Two definitions for $\mathcal F_{\infty}$

Question

I have just started learning about Brownian motion $(B_t)_{t \geq 0}$. The book I'm following defines a filtration $(\mathcal F_t)\_{t \geq 0}$ by $\mathcal F_t := \sigma(B_r, 0 \leq r \leq t)$. I think it's fairly clear that this is in fact a filtration.

What I'm confused about is that the author defines $\mathcal F_{\infty} := \sigma(B_t, t \geq 0)$. I was expecting a different definition, which I thought was standard, call it $\mathcal F_{\infty}' := \sigma\left( \bigcup_{t \geq 0} \mathcal F_t\right)$. Do we in fact have $\mathcal F_{\infty} = \mathcal F_{\infty}'$?

I think it should be true, but I have no idea how to show it. By definition, we have

\begin{align*} \mathcal F\_{\infty} = \sigma \left( \bigcup\_{t \geq 0} \{ B\_t^{-1}(A) \colon A \in \mathcal B(\mathbb R^d) \} \right), \end{align*}

and on the other hand,

\begin{align*} \mathcal F_{\infty}' = \sigma\left( \bigcup_{t \geq 0} \sigma\left( \bigcup_{s \leq t} \{ B_s^{-1}(A) \colon A \in \mathcal B(\mathbb R^d) \} \right) \right) \end{align*}

Thanks for any help showing that these two are equal

Answer 1

$\mathcal F_{\infty} \subseteq \mathcal F'_{\infty}$ because RHS is a $\sigma-$ field and each $B_t$ is measurable w.r.t. it.

$\mathcal F'_{\infty} \subseteq \mathcal F_{\infty}$ because each $\mathcal F_t$ is conatined in $\mathcal F_{\infty}$.