Brownian-filtration martingale representation

Product density on raw sigma-algebras. Fix $T\ge 0$, let $\mathcal{H}={\mathcal{F}}_T^0$ and ${\mathcal{G}}_m=\sigma ({\tilde{B}}_h:0\le h\le 1\mathrm{/}m)$. Finite sums of bounded products $UV$, with $U$ measurable in $\mathcal{H}$ and $V$ in ${\mathcal{G}}_1$, are dense in $L^2(\mathcal{H}\vee {\mathcal{G}}_1)$. Indeed, their closed linear span contains $1_{A\cap D}$ for $A\in \mathcal{H},D\in {\mathcal{G}}_1$. Sets whose indicators belong to the span form a Dynkin class: complements use the constant $1$, and disjoint countable unions follow by $L^2$ convergence of their finite indicator sums. The intersections form a generating pi-system. Pi-lambda therefore supplies all measurable indicators; simple approximation and truncation give density.

Bounded stopping for continuous integrable martingales. Let $L$ be such a martingale and let $\rho$ be a bounded stopping time. For fixed $s<t$ choose $K>t$ exceeding its bound and finite deterministic grids of $[0,K]$ containing $s,t$ with mesh tending to zero. Round $\rho$ upward to a grid stopping time ${\eta }_j$. The sampled process stopped at ${\eta }_j$ is a discrete martingale: each increment is an original martingale increment multiplied by the past-measurable indicator that stopping has not yet occurred. Thus, for $A\in {\mathcal{F}}_s$, $E[1_A{L}_{t\wedge {\eta }_j}]=E[1_A{L}_{s\wedge {\eta }_j}]\mathrm{.}$ Bounded optional sampling on the same grid writes each of these stopped values as a conditional expectation of the fixed variable $L_K$. They form a uniformly integrable family by [F12]. Continuity gives convergence almost surely to $L_{t\wedge \rho }$ and $L_{s\wedge \rho }$, hence convergence in $L^1$. Passing to the limit proves the martingale test. For each $t$, adaptation of $L_{t\wedge \rho }$ follows by rounding $t\wedge \rho$ upward on grids of $[0,t]$ and taking the continuous limit; all grid values and events are ${\mathcal{F}}_t$-measurable. Thus $L^{\rho }$ is a continuous integrable martingale.

Local integrand uniqueness, proved before the next patch. If two local integrals $H\cdot B,K\cdot B$ agree up to a stopping time $\tau$, stop also at $T$ and at level $N$ of the combined energy ${\int }_0^t(H^2+K^2)ds$, with time cap $N$. These are stopping times by the energy construction in [F1] (apply it to the predictable square root of $H^2+K^2$). Both stopped integrands have finite expected energy. The stopping identity and finite-energy linearity make the terminal integral of their stopped difference zero. Isometry gives $E{\int }_0^{T\wedge \tau \wedge {\sigma }_N}(H_s-K_s{)}^{2}ds=0.$ The levels exhaust each finite horizon almost surely; nonnegative convergence gives agreement $\mathrm{d}t\otimes P$-almost everywhere before $T\wedge \tau$. This proves the needed uniqueness independently of the representation to be constructed.

Removal of the right germ, before using Brownian integration in the augmented filtration. For bounded $U,V$ as in step 1.1, testing on $A\cap D$, $A\in \mathcal{H},D\in {\mathcal{G}}_m$, and using independence gives $E[UV\mid \mathcal{H}\vee {\mathcal{G}}_m]=UE[V\mid {\mathcal{G}}_m]\mathrm{.}$ The test extends to the generated sigma-algebra by pi-lambda. Downward convergence and Blumenthal's law give $E[V\mid {\mathcal{G}}_m]\to EV$ almost surely; the uniform bound yields $L^2$ convergence. Thus the displayed conditional expectation tends to $UEV=E[UV\mid \mathcal{H}]$. Density from step 1.1 and the $L^2$ contraction extend this to every $Y\in L^2(\mathcal{H}\vee {\mathcal{G}}_1)$. If $Z\in L^2({\mathcal{F}}_T)$, completion supplies a raw version in each $\mathcal{H}\vee {\mathcal{G}}_m$. Use the $m=1$ version in this convergence: every conditional expectation on the left is $Z$, so $Z=E[Z\mid {\mathcal{F}}_T^0]$ almost surely. In particular every ${\mathcal{F}}_T$ event agrees almost surely with a raw past event. The Brownian increment law and independence of the raw past therefore also hold for ${\mathcal{F}}_T$. Since $T$ was arbitrary, $B$ is Brownian relative to the usual filtration, as required by [F1] and [F2]. This argument uses only raw Brownian laws and conditional expectation, not martingale representation.

The range of terminal integrals: let $R_T:=\{{\int }_0^{T}HdB:H\text{ predictable},E{\int }_0^T{H}^{2}ds<\mathrm{\infty }\}\subseteq L^2(P)$. By [F1], $R_T$ is a closed linear subspace contained in the mean-zero subspace, and each of its elements is ${\mathcal{F}}_T$-measurable because elementary terminal integrals are finite combinations of Brownian increments and the $L^2$ limit of ${\mathcal{F}}_T$-measurable random variables is ${\mathcal{F}}_T$-measurable; hence $R_T\subseteq \{Z\in L^2({\mathcal{F}}_T):EZ=0\}$.

Orthogonality forces vanishing, step one: let $Z\in L^2({\mathcal{F}}_T)$ with $EZ=0$ and $Z\perp R_T$. For every deterministic step function $h$, [F2] puts $e^{q(T)\mathrm{/}2}\cos X_T-1$ and $e^{q(T)\mathrm{/}2}\sin X_T$ in $R_T$. Orthogonality and $EZ=0$ therefore give $E[Z\cos X_T]=E[Z\sin X_T]=0$, hence $E[Z\exp (i{\sum }_j{\lambda }_j(B_{t_j}-B_{t_{j-1}}))]=0$ for all real coefficients.

Cylinder Fourier transforms. Insert $t_0=0$ into any finite list of positive times $t_1<\cdots <t_n\le T$. Put ${\mu }_k={\sum }_{j=k}^n{\lambda }_j$; since $B_0=0$ almost surely, ${\sum }_j{\lambda }_j{B}_{t_j}={\sum }_k{\mu }_k(B_{t_k}-B_{t_{k-1}})$ almost surely. A coordinate at time zero contributes only an almost surely zero term. Push the two finite positive measures $Z^{+}dP$ and $Z^{-}dP$ forward by the coordinate vector. They are finite because $Z\in L^2(P)\subset L^1(P)$. Step 4.1 gives identical Fourier transforms, so [F5] makes the pushforwards equal. Hence ${\int }_{A}ZdP=0$ for every Brownian cylinder rectangle with times at most $T$.

Step three, pi-lambda: the cylinder rectangles with all times $\le T$ form a pi-system generating $\sigma (B_s:s\le T)={\mathcal{F}}_T^0$, and $A\mapsto {\int }_{A}ZdP$ is a finite signed measure vanishing there; the class of sets on which it vanishes is closed under complements, proper differences and increasing countable unions (continuity from below), so by the Dynkin pi-lambda theorem [F5] it vanishes on all of ${\mathcal{F}}_T^0$. Hence $E[Z\mid {\mathcal{F}}_T^0]=0$ almost surely.

Completion and the right germ. Step 2.1 applies to this $Z\in L^2({\mathcal{F}}_T)$ and gives $Z=E[Z\mid {\mathcal{F}}_T^0]$. Step 6.1 makes the latter zero. Thus every mean-zero $Z$ orthogonal to $R_T$ vanishes.

Conclusion of the $L^2$ stage. Set $S=R_T+\mathrm{span}\{1\}$ in the full real space $L^2({\mathcal{F}}_T)$. This subspace is closed: if $r_k+c_k\to Y$ in $L^2$, continuity of expectation gives $c_k\to EY$, and then $r_k\to Y-EY\in R_T$. A vector orthogonal to $S$ has mean zero and is orthogonal to $R_T$, hence vanishes by step 7.1. Apply [F6] on the probability space with sigma-algebra ${\mathcal{F}}_T$ to obtain $S=L^2({\mathcal{F}}_T)$. Thus $Z-EZ={\int }_0^{T}HdB$ for a finite-energy predictable $H$. The continuous integral martingale satisfies $EZ+{\int }_0^{t}HdB=E[Z\mid {\mathcal{F}}_t]$ for each $t\le T$. This assigns the conditional expectations their continuous version; continuous versions agree on rational times and at $T$, hence everywhere on a common full event. The isometry gives uniqueness of $H$ modulo $\mathrm{d}t\otimes P$.

Continuity before level stopping. Let $N$ be a cadlag integrable martingale and fix $T>0$. Truncate $Z=N_T$ to $Z^{(r)}=(-r)\vee (Z\wedge r)$. Step 8.1 gives continuous martingales $Y_t^{(r)}=E[Z^{(r)}\mid {\mathcal{F}}_t]$. At every time of the countable set $D=(\mathbb{Q}\cap [0,T])\cup \{T\}$, $\mathrm{\mid }{Y}_q^{(r)}-N_q\mathrm{\mid }\le E[\mathrm{\mid }{Z}^{(r)}-Z\mathrm{\mid }\mid {\mathcal{F}}_q]\mathrm{.}$ Apply the discrete Doob $L^1$ inequality to this nonnegative conditional-expectation martingale on increasing finite subsets of $D$ containing $0,T$. Their maxima increase to the supremum over $D$. On the common measurable full event of continuity of the $Y^{(r)}$ and cadlag paths of $N$, right continuity and inclusion of $T$ identify this with the supremum over $[0,T]$. Therefore $\epsilon P({\sup }_{0\le t\le T}\mathrm{\mid }{Y}_t^{(r)}-N_t\mathrm{\mid }>\epsilon )\le E\mathrm{\mid }{Z}^{(r)}-Z\mathrm{\mid }⟶0.$ Here the supremum is understood as its measurable countable-set version, agreeing with the path supremum on that full event. A subsequence converges uniformly almost surely by [F7]; thus $N$ itself has continuous paths on a measurable full event and is indistinguishable from a continuous version. Intersect these events over integer $T$. Their complement is an ambient null event in ${\mathcal{F}}_0$; replacing the process there by zero gives an everywhere continuous adapted version. Absolute value and powers of a martingale are submartingales Doob L1 maximal inequality

Continuity of the local martingale. By its definition, $N^{(n)}=M^{{\tau }_n}-M_0$ is a cadlag integrable martingale starting at zero. Step 9.1 supplies an everywhere continuous adapted version $C^{(n)}$ indistinguishable from it. Intersect their agreement events, the event of cadlag paths of $M$, and the event ${\tau }_n\uparrow \mathrm{\infty }$, obtaining a measurable full event $G\in {\mathcal{F}}_0$. On $G$, $C^{(n)}$ agrees with $M-M_0$ up to ${\tau }_n$; consequently $M-M_0$ is continuous on every finite interval. Define $C_t=M_t-M_0$ on $G$ and $C_t=0$ outside $G$. Then $C$ is everywhere continuous and adapted, starts at zero, and is indistinguishable from $M-M_0$. Each $C^{{\tau }_n}$ remains an integrable martingale. We use this centered process below; no integrability of $M_0$ was assumed or needed.

Bounded continuous pieces. For fixed $n$, set $C^{(n)}=C^{{\tau }_n}$ and ${\rho }_{n,m}=\inf \{t\ge 0:\mathrm{\mid }{C}_t^{(n)}\mathrm{\mid }\ge m\}\wedge m\mathrm{.}$ For $t<m$, its stopping event is $\{{\sup }_{u\in [0,t]}\mathrm{\mid }{C}_u^{(n)}\mathrm{\mid }\ge m\}$, measurable using rational times and $t$; compactness and continuity ensure the supremum is attained. For $t\ge m$ the event is all of $\mathrm{\Omega }$. Since $C_0^{(n)}=0$, continuity gives $\mathrm{\mid }(C^{(n)}{)}^{{\rho }_{n,m}}\mathrm{\mid }\le m$ and ${\rho }_{n,m}\uparrow \mathrm{\infty }$ pathwise. Step 1.2 makes these bounded processes martingales. On each positive integer horizon $T$, their terminal variables have mean zero and lie in $L^2$, so step 8.1 represents the entire processes by finite-energy predictable integrands. Their integral versions agree at all times by [F3], and isometry makes the integrands agree on overlaps. Choose the countably many representatives using [F8] and patch along deterministic unit intervals. This gives predictable $H^{(n,m)}$ of finite expected energy on every finite horizon representing $(C^{(n)}{)}^{{\rho }_{n,m}}$.

Agreement on overlaps: if $\rho \le {\rho }^{\mathrm{\prime }}$ and the bounded stopped martingales have representations $H,H^{\mathrm{\prime }}$ on $[0,T]$, then $H{1}_{[0,\rho ]}=H^{\mathrm{\prime }}{1}_{[0,\rho ]}$ $(\mathrm{d}t\otimes P)$-a.e. Indeed, stopping the two representations at $\rho$ gives the same continuous process, so ${\int }_0^T(H-H^{\mathrm{\prime }})1_{[0,\rho ]}dB=0$; the isometry then gives $E{\int }_0^T(H-H^{\mathrm{\prime }}{)}^2{1}_{[0,\rho ]}ds=0$.

Passing first in $m$. For fixed $n$, step 12.1 gives agreement of the integrands on the increasing stochastic intervals $[0,{\rho }_{n,m}]$. Put $H^{(n)}=H^{(n,1)}{1}_{[0,{\rho }_{n,1}]}+{\sum }_{m\ge 2}{H}^{(n,m)}{1}_{({\rho }_{n,m-1},{\rho }_{n,m}]}\mathrm{.}$ The indicators are predictable; the intervals are disjoint, so the sum is a pointwise limit of predictable finite sums with at most one nonzero summand. By [F13] and countable intersection, on one full event the patch agrees time-almost everywhere with $H^{(n,m)}$ before ${\rho }_{n,m}$ for every $m$ and integer horizon. Each path-horizon eventually lies there, and $H^{(n,m)}$ has finite energy almost surely, so $H^{(n)}$ has locally finite energy. Moreover its truncation at ${\rho }_{n,m}$ has finite expected energy on each horizon and its integral equals $(C^{(n)}{)}^{{\rho }_{n,m}}$. The localizing-sequence independence in [F1] therefore gives $H^{(n)}\cdot B=C^{(n)}$ up to indistinguishability.

Passing in $n$. For $n\le n^{\mathrm{\prime }}$, $(C^{(n^{\mathrm{\prime }})}{)}^{{\tau }_n}=C^{(n)}$. Step 1.3 makes $H^{(n^{\mathrm{\prime }})}$ and $H^{(n)}$ agree almost everywhere before ${\tau }_n$. Define $H=H^{(1)}{1}_{[0,{\tau }_1]}+{\sum }_{n\ge 2}{H}^{(n)}{1}_{({\tau }_{n-1},{\tau }_n]}\mathrm{.}$ As in step 13.1 this is predictable; [F13] and ${\tau }_n\uparrow \mathrm{\infty }$ show its energy is finite almost surely on every finite horizon. Before ${\tau }_n$ it agrees with $H^{(n)}$ in product measure. To compare their integrals, stop additionally at the combined-energy levels used in step 1.3; the isometry and stopping identity give equality there, and exhaustion gives $(H\cdot B{)}^{{\tau }_n}=C^{(n)}$. Finally ${\tau }_n\uparrow \mathrm{\infty }$ and countable intersection of the indistinguishability events yield $H\cdot B=C$. Consequently $M=M_0+H\cdot B$ up to indistinguishability.

Uniqueness in clause 2 follows from step 1.3 with $\tau =\mathrm{\infty }$, on every finite horizon. Its continuous integral version and step 14.1 give the asserted continuity.

Boundary and choice cases. At $T=0$, step 2.1 gives that every ${\mathcal{F}}_0$ event agrees almost surely with a $\sigma (B_0)$ event. Since $B_0=0$ almost surely, every such event has probability zero or one, and every real $L^2({\mathcal{F}}_0)$ variable is almost surely constant (apply this to its rational sublevel sets). The integral over the empty time interval is zero. Bounded Brownian cylinder variables satisfy clause 1 because they lie in $L^2({\mathcal{F}}_T)$; no explicit formula for their integrands is asserted. For $M=B$ the integrand is $1$, and for a constant process it is $0$. Clause 2 excludes nonzero jumps on a full event. Full AC covers all inherited interfaces and the countably many representative and integrand choices; these selections need not be canonical.