[proofplan] We verify the martingale property directly by computing the conditional expectation $\mathbb{E}[M_t^u \mid \mathcal{F}_s^+]$. The key step is factoring $M_t^u$ into an $\mathcal{F}_s^+$-measurable part and a part depending only on the independent increment $B_t - B_s$, then evaluating the moment generating [function](/page/Function) of the Gaussian increment. [/proofplan] [step:Factor $M_t^u$ into an $\mathcal{F}_s^+$-measurable term and an independent term] Fix $0 \leq s \leq t$ and $u \in \mathbb{R}^d$. Decompose the exponent using $\langle u, B_t \rangle = \langle u, B_s \rangle + \langle u, B_t - B_s \rangle$: \begin{align*} M_t^u &= \exp\!\left(\langle u, B_t \rangle - \frac{|u|^2 t}{2}\right) \\ &= \exp\!\left(\langle u, B_s \rangle - \frac{|u|^2 s}{2}\right) \cdot \exp\!\left(\langle u, B_t - B_s \rangle - \frac{|u|^2 (t-s)}{2}\right) \\ &= M_s^u \cdot \exp\!\left(\langle u, B_t - B_s \rangle - \frac{|u|^2 (t-s)}{2}\right). \end{align*} The first factor $M_s^u$ is $\mathcal{F}_s^+$-measurable. The second factor depends only on the increment $B_t - B_s$, which is independent of $\mathcal{F}_s^+$ by [Independence from the Germ Field](/theorems/1177). [/step] [step:Compute the conditional expectation using the Gaussian moment generating function] Taking conditional expectations and using the independence: \begin{align*} \mathbb{E}[M_t^u \mid \mathcal{F}_s^+] = M_s^u \cdot \mathbb{E}\!\left[\exp\!\left(\langle u, B_t - B_s \rangle - \frac{|u|^2(t-s)}{2}\right)\right]. \end{align*} The increment $B_t - B_s \sim \mathcal{N}(0, (t-s) I_d)$, so the projection $\langle u, B_t - B_s \rangle$ is a one-dimensional Gaussian with mean $0$ and variance $|u|^2(t-s)$. The moment generating function of $\mathcal{N}(0, \sigma^2)$ evaluated at $\lambda = 1$ gives $\mathbb{E}[e^X] = e^{\sigma^2/2}$ when $X \sim \mathcal{N}(0, \sigma^2)$. Therefore \begin{align*} \mathbb{E}\!\left[\exp\!\left(\langle u, B_t - B_s \rangle\right)\right] = \exp\!\left(\frac{|u|^2(t-s)}{2}\right), \end{align*} and hence \begin{align*} \mathbb{E}[M_t^u \mid \mathcal{F}_s^+] = M_s^u \cdot \exp\!\left(\frac{|u|^2(t-s)}{2} - \frac{|u|^2(t-s)}{2}\right) = M_s^u. \end{align*} Integrability: $\mathbb{E}[M_t^u] = M_0^u = 1 < \infty$ (by setting $s = 0$ in the martingale identity), so $M_t^u \in L^1(\mathbb{P})$ for all $t \geq 0$. [/step]