Consider Brownian motion $(B_t)_{t\geqslant 0}$ and define the Brownian bridge from $x$ to $y$ as $$ X_t:=x(1-t)+yt+ (B_t-tB_1) \, , \,\,\, t\in[0,1]. $$ Now if $(\mathcal{F}_t)_{t\geqslant 0}$ is the fitration for which the Brownian motion is adapted, write $$ X_t-x(1-t)-yt-(1-t)B_t=-t(B_1-B_t). $$ For a given $t$, The right side is independent of $\mathcal{F_t}$ by definition of the Brownian motion, and the terms on the left side besides $X_t$ are clearly $\mathcal{F}_t$-measurable.
Thus if $X_t$ $\textit{were}$ $\mathcal{F}_t$-measurable then it is also independent of $\mathcal{F}_t$, and hence almost surely constant. This is a contradiction since $(X_t)$ is a Guassian process.
So $(X_t)$ can't be adapted to $(\mathcal{F}_t)$, but can $(B_t)$ be adapted to $\mathcal{F}_t^X$?
I am struggling to figure this out. I have so far that \begin{align} \mathcal{F}_t^X=\sigma(X_s:s\geqslant t)&=\sigma(x(1-s)+ys +(B_s-sB_1): s\leqslant t) \\ &=\sigma(B_s-sB_1 : s\leqslant t) \\ &=\sigma((1-s)B_s-s(B_1-B_s): s\leqslant t). \end{align}
But at this stage I am stuck. I have tried to see if I can formulate some kind of containment $\mathcal{F}_t\subset \mathcal{F}_t^X$ to no avail, yet I can't see I way to show that the motion can't be adapted. It looks like it shouldn't be adaptable since any information we have on $B_t$ is shifted by a term independent of it in the filtration for the bridge.
Any help would be great!