In the proof of Girsanov's theorem i see the following: Let $$M_t=e^{ia^{tr}B_t+\frac12||a||^2t}$$ where B is the standard brownian motion and $a\in \mathbb R^d$.
We then know that $$\sup_{t\le T}|M_t|$$ is bounded for all $T\ge0$.
I'm not able to understand this last step, why it is bounded? I only see that $\sup_{t\le T}|M_t|\le\infty$ since $M_t$ is a.s. continuous.
Since $a^{\mathrm{tr}}B_t$ is a real number, we have that $$\tag{1} \lvert e^{ia^{\mathrm{tr}}B_t+\frac12\lVert a\rVert^2t}\rvert = e^{\frac12\lVert a\rVert^2t}. $$ We have used a standard property of the complex exponential; $$ \lvert e^{x+iy}\rvert = e^{x}, \qquad \forall (x, y)\in \mathbb R^2.$$ Now, $$ \sup_{t\in [0, T]} e^{\frac12\lVert a\rVert^2t}= e^{\frac12\lVert a\rVert^2 T}.$$ So the left-hand side in (1) is bounded by $e^{\frac12\lVert a\rVert^2 T}$.