Statement: The $σ$-algebra $\mathcal{F}_{0+}$ is trivial in the sense that $\mathbb{P}[A] = 0$ or $1$ for every $A ∈ \mathcal{F}_{0+}$, where $\mathcal{F}_{0+}=\bigcap_{s>0}\mathcal{F}_s$
The proof I'm looking at starts like this:
Proof: Let $0 < t_1 < t_2 ··· < t_k$ and let $g : \mathbb{R}^k → \mathbb{R}$ be a bounded continuous function. Also, fix $A ∈ \mathcal{F}_{0+}$. Then by a continuity argument, $$\mathbb{E}[1_Ag(B_{t_1} ,...,B_{t_k} )] = \lim_{ ε↓0} \mathbb{E}[1_Ag(B_{t_1} −B_ε ,...,B_{t_k} −B_ε )]$$
What I don't understand is the continuity argument - we are really interchanging an integral with a limit without explicit enough conditions to use MCT, DCT etc.
Since $g$ is bounded, the Bounded Convergence Theorem allows the swapping of $\Bbb E$ and $\lim\limits_{\epsilon \downarrow 0}$. $$\lim_{ ε↓0} \mathbb{E}[1_Ag(B_{t_1} −B_ε ,...,B_{t_k} −B_ε )] = \Bbb{E}[\lim\limits_{\epsilon \downarrow 0} 1_A g(B_{t_1} - B_\epsilon, \dots, B_{t_k} - B_\epsilon)]$$ Pass the limit sign into the bracket as $g$ is continuous. $$\Bbb{E}[\lim\limits_{\epsilon \downarrow 0} 1_A g(B_{t_1} - B_\epsilon, \dots, B_{t_k} - B_\epsilon)] = \Bbb{E}[1_A g(\lim\limits_{\epsilon \downarrow 0} (B_{t_1} - B_\epsilon, \dots, B_{t_k} - B_\epsilon))]$$ Since the brownian motion $B_t$ has continous trajectory, $B_\epsilon \to B_0 = 0$ as $\epsilon \downarrow 0$. $$\Bbb{E}[1_A g(\lim\limits_{\epsilon \downarrow 0} (B_{t_1} - B_\epsilon, \dots, B_{t_k} - B_\epsilon))] = \mathbb{E}[1_Ag(B_{t_1},\dots,B_{t_k})]$$ Combine the equalities to end the proof.