For Brownian motion $B_s$, $\forall F \in \mathcal F_t \ : E[1_F(B_u-B_t)]=P[F]E[B_u-B_t]$ imply $B_u-B_t $ is independent of $\mathcal F_t $? I wonder if the claim in the title is correct? The full context is this: I let $\mathcal F_t ^B:=\sigma(B_s,s \le t )$. I know that for every $u>t \ : B_u-B_t$ is independent of $\mathcal F_t^B $. 
We "enlarge" $\mathcal F_t^B $ with all the null sets $\mathcal N $, that is we let $\mathcal F_t :=\sigma(\mathcal F_t^B ,\mathcal N)$ - and I want to show that or every $u>t \ : B_u-B_t$ is independent of $\mathcal F_t$ also holds.
Given $F \in \mathcal F_t $ we have by the definition of completion that $F=\bar F \cup N $ where $\bar F \in \mathcal F_t^B $ and $N \in \mathcal N $. For this we easily se that $E[1_F(B_u-B_t)]=P[F]E[B_u-B_t]$. Which says that $1_F $ and $B_u-B_t $ are uncorrelated, does it also imply independence?
Thanks in advance!
 A: Claim. If $\mathscr{X}$ and $\mathscr{Y}$ are two sigma fields, and $\bar{\mathscr{X}}$ and $\bar{\mathscr{Y}}$ denote their completions, then the condition $\mathscr{X}$ and $\mathscr{Y}$ are mutually independent is necessary and sufficient for the mutual independence of $\bar{\mathscr{X}}$ and $\bar{\mathscr{Y}}.$
Proof. The necessity is obvious, as for the sufficency, consider events $\mathrm{X} \cup \mathrm{M}$ and $\mathrm{Y} \cup \mathrm{N}$ in $\bar{\mathscr{X}}$ and $\bar{\mathscr{Y}},$ respectively. Then, $(\mathrm{X} \cup \mathrm{M}) \cap (\mathrm{Y} \cup \mathrm{N}) \subset (\mathrm{X} \cap \mathrm{Y}) \cup \mathrm{M} \cup \mathrm{N}$ and so $$\mathbf{P}(\mathrm{X} \cap \mathrm{Y}) \leq \mathbf{P}((\mathrm{X} \cup \mathrm{M}) \cap (\mathrm{Y} \cup \mathrm{N})) \leq \mathbf{P}(\mathrm{X} \cap \mathrm{Y}) + \mathbf{P}(\mathrm{M}) + \mathbf{P}(\mathrm{N}),$$
proving the claim.
Proof of exercise. Since $\sigma(B_u - B_t)$ and $\mathscr{F}^B_t$ are independent so are their completions and so are $\sigma(B_u - B_t)$ and $\mathscr{F}_t.$ Q.E.D.
