Why does $XE_{B_{s}}Y \in \mathcal{F}_{s}^{\circ} \Rightarrow E(Z \mid \mathcal{F_{s}}^{+})=E(Z \mid \mathcal{F_{s}}^{\circ})$?
This is in the context of a Theorem, explained in Durret's textbook, that:
If $Z$ is a bounded RV, then for all $s \geq 0$ and $x \in \mathbb R^{d}$, then:
$$E(Z\mid \mathcal{F_{s}}^{+})=E(Z\mid \mathcal{F_{s}}^{\circ}).$$
Note that $\mathcal{F_{s}}^{\circ}=\sigma(\{B_{i}\vert 0\leq i\leq s\})$ and $\mathcal{F_{s}}^{+}=\bigcap\limits_{s < t}\mathcal{F_{t}}^{\circ}$
I do not understand the last part of the proof:
Let $Z=\prod\limits_{1\leq i\leq m}f_{m}(B_{t_{m}})$ where $f_{m}$ are bounded and measurabe and $(B_{t})_{t \geq 0}$ a Brownian motion. It follows that: for $Z=X(Y\circ\theta_{s})$ where $X$ is $\in \mathcal{F}_{s}^{\circ}$ and $Y$ measurable
$$E_{x}(Z\mid \mathcal{F_{s}}^{+})=XE_{x}(Y\circ\theta_{s}\mid \mathcal{F_{s}}^{+})=XE_{B_{s}}Y\in \mathcal{F}_{s}^{\circ} \tag{$\star$}$$
I do not understand why the last line renders $E(Z\mid \mathcal{F_{s}}^{+})=E(Z\mid \mathcal{F_{s}}^{\circ})$
The proof shows that $$E_x(Z \mid \mathcal{F}_s^+) = X E_{B_s}(Y). \tag{1} $$ Since $\mathcal{F}_s^{\circ}$ is a sub-$\sigma$-algebra of $\mathcal{F}_s^+$, it follows from the tower property of conditional expectation that
$$E_x(Z \mid \mathcal{F}_s^{\circ}) = E_x \big[ E_x(Z \mid \mathcal{F}_s^+)\mid \mathcal{F}_s^{\circ} \big].$$
Hence, by $(1)$,
$$E_x(Z \mid \mathcal{F}_s^{\circ}) = E_x \big[ X E_{B_s}(Y) \mid \mathcal{F}_s^{\circ} \big].$$
As $X E_{B_s}(Y)$ is $\mathcal{F}_s^{\circ}$-measurable, this gives
$$E_x(Z \mid \mathcal{F}_s^{\circ}) = X E_{B_s}(Y). \tag{2}$$
Combining $(1)$ and $(2)$, we conclude that
$$E_x(Z \mid \mathcal{F}_s^{\circ}) \stackrel{(2)}{=} X E_{B_s}(Y) \stackrel{(1)}{=} E_x(Z \mid \mathcal{F}_s^+)$$
... as claimed.