How can I show that $(\|B_t\|^2-dt)_{t\geq 0}$ is a martingale?

Question

Let $B$ be a $d$-dimensional $(\Bbb{F}, \Bbb{P})$-Brownian motion where $\Bbb{F}=(\mathcal{F}_t)_{t\geq 0}$. Then consider $X:=(X_t)_{t\geq 0}=(\|B_t\|^2-dt)_{t\geq 0}$. I want to check that it is a martingale.

To do so I need to check the following three points:

1. $\Bbb{E}(\|X_t\|)<\infty$ for all $t\geq 0$
2. $X$ is $\Bbb{F}$-adapted and measurable
3. $\Bbb{E}(X_t|\mathcal{F}_s)=X_s$ for all $0\leq s\leq t$.

Now let me prove the first point. Fix $t\geq 0$ then $$\begin{align}\Bbb{E}(\|X_t\|)=\Bbb{E}(\|B_t\|^2-dt)=\Bbb{E}(\|B_t\|^2)-dt=0<\infty\end{align}$$

The second point causes a problem, since I don't see how to prove adaptedness and measurability. I thought that I can take $A\in \mathcal{B}(\Bbb{R})$ and show that $X^{-1}(A)\in \mathcal{F}_s$ but then I struggle a bit with the measurability. Can someone explain me this?

For the third point let $0\leq s\leq t$ then $$\begin{align}\Bbb{E}(X_t|\mathcal{F}_s)&=\Bbb{E}(\|B_t\|^2-dt|\mathcal{F}_s)\\&=\Bbb{E}(\|B_t\|^2|\mathcal{F}_s)-dt\\&\end{align}$$ But also here I don't see further. Can someone explain me this also?

Answer 1

$X_t$ is the composition of $B_t$ with the continuous map $x \mapsto \|x\|^{2}-dt$ so it is $\mathcal F_t$ measurable. $X$ has continuous paths also so it is a measuarble process. For the last part a hint is $B_t-B_s$ and $B_s$ are orthogonal in $L^{2}$: $E(\|B_t\|^{2}|\mathcal F_s)=E([\|(B_t-B_s)+B_s\|^{2}]\mathcal F_s)=E([\|(B_t-B_s)\|^{2}]\mathcal F_s)+E([\|B_s\|^{2}]\mathcal F_s)$. Can you finish?