= Solution
Fix $s<t$ and an event $A\in\mathcal F_s$. Multiplying the conditional <characteristic function> identity by $1_A$ and taking expectation gives
$$
\mathbb E[1_Ae^{i\theta(X_t-X_s)}]
=\mathbb P(A)e^{-\theta^2(t-s)/2}.
$$
The left side is the Fourier transform of the finite measure
$$
\mu_A(D)=\mathbb P(A\cap\{X_t-X_s\in D\}).
$$
By the <uniqueness theorem for characteristic functions>, this measure equals $\mathbb P(A)$ times the $N(0,t-s)$ distribution. Thus, for every Borel set $D$,
$$
\mathbb P(A\cap\{X_t-X_s\in D\})
=\mathbb P(A)\,\mathbb P(N(0,t-s)\in D).
$$
Taking $A$ to be the whole sample space identifies the increment's law, and the full identity proves independence from $\mathcal F_s$. Together with the assumed adaptation, continuity, and initial value, these are exactly the defining Brownian properties. This proves the <conditional characteristic-function criterion for Brownian increments>.
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