Martingale Representation Theorem

Proof.

If \(Y = g(\mathbf{W})\) with \(g\) Borel, then for Borel \(E \subseteq \mathbb{R}\) the preimage \(Y^{-1}(E) = \mathbf{W}^{-1}(g^{-1}(E))\) lies in \(\sigma(\mathbf{W})\), which is the easy direction. For the converse, climb from indicators. If \(Y = \mathbf{1}_A\) with \(A \in \sigma(\mathbf{W})\), then \(A = \mathbf{W}^{-1}(B)\) for some Borel \(B\), and \(Y = \mathbf{1}_B(\mathbf{W})\). If \(Y\) is a simple function, a finite linear combination of such indicators, take the same combination of the corresponding \(\mathbf{1}_B\). For general \(\sigma(\mathbf{W})\)-measurable \(Y\), the standard staircase truncations

\[ Y_n = \Bigl( \bigl( 2^{-n} \lfloor 2^{n} Y \rfloor \bigr) \vee (-n) \Bigr) \wedge n \]

are simple, \(\sigma(\mathbf{W})\)-measurable, and converge to \(Y\) pointwise, since \(|Y_n - Y| \leq 2^{-n}\) once \(|Y| \lt n\). Write \(Y_n = g_n(\mathbf{W})\) with \(g_n\) Borel by the simple case, and set \(g(\mathbf{x}) = \limsup_n g_n(\mathbf{x})\) where the limit superior is finite and \(g(\mathbf{x}) = 0\) elsewhere, a Borel function. At every \(\omega\), \(g_n(\mathbf{W}(\omega)) = Y_n(\omega) \to Y(\omega)\), so the limit superior at \(\mathbf{W}(\omega)\) is finite and equals \(Y(\omega)\). Hence \(Y = g(\mathbf{W})\).

Proof.

Step 1 (Countably many samples suffice).
Let \(t_1, t_2, \ldots\) enumerate a dense subset of \([0, T]\) and let \(\mathcal{H}_k = \sigma(w_{t_1}, \ldots, w_{t_k})\), an increasing family with union generating a \(\sigma\)-algebra \(\mathcal{H}_\infty = \sigma\bigl( \bigcup_k \mathcal{H}_k \bigr)\). We claim \(\mathcal{F}_T\) exceeds \(\mathcal{H}_\infty\) only by null sets. For any \(s \in [0, T]\), pick \(t_{k_i} \to s\) from the dense set. Paths are continuous outside a null set, so \(w_s = \lim_i w_{t_{k_i}}\) almost surely, and a pointwise limit of \(\mathcal{H}_\infty\)-measurable functions, adjusted on the exceptional null set, is measurable with respect to \(\mathcal{H}_\infty \vee \mathcal{N}\), the augmentation by null sets. Since \(\mathcal{F}_T\) is generated by exactly these variables \(w_s\) and the null sets, \(\mathcal{F}_T \subseteq \mathcal{H}_\infty \vee \mathcal{N}\).

Step 2 (Augmentation is invisible to \(L^2\)).
Consider the family \(\mathcal{G}\) of sets \(A \subseteq \Omega\) differing from some \(A' \in \mathcal{H}_\infty\) by a null set, meaning \(A \,\triangle\, A' \in \mathcal{N}\). It contains \(\mathcal{H}_\infty\) and every null set, and it is a \(\sigma\)-algebra. Complements preserve the symmetric difference, and for countable unions \(\bigl( \bigcup A_i \bigr) \triangle \bigl( \bigcup A_i' \bigr) \subseteq \bigcup ( A_i \triangle A_i' )\), a countable union of null sets. Hence \(\mathcal{H}_\infty \vee \mathcal{N} \subseteq \mathcal{G}\), and every \(\mathcal{F}_T\)-measurable indicator equals an \(\mathcal{H}_\infty\)-measurable indicator almost surely. Climbing through simple functions and pointwise limits as in the factorization proof shows that every \(\mathcal{F}_T\)-measurable variable agrees almost surely with an \(\mathcal{H}_\infty\)-measurable one, so it suffices to approximate \(g \in L^2(\mathcal{H}_\infty)\).

Step 3 (Down to finitely many samples).
Let \(\mathcal{S}\) be the closure in \(L^2(\mathbb{P})\) of \(\bigcup_k L^2(\mathcal{H}_k)\), a closed subspace. The family \(\{ A \in \mathcal{H}_\infty : \mathbf{1}_A \in \mathcal{S} \}\) contains the algebra \(\bigcup_k \mathcal{H}_k\) and is stable under complements, by linearity and \(\mathbf{1}_{A^c} = 1 - \mathbf{1}_A\). Stability under increasing countable unions holds because \(\mathbf{1}_{A_i} \to \mathbf{1}_{\cup A_i}\) pointwise with the constant dominator one, hence in \(L^2\) by the dominated convergence theorem.

The passage from an algebra with these closure properties to the full generated \(\sigma\)-algebra is the monotone-class step this track has taken on faith since the construction of the integral, and we extend the same trust to this one application. Simple functions are then in \(\mathcal{S}\) by linearity, and general \(g \in L^2(\mathcal{H}_\infty)\) by the staircase truncations of the factorization proof, the squared differences being dominated by the integrable \(g^2 + 1\), since each truncation error is at most \(1\) where \(|g| \lt n\) and at most \(|g|\) elsewhere. So \(\mathcal{S} = L^2(\mathcal{H}_\infty)\), and it suffices to approximate \(g \in L^2(\mathcal{H}_k)\) for fixed \(k\).

Step 4 (Bounding the dependence).
Fix \(k\) and \(g \in L^2(\mathcal{H}_k)\). By the factorization theorem, \(g = g_k(\mathbf{W})\) for a Borel \(g_k : \mathbb{R}^k \to \mathbb{R}\), where \(\mathbf{W} = (w_{t_1}, \ldots, w_{t_k})\). The truncations \(\varphi_n = ( g_k \wedge n ) \vee ( -n )\) are bounded and Borel, so each \(\varphi_n(\mathbf{W}) = ( g \wedge n ) \vee ( -n )\) is a cylinder function. They converge to \(g\) pointwise, and \(( g - \varphi_n(\mathbf{W}) )^2 \leq g^2\), an integrable dominator, so the dominated convergence theorem gives \(\varphi_n(\mathbf{W}) \to g\) in \(L^2\). Chaining the four steps completes the proof.

Proof.

Step 1 (Reduction to a vanishing pairing).
Each \(\mathcal{E}(h)\) with \(h\) a step function lies in \(L^2(\mathcal{F}_T)\), since \(\int_0^T h\, dw\) is then a finite linear combination of samples of the path, a centered Gaussian variable whose exponential has a finite second moment by the moment generating function computed in Step 2. If the closed span were a proper subspace of the Hilbert space \(L^2(\mathcal{F}_T)\), the projection theorem would produce a nonzero \(g\) orthogonal to every \(\mathcal{E}(h)\). It therefore suffices to show that such orthogonality forces \(g = 0\). We will show it forces \(\mathbb{E}[\varphi(\mathbf{W})\, g] = 0\) for every cylinder function \(\varphi(\mathbf{W})\), where \(\mathbf{W} = (w_{t_1}, \ldots, w_{t_k})\) is the vector of samples that \(\varphi\) reads. Then the density of cylinder functions, proved above, supplies a sequence \(\varphi_n(\mathbf{W}_n) \to g\) in \(L^2\), with \(\mathbf{W}_n\) the sample vector that \(\varphi_n\) reads, and \(0 = \mathbb{E}[\varphi_n(\mathbf{W}_n)\, g] \to \mathbb{E}[g^2]\) yields \(g = 0\).

Step 2 (From exponentials of integrals to exponentials of samples).
Fix times \(0 \leq t_1 \lt \cdots \lt t_k \leq T\), write \(\mathbf{W} = (w_{t_1}, \ldots, w_{t_k})\), and take \(\lambda \in \mathbb{R}^k\). The step function \(h = \sum_i c_i \mathbf{1}_{[0, t_i)}\) has \(\int_0^T h\, dw = \sum_i c_i\, w_{t_i}\), by linearity of the integral and the identity \(\int_0^T \mathbf{1}_{[0, t_i)}\, dw = w_{t_i}\). With \(c_i = \lambda_i\), the weight on each \(w_{t_i}\) is exactly the prescribed \(\lambda_i\). Orthogonality to \(\mathcal{E}(h)\), with the deterministic factor \(\exp(\tfrac{1}{2} \int h^2)\) multiplied back in, therefore gives

\[ \begin{align*} G(\lambda) &= \mathbb{E}\Bigl[ \exp\bigl( \lambda_1 w_{t_1} + \cdots + \lambda_k w_{t_k} \bigr)\, g \Bigr] \\\\ &= 0 \quad \forall\, \lambda \in \mathbb{R}^k . \end{align*} \]

Every expectation here is finite. The exponent \(\lambda \cdot \mathbf{W}\) is a linear combination of values of the Gaussian process, hence centered Gaussian with some variance \(\sigma^2\). Completing the square against the density of a centered Gaussian variable \(Z\) with variance \(\sigma^2 \gt 0\) gives, for every \(s \in \mathbb{R}\), the Gaussian moment generating function,

\[ \begin{align*} \mathbb{E}\bigl[ e^{s Z} \bigr] &= \int_{-\infty}^{\infty} e^{s z}\, \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-z^2 / 2 \sigma^2}\, dz \\\\ &= e^{s^2 \sigma^2 / 2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(z - s \sigma^2)^2 / 2 \sigma^2}\, dz \\\\ &= e^{s^2 \sigma^2 / 2} \end{align*} \]

The case \(\sigma^2 = 0\) is trivial, and the shifted density integrates to one by the Gaussian integral. With the Cauchy-Schwarz inequality, \(\mathbb{E}[e^{\lambda \cdot \mathbf{W}} |g|] \leq \mathbb{E}[e^{2 \lambda \cdot \mathbf{W}}]^{1/2}\, \|g\|_{L^2} \lt \infty\), so \(G\) is well defined. This moment generating function returns twice more on this page.

Step 3 (Two measures with one generating function).
Split \(g = g^+ - g^-\) and define two finite Borel measures on \(\mathbb{R}^k\) as the \(g^\pm\)-weighted laws of the sample vector,

\[ \nu^\pm(B) = \mathbb{E}\bigl[ \mathbf{1}_B(\mathbf{W})\, g^\pm \bigr] , \quad B \in \mathcal{B}(\mathbb{R}^k) , \]

countably additive by the monotone convergence theorem and finite since \(\mathbb{E}[g^\pm] \leq \mathbb{E}[|g|] \leq \|g\|_{L^2}\). The vanishing of \(G\) says exactly that the two measures have the same moment generating function everywhere,

\[ \int_{\mathbb{R}^k} e^{\lambda \cdot x}\, d\nu^+(x) = \int_{\mathbb{R}^k} e^{\lambda \cdot x}\, d\nu^-(x) \quad \forall\, \lambda \in \mathbb{R}^k . \]

The passage from expectations of \(e^{\lambda \cdot \mathbf{W}} g^\pm\) to integrals against \(\nu^\pm\) is the image-measure substitution, run through indicators, simple functions, and monotone limits. We now invoke one fact at statement level. Two finite Borel measures on \(\mathbb{R}^k\) whose moment generating functions, here the integrals displayed above as functions of \(\lambda\), exist and agree on all of \(\mathbb{R}^k\) are equal. This uniqueness theorem sits at the same epistemic tier as the continuity theorem already recorded without proof on the convergence page. Its standard proofs run through analytic continuation or Fourier methods that this track has not built, and it is the single trusted input of this lemma. Granting it, \(\nu^+ = \nu^-\).

Step 4 (Back up the ladder).
Equal measures integrate every bounded Borel function equally, so for every bounded Borel \(\varphi : \mathbb{R}^k \to \mathbb{R}\),

\[ \begin{align*} \mathbb{E}\bigl[ \varphi(\mathbf{W})\, g \bigr] &= \int \varphi\, d\nu^+ - \int \varphi\, d\nu^- \\\\ &= 0 . \end{align*} \]

A cylinder function with unordered or repeated times is a bounded Borel function of the samples at the distinct times, listed in increasing order, so the displayed identity covers it as well. The times and \(k\) were arbitrary, so \(g\) is orthogonal to every cylinder function, and Step 1 closes the argument. For general deterministic \(h \in L^2([0, T])\), the exponentials lie in \(L^2(\mathcal{F}_T)\) by the Gaussian law of the next section, and the larger family has a closed span at least as large.

Proof.

Step 1 (Step integrands, exactly).
Suppose first that \(h\) is a step function. Then \(I_t = \sum_i c_i (w_{r_{i+1} \wedge t} - w_{r_i \wedge t})\) is a finite linear combination of increments over disjoint intervals. Its moment generating function factors over the increments, which are independent of the past, so the expectation of the product peels one factor at a time from the last increment backwards, the product of the earlier factors being measurable with respect to the filtration at the start of the last increment and hence independent of it, by the product rule. Each peeled factor is a Gaussian moment generating function, computed by the completed square of the previous section. Multiplying,

\[ \begin{align*} \mathbb{E}\bigl[ e^{s I_t} \bigr] &= \prod_i \exp \Bigl( \tfrac{s^2 c_i^2}{2} \bigl( r_{i+1} \wedge t - r_i \wedge t \bigr) \Bigr) \\\\ &= \exp \Bigl( \tfrac{s^2}{2} \int_0^t h(r)^2\, dr \Bigr) . \end{align*} \]

Step 2 (General integrands by \(L^1\) transfer).
For general deterministic \(h \in L^2([0, T])\), extend the restriction of \(h\) to \([0, t]\) by zero to a function in \(L^2(\mathbb{R})\). The density of the Schwartz class provides Schwartz approximants of it in \(L^2(\mathbb{R})\), and their real parts, restricted to \([0, t]\), approximate \(h\) in \(L^2([0, t])\) at least as well, since \(h\) is real. Each approximant is continuous, hence uniformly continuous on \([0, t]\), so left-endpoint discretization over a fine enough partition of \([0, t]\) produces a step function uniformly close to it on \([0, t]\), and therefore close in \(L^2([0, t])\). So there are step functions \(h_n \to h\) in \(L^2([0, t])\). The Itô isometry gives \(I^{(n)} = \int_0^t h_n\, dw \to I_t\) in \(L^2(\mathbb{P})\). Fix \(s \in \mathbb{R}\). The elementary bound \(|e^{a} - e^{b}| \leq |a - b|\, (e^{a} + e^{b})\), from the mean value of the exponential between \(a\) and \(b\), and the Cauchy-Schwarz inequality give

\[ \Bigl| \mathbb{E}\bigl[ e^{s I^{(n)}} \bigr] - \mathbb{E}\bigl[ e^{s I_t} \bigr] \Bigr| \leq |s|\, \bigl\| I^{(n)} - I_t \bigr\|_{L^2} \Bigl( \bigl\| e^{s I^{(n)}} \bigr\|_{L^2} + \bigl\| e^{s I_t} \bigr\|_{L^2} \Bigr) . \]

The first factor tends to zero. The norms are bounded along the sequence, since \(\mathbb{E}[ e^{2 s I^{(n)}} ] = \exp( 2 s^2 \int_0^t h_n^2 )\) converges by Step 1. Moreover \(\mathbb{E}[ e^{2 s I_t} ] \lt \infty\), because along an almost surely convergent subsequence Fatou's lemma bounds it by the limit of the exact values. So \[ \begin{align*} \mathbb{E}[ e^{s I_t} ] &= \lim_n \exp( \tfrac{s^2}{2} \int_0^t h_n^2 ) \\\\ &= \exp( \tfrac{s^2}{2} \int_0^t h^2 ) . \end{align*} \]

Step 3 (Identifying the law).
The law of \(I_t\) and the centered Gaussian law with variance \(\sigma^2 = \int_0^t h^2\, dr\) are two finite Borel measures whose moment generating functions exist and agree everywhere, the Gaussian side by the completed square. The uniqueness of finite measures with a common moment generating function, the input already trusted in the density lemma for exponentials, identifies the two laws, and \(I_t\) is centered Gaussian with variance \(\sigma^2\).

Proof.

Step 1 (Membership).
The process \(h(s) Z_s\) is progressively measurable, being a product of a deterministic step function and a continuous adapted process, and by the Gaussian theorem, with \(I_s = \int_0^s h\, dw\),

\[ \begin{align*} \mathbb{E}\bigl[ Z_s^2 \bigr] &= e^{- \int_0^s h^2}\, \mathbb{E}\bigl[ e^{2 I_s} \bigr] \\\\ &= e^{- \int_0^s h^2}\, e^{2 \int_0^s h^2} \\\\ &= e^{\int_0^s h^2} \\\\ &\leq e^{\|h\|_{L^2}^2} , \end{align*} \]

so \(\mathbb{E}\bigl[ \int_0^T h^2 Z_s^2\, ds \bigr] \leq \|h\|_{L^2}^2\, e^{\|h\|_{L^2}^2} \lt \infty\), and \(h Z \in \mathcal{V}(0, T)\).

Step 2 (The identity, one piece at a time).
Extend \(h\) by \(h = 0\) on \((T, \infty)\), so that \(I_t = \int_0^t h\, dw\) is defined for every \(t \geq 0\) and is unchanged on \([0, T]\). The extended \(h\) is deterministic, bounded, and Borel, hence in \(\mathcal{V}(0, T')\) for every \(T' \gt 0\), and \(X_t = I_t\) is an Itô process with \(u = 0\) and \(v = h\). The natural map \(g(t, x) = \exp( x - \tfrac{1}{2} \int_0^t h^2 )\) need not be \(C^2\) in \(t\), its time derivative jumping where \(h^2\) jumps, so we work one interval of the partition at a time with a smooth stand-in. Fix a piece \([r_i, r_{i+1})\), on which \(h \equiv c_i\), and define

\[ \begin{align*} g_i(t, x) &= \exp \Bigl( x - A(r_i) - \tfrac{c_i^2}{2} ( t - r_i ) \Bigr) , \\\\ A(r) &= \tfrac{1}{2} \int_0^{r} h^2\, ds , \end{align*} \]

a \(C^2\) function on all of \([0, \infty) \times \mathbb{R}\) that agrees with the natural map for \(t \in [r_i, r_{i+1}]\), where \(g_i(t, I_t) = Z_t\). The membership condition of \(g_i\) holds for every horizon, since the diffusion entry \(h(s)\, g_i(s, I_s)\) vanishes for \(s \gt T\) and on \([0, T]\) differs from \(h Z\) by the bounded deterministic factor \(e^{A(s) - A(r_i) - c_i^2 (s - r_i)/2}\), so the Itô formula applies to \(g_i\). Its drift integrand is

\[ \frac{\partial g_i}{\partial t} + \tfrac{1}{2}\, h(s)^2\, \frac{\partial^2 g_i}{\partial x^2} = \Bigl( - \tfrac{c_i^2}{2} + \tfrac{h(s)^2}{2} \Bigr)\, g_i(s, I_s) , \]

which vanishes identically for \(s \in [r_i, r_{i+1})\), where \(h(s) = c_i\). This is the discount doing its work. The half squared integrand subtracted in the exponent is consumed, to the last drop, by the half second derivative the chain rule inserts. We apply the formula at times \(r_i\) and \(t \in [r_i, r_{i+1}]\) and subtract. The \(ds\)-integrals over \([r_i, t]\) then carry only the vanishing drift, and

\[ \begin{align*} Z_t - Z_{r_i} &= \int_{r_i}^{t} h(s)\, g_i(s, I_s)\, dw_s \\\\ &= \int_{r_i}^{t} h(s)\, Z_s\, dw_s . \end{align*} \]

The integrands agree on the interval of integration, and the interval splitting is the additivity of the integral. Telescoping over the pieces up to \(t\), with \(Z_0 = 1\), yields the identity, one null set per piece and finitely many pieces.

Step 3 (Martingale).
With \(h Z \in \mathcal{V}(0, T)\), the process \(Z_t - 1\) is an Itô integral process and hence a martingale with zero mean, so \(Z\) is a martingale with \(\mathbb{E}[Z_t] = 1\).

Proof.

Step 1 (The dense subspace is represented).
For a single exponential \(\mathcal{E}(h)\) with \(h\) a step function, the previous section gives \(\mathcal{E}(h) = 1 + \int_0^T h\, Z^{h}_s\, dw_s\) with \(h Z^{h} \in \mathcal{V}(0, T)\) and \(\mathbb{E}[\mathcal{E}(h)] = 1\), writing \(Z^h\) for the exponential martingale of \(h\). For a finite linear combination \(F_0 = \sum_j a_j\, \mathcal{E}(h_j)\) with step functions \(h_j\), linearity of the integral gives

\[ \begin{align*} F_0 &= \sum_j a_j + \int_0^T \Bigl( \sum_j a_j\, h_j(s)\, Z^{h_j}_s \Bigr) dw_s \\\\ &= \mathbb{E}[F_0] + \int_0^T f_0(s, \omega)\, dw_s , \end{align*} \]

the mean being identified by \(\mathbb{E}[\mathcal{E}(h_j)] = 1\). The integrand \(f_0\), a finite combination of progressively measurable members of \(\mathcal{V}(0, T)\), lies in \(\mathcal{V}(0, T)\) and is progressively measurable.

Step 2 (The isometry transports the Cauchy property).
Let \(F \in L^2(\mathcal{F}_T)\) and, by the density theorem, choose span elements \(F_n \to F\) in \(L^2(\mathbb{P})\), each represented as in Step 1 by a progressively measurable \(f_n \in \mathcal{V}(0, T)\). Means converge, \(|\mathbb{E}[F_n] - \mathbb{E}[F]| \leq \mathbb{E}[|F_n - F|] \leq \|F_n - F\|_{L^2}\) by the Cauchy-Schwarz inequality against the constant one. We subtract two representations and apply the Itô isometry, with a Tonelli exchange writing the squared norm as an iterated integral, to obtain

\[ \begin{align*} \int_0^T \mathbb{E}\bigl[ ( f_n - f_m )^2 \bigr]\, ds &= \mathbb{E}\Bigl[ \Bigl( \int_0^T ( f_n - f_m )\, dw_s \Bigr)^{2} \Bigr] \\\\ &= \mathbb{E}\Bigl[ \bigl( F_n - F_m - \mathbb{E}[ F_n - F_m ] \bigr)^{2} \Bigr] \\\\ &\leq \mathbb{E}\bigl[ ( F_n - F_m )^2 \bigr] , \end{align*} \]

the last inequality because subtracting the mean can only decrease the second moment, \(\mathbb{E}[(X - \mathbb{E}X)^2] = \mathbb{E}[X^2] - (\mathbb{E}X)^2\). The right side tends to zero as \(n, m \to \infty\), so \((f_n)\) is a Cauchy sequence in \(L^2([0, T] \times \Omega,\ \lambda \otimes \mathbb{P})\), and by completeness it converges to some \(f\) in that space.

Step 3 (The limit is a legitimate integrand).
Along a subsequence, \(f_{n_k} \to f\) at \(\lambda \otimes \mathbb{P}\)-almost every point of \([0, T] \times \Omega\). Define

\[ \tilde{f}(s, \omega) = \limsup_{k}\, f_{n_k}(s, \omega) , \]

set to zero where the limit superior fails to be finite. Each \(f_{n_k}\) restricted to \([0, t] \times \Omega\) is \(\mathcal{B}([0, t]) \otimes \mathcal{F}_t\)-measurable, limit superiors preserve measurability, and the finiteness set is measurable, so \(\tilde{f}\) is progressively measurable. At almost every point \(\tilde{f} = f\), so \(\tilde{f}\) lies in the same \(L^2\)-class, is jointly measurable and adapted through progressivity, and is square integrable. Hence \(\tilde{f} \in \mathcal{V}(0, T)\), and we rename it \(f\). The class \(\mathcal{V}(0, T)\) is, in this sense, closed in \(L^2(\lambda \otimes \mathbb{P})\).

Step 4 (Passing to the limit).
By the isometry once more, \(\int_0^T f_n\, dw \to \int_0^T f\, dw\) in \(L^2(\mathbb{P})\), and combining the three convergences gives

\[ \begin{align*} F &= \lim_n F_n \\\\ &= \lim_n \Bigl( \mathbb{E}[F_n] + \int_0^T f_n\, dw_s \Bigr) \\\\ &= \mathbb{E}[F] + \int_0^T f\, dw_s , \end{align*} \]

the limits taken in \(L^2(\mathbb{P})\) and the identity holding almost surely.

Step 5 (Uniqueness).
If \(f_1, f_2 \in \mathcal{V}(0, T)\) both represent \(F\), subtracting the two identities gives \(\int_0^T (f_1 - f_2)\, dw_s = 0\) almost surely, and the isometry, with the same Tonelli exchange, turns the vanishing of the integral into the vanishing of the integrand,

\[ \begin{align*} 0 &= \mathbb{E}\Bigl[ \Bigl( \int_0^T ( f_1 - f_2 )\, dw_s \Bigr)^{2} \Bigr] \\\\ &= \int_0^T \mathbb{E}\bigl[ ( f_1 - f_2 )^2 \bigr]\, ds , \end{align*} \]

so \(f_1 = f_2\) at \(\lambda \otimes \mathbb{P}\)-almost every point, the nonnegative integrand with vanishing integral being zero almost everywhere.

Proof.

Step 1 (A representation at every horizon).
Fix \(t \gt 0\). The variable \(M_t\) lies in \(L^2(\mathcal{F}_t)\), so the representation theorem with horizon \(t\) supplies a progressively measurable \(f^{(t)} \in \mathcal{V}(0, t)\), unique almost everywhere, with

\[ M_t = \mathbb{E}[M_t] + \int_0^t f^{(t)}(s, \omega)\, dw_s . \]

The mean is constant in \(t\). By the tower property and the martingale identity, \(\mathbb{E}[M_t] = \mathbb{E}\bigl[ \mathbb{E}[M_t \mid \mathcal{F}_0] \bigr] = \mathbb{E}[M_0]\).

Step 2 (Consistency of the recipes).
Let \(0 \lt t_1 \lt t_2\). Condition the horizon-\(t_2\) representation on \(\mathcal{F}_{t_1}\). The martingale identity for \(M\) on the left and the martingale property of the Itô integral on the right give, almost surely,

\[ \begin{align*} M_{t_1} &= \mathbb{E}\bigl[ M_{t_2} \mid \mathcal{F}_{t_1} \bigr] \\\\ &= \mathbb{E}[M_0] + \mathbb{E}\Bigl[ \int_0^{t_2} f^{(t_2)}\, dw_s \,\Bigm|\, \mathcal{F}_{t_1} \Bigr] \\\\ &= \mathbb{E}[M_0] + \int_0^{t_1} f^{(t_2)}\, dw_s . \end{align*} \]

The restriction of \(f^{(t_2)}\) to \([0, t_1]\) lies in \(\mathcal{V}(0, t_1)\) and represents \(M_{t_1}\), as does \(f^{(t_1)}\), so the uniqueness clause of the representation theorem forces, at \(\lambda \otimes \mathbb{P}\)-almost every point of \([0, t_1] \times \Omega\),

\[ f^{(t_2)} = f^{(t_1)} . \]

Recipes for longer horizons extend recipes for shorter ones. Nothing needs to be recomputed as time grows.

Step 3 (One process for all horizons).
Define \(g\) by gluing along integer horizons,

\[ g(s, \omega) = f^{(n)}(s, \omega) , \quad s \in (n - 1, n] ,\ n \in \mathbb{N} , \]

and \(g(0, \cdot) = f^{(1)}(0, \cdot)\). Each piece is the product of a progressively measurable process and a deterministic indicator, and the pieces live on disjoint time intervals, so \(g\) is progressively measurable. Fix \(t \gt 0\) and let \(N = \lceil t \rceil\). By Step 2, chained from each integer \(n \leq N\) up to \(N\), the process \(g\) agrees with \(f^{(N)}\) at almost every point of \([0, t] \times \Omega\), so \(g \in \mathcal{V}(0, t)\) and, since integrands equal almost everywhere have equal integrals by the isometry,

\[ \begin{align*} \int_0^t g\, dw_s &= \int_0^t f^{(N)}\, dw_s \\\\ &= M_t - \mathbb{E}[M_t] \\\\ &= M_t - \mathbb{E}[M_0] \end{align*} \]

almost surely, where the middle equality is Step 1 when \(t = N\) and otherwise the computation of Step 2 with \(t_1 = t\) and \(t_2 = N\). This is the claimed identity for \(t \gt 0\). At \(t = 0\) the identity reads \(M_0 = \mathbb{E}[M_0]\), which holds almost surely because \(\mathcal{F}_0\), being generated by \(w_0 = 0\) and the null sets, consists of null sets and their complements, so \(\mathcal{F}_0\)-measurable variables are almost surely constant.

Step 4 (Uniqueness).
If \(g\) and \(g'\) both work, then for every \(t\) they both represent \(M_t\) over \([0, t]\), so they agree \(\lambda \otimes \mathbb{P}\)-almost everywhere on \([0, t] \times \Omega\) by the representation theorem's uniqueness. Letting \(t \to \infty\) through the integers, we conclude that they agree almost everywhere on \([0, \infty) \times \Omega\).