Why is $\sigma(B_{s}-s: 0\leq s\leq t)$ strictly included in $\mathcal{F}_{t}^{B}$ Why is $\sigma(B_{s}-s: 0\leq s\leq t)$ strictly included in $\mathcal{F}_{t}^{B}$ , i.e. the completed natural filtration of the standard Brownian motion $B$?
It is clear to me that $B_{s}-s$ for $0\leq s\leq t$ is contained in $\mathcal{F}_{t}^{B}$, since $f_{s}(B_{s})=B_{s}-s$ is a composition of continuous map $f_{s}$ and $B_{s}$, hence $B_{s}$ measurable. Why is the inclusion strict?
I am struggling to find sets that are $\mathcal{F}^{B}_{t}$- measurable but not in $\sigma(B_{s}-s: 0\leq s\leq t)$.
 A: Let $\mathcal G^B_t$ be the usual incomplete filtration generated by $(B_s)_{0 \leq s \leq t}$. Then the claim is that $\mathcal G^B_t = \sigma(B_s -s : 0 \leq s \leq t)$.
Indeed, recall that $$\sigma(B_s - s : 0 \leq s \leq t) = \sigma\left(\bigcup_{n=1}^\infty \bigcup_{0 \leq t_1 \leq ... \leq t_n \leq t} \sigma(B_{t_1}-t_1,...,B_{t_n} - t_n)\right)$$
and similarly for $\mathcal G^B_t$. It is sufficient to prove that for all $n$ and $0 \leq t_1 \leq ... \leq t_n \leq t$ we have $\sigma(B_{t_1}-t_1,...,B_{t_n} - t_n) = \sigma(B_{t_1},...,B_{t_n})$.
But now we use a result : a function is measurable w.r.t a sigma-algebra generated by finitely many random variables, if and only if it is a Borel function in those variables.
That is, let $f$ be measurable w.r.t. $\sigma(B_{t_1}-t_1,...,B_{t_n} - t_n)$. Then $$f = g(B_{t_1}- t_1, B_{t_2} - t_2,....,B_{t_n}-t_n)$$ for a measurable function $g$. Define $g'(x_1,...,x_n) = g(x_1-t_1,...,x_n - t_n)$. Then note that $g'$ is measurable and $f = g(B_{t_1},...,B_{t_n})$.
Therefore $f$ is measurable w.r.t $\sigma(B_{t_1},...,B_{t_n})$. I am sure you can go the other way as well.

Therefore, the strictness of the containment comes from the fact that one is complete while the other is apparently not. If they are both complete, then the filtrations will be the same as the null sets of both filtrations are the same.
However, we can show that the natural filtration is not complete, using a characterisation of the sets that are measurable w.r.t  the sigma algebras in the natural filtration.
Recall that the one-dimensional Brownian motion on $[0,t]$ sits on the probability space
$\Omega = \{f : [0,t] \to \mathbb R\}$, the sigma algebra being $\mathcal B^{[0,t]}(\mathbb R)$ i.e. the sigma-algebra generated by $\{\pi_s^{-1}(B) : B \in \mathcal B(\mathbb R) , 0 \leq s \leq t\}$ where $\pi_s (w) = w(s)$ is the $s$th coordinate projection , and the measure being the Wiener measure i.e. that whose finite dimensional distributions are given by $\mathbb P_{t_1,...,t_n}(C_1 \times ... \times C_n) = \mathbb P(B_{t_i} \in C_i , 1 \leq i \leq n)$ (which has an explicit formula using the normal densities and independent increments assumption).
Note that $\mathcal B^{[0,t]}(\mathbb R)$ is the same as $\mathcal F^B_t$.

Using this, we can prove the following result on sets which are in $\mathcal B^{[0,t]}(\mathbb R)$.

Let $\Gamma \in \mathcal B^{[0,t]}(\mathbb R)$. Then there exists a countable set $S$ such that if $f \in \Gamma$, $w \in \Omega$ and $w|_S = f|_S$ then $w \in \Gamma$.

In other words, every measurable subset of functions  w.r.t the usual sigma-algebra is determined by countably many coordinates.
To prove this, let $$\Sigma = \{\Gamma \in \mathcal B^{[0,t]}(\mathbb R) : \exists S_{\Gamma} \textrm{ countable } , \forall w \in \Omega,f \in \Gamma, w|_{S_\Gamma} = f|_{S_\Gamma} 
 \implies w \in \Gamma\}$$
i.e. those $\Gamma$ which do satisfy what we need. We note that if $C \in \mathcal B(\mathbb R)$ and $0\leq s \leq t$ then $\{w : \pi_s(w) \in C\}$ belong in $\Sigma$ because you can take $S_{\Gamma} = \{s\}$, it is the single coordinate at which information will tell you if a function should belong in this set or not.
On the other hand, one can show that $\Sigma$ is a sigma-algebra (left as an exercise, you can see the hidden block for hints).

 $\phi, \Omega \in \Sigma$ obviously. For $\Gamma \in \Sigma$, we can show that $S_{\Gamma^c} = S_{\Gamma}$ works out so $\Gamma^c \in \Sigma$. For the infinite union, just take the infinite union of the $S_i$ as well.

Therefore $\Sigma$ is a sigma-algebra which contains the generating set for $\mathcal B^{[0,t]}(\mathbb R)$. It follows that $\Sigma = \mathcal B^{[0,t]}(\mathbb R)$.

This can be used to find sets which are not elements of $\mathcal B^{[0,t]}(\mathbb R)$ but are contained in elements of $\mathcal B^{[0,t]}(\mathbb R)$, which show incompleteness of the natural filtration.
Indeed, let $C$ denote the Cantor set and let $E = \{w : w_{i} \in [0,1] \text{ for all } i \in C$. Then I claim that $E$ is not an event which is measurable with respect to $\mathcal F^B_t$ for $t \geq 1$.
If not, let $S$ be a set such that if $f \in E$ and $w|_S = f|_S$ then $w \in E$. Clearly the Cantor function belongs in $E$ (I don't actually need continuity of the function so I could have gone with just the indicator of the Cantor set. I'm just doing this because usually we associate Brownian paths with continuity so I thought the function could be continuous as well!).
Now, because $S$ is countable and $C$ is not, we know that there is a point in $C \setminus S$, call it $s'$. Define $w(s) = f(s)$ except at $s'$ where $w(s') = 2$. Then $f|_{S} = w|_{S}$ and yet $w \notin E$ because $w(s') \notin [0,1]$.
Thus, no such countable set can exist, so $E$ is not in $\mathcal F^B_t$. However, you can try to find a measurable $\mathbb P$-null set by yourself such that $E$ is contained in that null set.

Thus, the natural filtration is very much extended by completion. Completion of the filtration allows us to do one very important thing : it allows us to ensure that we can modify processes to retain measurability and adaptedness. For example, constructing continuous modifications of a stochastic process like a martingale will require us to modify the martingale on sets which are not null sets but are clearly contained in them. Thus we require completeness to ensure measurability and adaptability are not affected as we make these changes.
