but I don't find the right argument to conclude the independence of $\left(B_{s_1}',\ldots,B_{s_m}'\right)$ and $$\mathcal F_t^+=\bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}\;.$$ So, how do I need to argue?
 A: If $X_n\to X$ almost surely and $X_n$ is independent of $\mathcal G$, then $X$ is independent of $\mathcal G$. We use actually this. 
Call $X$ the left hand side of (2). We can show that for each continuous and bounded function $f\colon\mathbf R^n\to\mathbf R$ and $E\in\mathcal F_t$, we have $\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$ (this is true from your work when $X$ is replaced by $X_n$; then use dominated convergence). You can approximate the characteristic function of a closed set by continuous functions bounded by $1$, hence we have for each closed set $F\subset \mathbf R^n$ that $\mu(\{X\in F\}\cap E )=\mu(\{X\in F\})\mu(E)$. This extends by a monotone class argument to any Borel-measurable set instead of $F$. 
For the second question, it suffices to show that $\mathcal F_t^+\supset\bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}$; the other inclusion is obvious. Pick $s\gt t$  and $A\in \bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}$. For some $n$, $t\lt t_n\lt s$, and by definition of $\mathcal F_s$, we have $A\in \mathcal F_{t_n}   \subset  \mathcal F_s$. Since $s$ is arbitrary, it follows that $A\in \mathcal F_t^+$. 
A: As $\mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}$, we know that $X_n$ is independent of $\mathcal{F}_{t+}$. Therefore,
$$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$
is also independent of $\mathcal{F}_{t+}$.

Lemma: Let $\mathcal{F}$ be a $\sigma$-algebra and $(X_n)_{n \in \mathbb{N}}$ such that $X_n$ and $\mathcal{F}$ are independent for each $n \in \mathbb{N}$ and $X_n \to X$ almost surely. Then $X$ and $\mathcal{F}$ are independent.

Proof. Fix $F \in \mathcal{F}$. For any $\xi, \eta \in \mathbb{R}$, we have by the dominated convergence theorem
$$\begin{align*} \mathbb{E} \exp(\imath \, (\xi X+\eta 1_F)) &= \lim_{n \to \infty} \mathbb{E}\exp(\imath \, (\xi X_n+\eta 1_F)) \\ &= \lim_{n \to \infty} \mathbb{E}\exp(\imath \, \xi X_n) \mathbb{E}\exp(\imath \, \eta 1_F) \\ &= \mathbb{E}\exp(\imath \, \xi X) \mathbb{E}\exp(\imath \, \eta 1_F). \end{align*}$$
By the uniqueness of the Fourier transform, this shows that the joint distribution $(X,1_F)$ equals the product of the distribution of $X$ and $1_F$; hence, $X$ and $1_F$ are independent (see also this question).
