[proofplan]
We show that every event $A \in \mathcal{F}_0^+$ is independent of itself, forcing $\mathbb{P}(A) \in \{0,1\}$. The key input is that $A$ is simultaneously measurable with respect to $\sigma(B_t : t \geq 0)$ and belongs to $\mathcal{F}_0^+$, while the [Independence from the Germ Field](/theorems/1177) guarantees that the post-zero process $(B_t - B_0)_{t \geq 0} = (B_t)_{t \geq 0}$ is independent of $\mathcal{F}_0^+$.
[/proofplan]
[step:Show $A$ is independent of itself]
Let $A \in \mathcal{F}_0^+$. Since $\mathcal{F}_0^+ \subset \sigma(B_t : t \geq 0)$, the event $A$ is measurable with respect to $\sigma(B_t : t \geq 0) = \sigma(B_t - B_0 : t \geq 0)$ (using $B_0 = 0$ a.s.). By [Independence from the Germ Field](/theorems/1177) applied at $s = 0$, the process $(B_{t+0} - B_0)_{t \geq 0} = (B_t)_{t \geq 0}$ is independent of $\mathcal{F}_0^+$. Since $A \in \sigma(B_t : t \geq 0)$ and every event in $\sigma(B_t : t \geq 0)$ is independent of every event in $\mathcal{F}_0^+$, the event $A$ is in particular independent of itself:
\begin{align*}
\mathbb{P}(A) = \mathbb{P}(A \cap A) = \mathbb{P}(A) \cdot \mathbb{P}(A) = \mathbb{P}(A)^2.
\end{align*}
The only solutions to $p = p^2$ with $p \in [0,1]$ are $p = 0$ and $p = 1$. Therefore $\mathbb{P}(A) \in \{0,1\}$.
[/step]