How to prove Brownian is a Martingale as to the part of measurability

Question

How can I prove in an exhaustive way that a Brownian Motion $B_t$ is a martingale with respect to its natural filtration as to the part in which one has to show that $B_t$ is $\mathcal{F}_t$-measurable, that is the fact that: $$B_t^{-1}(B)\subset\mathcal{F}_t, \forall B\in\mathcal{B}(\mathcal{R})$$ I know that $$\sigma[B_t]\subset\sigma[B_s|0\leq s\leq t]=\mathcal{F}_t$$ But how can this result imply the measurability?

Answer 1

Since $\sigma(B_t)\subset\sigma(B_s:s\le t)$, for any Borel set $C$, $B_t^{-1}(C)\in \sigma(B_s:s\le t)$. Then using the Markov property, for $t>s$, $$ \mathsf{E}[B_t\mid \mathcal{F}_s]=\mathsf{E}[B_s+(B_t-B_s)\mid \mathcal{F}_s]=B_s \quad\text{a.s.} $$

Answer 2

There is nothing to prove.

$B_t$ is $\mathcal{F}_t$ measurable by definition of $\mathcal{F}_t$

$\mathcal{F}_t$ is defined as the smallest $\sigma$-algebra s.t. $B_s$ is $\mathcal{F}_t$-measurable for all $s \in [0,t]$, so especially $B_t$ is $\mathcal{F}_t$ measurable.