# A proof of Lévy's theorem: how to obtain the independence $X_t-X_s \perp \mathcal{F}_s$?

I'm reading about Lévy's theorem at page $$42$$ of these notes, i.e.,

Let $$X$$ be a continuous local martingale such that $$X_0=0$$ a.s. and $$\langle X\rangle_t=t$$ a.s., $$\forall t \in \mathbb{R}_{+}$$. Then $$X$$ is a standard Brownian motion.

Proof. $$\forall c \in \mathbb{R}, c X$$ is a continuous local martingale such that $$\langle c X\rangle_t=c^2 t$$. Therefore, by Theorem 7.10, the process $$\left(Y_t=\exp \left(c X_t-\frac{c^2 t}{2}\right), t \in \mathbb{R}_{+}\right)$$is a martingale, i.e., $$\mathbb{E}\left(\exp \left(c X_t-\frac{c^2 t}{2}\right) \mid \mathcal{F}_s\right)=\exp \left(c X_s-\frac{c^2 s}{2}\right), \quad \forall c \in \mathbb{R}$$ or $$\mathbb{E}\left(\exp \left(c\left(X_t-X_s\right)\right) \mid \mathcal{F}_s\right)=\exp \left(\frac{c^2(t-s)}{2}\right), \quad \forall c \in \mathbb{R} . \quad (\star)$$

Fact 1: since the right-hand side is deterministic, $$X_t-X_s \perp \mathcal{F}_s$$. Moreover, by taking expectations, we obtain $$\mathbb{E}\left(\exp \left(c\left(X_t-X_s\right)\right)\right)=\exp \left(\frac{c^2(t-s)}{2}\right), \quad \forall c \in \mathbb{R} .$$

Fact 2: this implies that $$X_t-X_s \sim \mathcal{N}(0, t-s)$$. Therefore, $$X$$ is a standard Brownian motion.

My question Could you please elaborate on the proof of Fact 1, i.e., how $$(\star)$$ implies $$X_t-X_s$$ is independent of $$\mathcal F_s$$.

You can find another proof of Lévy's characterisation in Schilling, Partszch, Boettcher; 9.12 p. 148; the use of characteristic functions is less troublesome, in my opinion, than MGFs. In the reference, we get $$E[e^{i\xi (X_t-X_s)}|\mathscr{F}_s]=e^{-(t-s)\xi^2/2}$$; then multiplying by $$\mathbf{1}_F$$ for any $$\mathscr{F}_s$$-measurable set $$F$$ and then taking expectations we obtain the relation $$E[e^{i\xi (X_t-X_s)}\mathbf{1}_F]=E[e^{i\xi (X_t-X_s)}]P(F),\forall F \in \mathscr{F}_s$$ which the authors then claim to be sufficient for $$X_t-X_s$$ to be independent from $$\mathscr{F}_s$$. This, however, is not really immediate. We may also choose any rv $$Z$$ s.t. $$\sigma(Z)\subseteq \mathscr{F}_s$$; in this case we obtain $$E[e^{i\xi_1 (X_t-X_s)}e^{i\xi_2 Z}]=E[e^{i\xi_1 (X_t-X_s)}]E[e^{i\xi_2 Z}],\forall \xi_1,\xi_2\in \mathbb{R}$$ This actually proves independence of $$X_t-X_s$$ and $$Z$$ for any such $$Z$$ $$\mathscr{F}_s$$-measurable due to an important, but elusive theorem known as Kac's theorem. At this point choose the $$\mathscr{F}_s$$-measurable function $$\mathbf{1}_F$$ for some arbitrary $$F \in \mathscr{F}_s$$, we get for any Borel $$B$$ \begin{aligned}P(\{X_t-X_s\in B\}\cap F)&=P(X_t-X_s\in B,\mathbf{1}_F=1)=\\ &=P(X_t-X_s\in B)P(\mathbf{1}_F=1)=\\ &=P(X_t-X_s\in B)P(F) \end{aligned} But this proves independence of $$\sigma(X_t-X_s)$$ and $$\mathscr{F}_s$$, since this holds for all $$F\in \mathscr{F}_s$$.