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For deterministic continuous functions $h_j$ on a finite interval, their stochastic exponentials $M(h_j)$ are square-integrable martingales and satisfy $dM(h_j)=h_{j}M(h_j)dB$. Thus $F={\sum }_j{c}_{j}M(h_j{)}_T$ has mean ${\sum }_j{c}_j$ and stochastic-integral representation with the displayed integrand. Itô isometry proves uniqueness up to time-probability null sets among square-integrable predictable integrands. It does not give pointwise uniqueness, since changing a value at one deterministic time changes no integral.

Apply Itô's lemma to $F(t,x)=e^{i\xi x+{\xi }^{2}t/2}$. Since $F_t={\xi }^{2}F/2$, $F_x=i\xi F$ and $F_{xx}=-{\xi }^{2}F$, the drift terms cancel:This is initially a local martingale. On every finite interval $[0,T]$,so the stochastic integral is a true square-integrable martingale. Conditional expectation gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$; divide by the nonzero ${\mathcal{F}}_s$-measurable factor $e^{i\xi B_s+{\xi }^{2}t/2}$ to obtain

Fix a finite horizon $T$ and use the completed natural Brownian filtration. Martingale representation theorem states that every $F\in L^2({\mathcal{F}}_T)$ can be writtenwhere $H$ is predictable and is unique up to equality for $dsd\mathbb{P}$ almost every $(s,\omega )$.

The Martingale representation theorem states that every square-integrable martingale $M$ in this filtration has a continuous version of the formwith the same integrability and uniqueness condition. The completed natural filtration has only null-set randomness at time zero, so $M_0$ is almost surely constant. The first statement gives the second by representing $M_T$ and taking conditional expectations; the second gives the first by taking $M_t=\mathbb{E}(F\mid {\mathcal{F}}_t)$.

There is also the local version: every local martingale in the Brownian filtration has a continuous version representable by a predictable, locally square-integrable integrand. The filtration restriction matters; an enlarged filtration with independent additional martingales need not have representation using $B$ alone.