Let $(\Omega,\mathscr{F},(\mathscr{F}_t)_{t\geq 0},P)$ be a filtered probability space. That is $(\Omega,\mathscr{F},P)$ is a probability space and $(\mathscr{F}_t)_{t\geq 0}$ is a filtration in $\mathscr{F}$, i.e. $\mathscr{F}_t\subseteq \mathscr{F}$ is a sigma-algebra for every $t\geq 0$ and for $0\leq s<t$ we have $\mathscr{F}_s\subseteq\mathscr{F}_t$. Then a mapping $X=(X_t)_{t\geq 0}:[0,\infty)\times\Omega\to\mathbb{R}^n$ is called a stochastic process if for every $t\geq 0$
$$
\Omega\ni \omega\mapsto X(t,\omega)=X_t(\omega)
$$
is $(\mathscr{F},\mathscr{B}(\mathbb{R}^n))$-measurable.
Since, for a fixed $t\geq 0$, $\mathscr{F}_t\subseteq\mathscr{F}$, the mapping $\omega\mapsto X(t,\omega)$ is not necessarily $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}^n))$-measurable (why?), and hence we need to make this a definition.
Therefore, we say that $(X_t)_{t\geq 0}$ is $(\mathscr{F}_t)_{t\geq 0}$-adapted if the mapping
$$
\Omega\ni \omega\mapsto X(t,\omega)
$$
is $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}^n))$-measurable for every $t\geq 0$.
I believe that in your setup, we have $(\mathscr{F}_t)_{t\geq 0}$ being the natural filtration induced by the Brownian motion $(B_t)_{t\geq 0}$. That is
$$
\mathscr{F}_t=\sigma(B_s\mid 0\leq s\leq t),\quad\text{for all }\;t\geq 0,
$$
i.e. $\mathscr{F}_t$ is the smallest sigma-algebra making all $B_s$, $s\leq t$, measurable. Now, it is obvious that $(B_t)_{t\geq 0}$ is adapted to $(\mathscr{F}_t)_{t\geq 0}$ (right?).
To see why, e.g. $h_1(t,\omega)=B_{t/2}(\omega)$ is $(\mathscr{F}_t)_{t\geq 0}$-adapted, we need to show that $\omega\mapsto h_1(t,\omega)$ is $(\mathscr{F}_t,\mathscr{B}(\mathbb{R}))$-measurable for every $t\geq 0$. So let $t\geq 0$ be given. Then
$$
\omega\mapsto B_{t/2}(\omega)
$$
is $(\mathscr{F}_{t/2},\mathscr{B}(\mathbb{R}))$-measurable and since $\mathscr{F}_{t/2}\subseteq\mathscr{F}_t$ it is also $(\mathscr{F}_{t},\mathscr{B}(\mathbb{R}))$-measurable.
Best Answer
Yes, and this is quite general since, for any collection $(X_s)_{0\leqslant s\leqslant t}$ of random variables, $$\sigma(X_s;0\leqslant s\leqslant t)=\sigma\left(\bigcup_{0\leqslant s\leqslant t}\sigma(X_s)\right).$$ Proof: Double inclusion.