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

Question

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

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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.

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

Answer 1

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 $F\in \mathscr{F}_s$.