Proof that a $\mathbb{F}$-Wiener-martingale is an $\mathbb{F}$ martingale

Question

I am confused with the following proof,

Claim: Let $W$ be a Wiener process adapted to a filtration $\mathbb{F}=\{\mathcal{F}_t\}_{t \geq 0}$, and suppose that $W_t-W_s$ is independent of $\mathcal{F}_s$ for all $s<t$. Show that W is an $\mathbb{F}$-martingale.

Proof:

1. $W$ is adapted to $\mathbb{F}$ by definition
2. We use the Cauchy Schwartz inequality to see that $\mathbb{E} |W_t| \leq \mathbb{E}(|W_t^2|)^{\frac{1}{2}} = \sqrt{t} < \infty$. So $W_t$ is integrable
3. We see that, $\mathbb{E}(W_t|\mathcal{F}_s)=\mathbb{E}(W_t + W_s - W_s|\mathcal{F}_s) = \mathbb{E}(W_t-W_s|\mathcal{F}_s) + \mathbb{E}(W_s|\mathcal{F}_s) = W_s$

My questions are:

1. In step 2, since $W$ is Wiener, is it not just the case that by definition $\mathbb{E}(W_t) = 0, \forall t$? I do not see a need to use Cauchy Schwartz here.
2. Why does independent increments imply that $\mathbb{E}(W_t-W_s|\mathcal{F}_s) = 0$?

Answer 1

You are right about step 2).

$E(X|\mathcal G)=EX$ whenever $X$ is independent of $\mathcal G$. This basic property of conditional expectations follows immediately from the definition of conditional expectations. Hence $E(W_t-W_s|\mathcal F_s)=E(W_t-W_s)=EW_t-EW_s=0-0=0$.