From Square-Integrable to Local Representation
The Brownian martingale representation theorem applies directly to square-integrable martingales: their fluctuations can be written as Ito integrals against Brownian motion. A continuous local martingale need not be square-integrable on the whole time interval. Localization makes the extension possible: stop the process so that each stopped martingale is square-integrable, represent each one, and show that the resulting integrands agree wherever the stopped processes overlap.
Fix a finite horizon \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Assume the filtration has the square-integrable Brownian representation property on \([0,T]\): every square-integrable martingale \(N\) on this interval has a representation \(N_t=N_0+\int_0^t H_s\,dB_s\) for a progressively measurable integrand with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). This is the earlier L2 Brownian martingale representation result.
A continuous local martingale \(M\) is a continuous adapted process for which there are stopping times increasing to the horizon such that the stopped processes are martingales. We use the usual convention that its values at fixed times are integrable. The representation is for its increment: the initial value \(M_0\) need not be zero.
Stopping and Matching the Representations
Proof. Set \(X_t=M_t-M_0\), so \(X_0=0\). Choose increasing localizing times \(\sigma_n\uparrow T\) for \(X\), and define \(\rho_n=\inf\{t\in[0,T]:|X_t|\ge n\}\wedge T\), with the infimum interpreted as \(T\) if the set is empty. Put \(\tau_n=\sigma_n\wedge\rho_n\). Both sequences of stopping times are increasing, so \(\tau_n\) is increasing. Each continuous sample path of \(X\) is bounded on \([0,T]\); hence \(\rho_n=T\) for all sufficiently large \(n\), path by path. It follows that \(\tau_n\uparrow T\) almost surely.
The stopped process \(X^{\tau_n}_t=X_{t\wedge\tau_n}\) is a local martingale bounded in absolute value by \(n\), and is therefore a martingale. To justify this fact, localize \(X^{\tau_n}\) further by stopping times \(\eta_k\) so that \(X^{\tau_n}_{t\wedge\eta_k}\) is a martingale. These further-stopped processes remain bounded by \(n\). For \(s\le t\), their conditional-expectation identities pass to the limit as \(k\to\infty\) by bounded convergence, since \(\eta_k\uparrow T\) and the paths are continuous. Thus \(\mathbb{E}[X^{\tau_n}_t\mid\mathcal{F}_s]=X^{\tau_n}_s\). Boundedness also gives \(\mathbb{E}[(X^{\tau_n}_T)^2]\le n^2\), so this martingale is square-integrable.
By the assumed L2 representation property, for each \(n\) there is a progressively measurable \(H^n\) with \(\mathbb{E}\int_0^T(H^n_s)^2\,ds<\infty\) and $$ X_{t\wedge\tau_n}=\int_0^t \mathbf{1}_{\{s\le\tau_n\}}H^n_s\,dB_s. $$ Equivalently, we may write the right side as the stopped integral \(\int_0^{t\wedge\tau_n}H^n_s\,dB_s\). For \(m\ge n\), stopping the representation at \(\tau_n\) gives $$ \int_0^{t\wedge\tau_n}H^m_s\,dB_s =X_{t\wedge\tau_n} =\int_0^{t\wedge\tau_n}H^n_s\,dB_s. $$ The uniqueness of square-integrable Brownian integrands, applied to the stopped integrands, implies $$ \mathbb{E}\int_0^{T\wedge\tau_n}|H^m_s-H^n_s|^2\,ds=0. $$ Thus \(H^m=H^n\) for \(ds\,d\mathbb{P}\)-almost every \((s,\omega)\) with \(s\le\tau_n(\omega)\). This is the compatibility needed to join the stopped integrands.
Choose progressively measurable versions and define \(H\) by using \(H^1\) on \([0,\tau_1]\) and \(H^n\) on \((\tau_{n-1},\tau_n]\) for \(n\ge2\); set \(H_T=0\) if needed. This countable pasting is progressively measurable. Compatibility implies that, for every \(n\), \(H=H^n\) almost everywhere on the random interval \([0,\tau_n]\). Consequently, $$ \mathbb{E}\int_0^{T\wedge\tau_n}H_s^2\,ds =\mathbb{E}\int_0^{T\wedge\tau_n}(H^n_s)^2\,ds \le \mathbb{E}\int_0^T(H^n_s)^2\,ds<\infty. $$ So \(H\) is locally square-integrable. The stopped representations now give \(X_t=\int_0^t H_s\,dB_s\) on every interval \([0,\tau_n]\). Since \(\tau_n\uparrow T\), these identities imply the representation for all \(t<T\), simultaneously outside one null set.
It remains to justify the endpoint. For \(t<T\), the equality with \(X\) and the quadratic-variation identity for Ito integrals give $$ \int_0^t H_s^2\,ds=[X]_t. $$ A continuous local martingale has finite quadratic variation \([X]_T\) almost surely. Letting \(t\uparrow T\), monotone convergence of the pathwise energy therefore gives \(\int_0^T H_s^2\,ds=[X]_T<\infty\) almost surely. The locally defined Ito integral has a finite terminal value, obtained as its limit from times below \(T\); its continuity and the continuity of \(X\) show that this value is \(X_T\). Thus the representation holds at \(T\) as well. \(\square\)
The terminal argument matters: local square-integrability by itself only bounds expected energy up to each localizing time. It does not assert finite expected energy on all of \([0,T]\), or even finite pathwise energy there. Here it is the equality before \(T\), followed by finiteness of \([X]_T\), that supplies the required terminal control.
Worked Applications
Worked Example: A Brownian Motion Stopped at a Level
Let \(\tau=\inf\{t\ge0:|B_t|\ge1\}\), and on \([0,T]\) set \(M_t=B_{t\wedge\tau}\). Identify its representing integrand and check its energy.
The stopped Brownian integral identity gives $$ M_t=\int_0^t \mathbf{1}_{\{s\le\tau\}}\,dB_s. $$ Thus \(H_s=\mathbf{1}_{\{s\le\tau\}}\), a progressively measurable integrand. Since \(0\le\mathbf{1}_{\{s\le\tau\}}\le1\), $$ \mathbb{E}\int_0^T H_s^2\,ds =\mathbb{E}[T\wedge\tau]\le T<\infty. $$ So in this example the integrand is globally square-integrable, not merely locally square-integrable.
Check: The quadratic variation of the stopped integral is \([M]_t=\int_0^t\mathbf{1}_{\{s\le\tau\}}\,ds=t\wedge\tau\), which is also the quadratic variation of stopped Brownian motion. The indicator at the single time \(s=\tau\) does not change the time integral.
Worked Example: An Integral with a Random Integrand
On \([0,T]\), let \(M_t=\int_0^t B_s\,dB_s\). Verify its representation, check the integrability condition, and compute its terminal mean and variance.
The integrand \(H_s=B_s\) is progressively measurable. Since \(\mathbb{E}[B_s^2]=s\), $$ \mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=\frac{T^2}{2}<\infty. $$ Thus the Ito integral is square-integrable. Ito's formula for \(f(x)=x^2\) has \(f'(x)=2x\), \(f''(x)=2\), and correction term \(\tfrac12 f''(B_t)\,dt=dt\). It gives \(d(B_t^2)=2B_t\,dB_t+dt\), hence $$ M_t=\frac12(B_t^2-t). $$ The integral has mean zero because it is square-integrable and its martingale increments have conditional expectation zero; in particular, \(\mathbb{E}[M_T]=0\).
For the variance, \(B_T\) is normal with mean zero and variance \(T\), so \(\mathbb{E}[B_T^4]=3T^2\). Therefore $$ \operatorname{Var}(M_T)=\mathbb{E}[M_T^2] =\frac14\mathbb{E}\big[(B_T^2-T)^2\big] =\frac14(3T^2-2T^2+T^2)=\frac{T^2}{2}. $$ Check: the Ito isometry gives \(\mathbb{E}[M_T^2]=\mathbb{E}\int_0^T B_s^2\,ds=T^2/2\), agreeing with the direct calculation.
Worked Example: Local Energy Need Not Be Finite at the Endpoint
On \([0,1)\), consider the deterministic integrand \(H_s=(1-s)^{-1/2}\). For \(\tau_n=1-1/n\), with \(n\ge2\), compute its expected energy up to \(\tau_n\), then compare it with its energy up to time \(1\).
The integrand is progressively measurable, and for each \(n\ge2\), $$ \mathbb{E}\int_0^{\tau_n}H_s^2\,ds =\int_0^{1-1/n}\frac{1}{1-s}\,ds =\big[-\log(1-s)\big]_0^{1-1/n} =\log n<\infty. $$ But the total energy is $$ \int_0^1 H_s^2\,ds=\int_0^1\frac{1}{1-s}\,ds=\infty. $$ Thus the integrals are well-defined up to every \(\tau_n\), but this alone does not produce a finite continuous integral at time \(1\).
Check: the quadratic variation of the integral up to \(t<1\) is \(\int_0^t H_s^2\,ds=-\log(1-t)\), which diverges as \(t\uparrow1\). A continuous local martingale defined on the closed interval \([0,1]\) has finite quadratic variation at \(1\) almost surely. Therefore this process cannot be extended as such a martingale to time \(1\). In the theorem, terminal finiteness follows instead from matching the integral to the given continuous local martingale before the endpoint.
Why the Compatibility Step Matters
Each stopped martingale has a representation, but an unrelated choice of integrand for every stopping time would not yet represent the original process. The overlap argument is what makes a single integrand possible: two stopped integrals agree as processes before the earlier stopping time, and the Ito isometry converts that process equality into equality of the integrands almost everywhere on that interval. Pasting is then legitimate because the choices coincide wherever both are used.
A common error is to conclude that \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) from the local bounds \(\mathbb{E}\int_0^{T\wedge\tau_n}H_s^2\,ds<\infty\). The singular integrand above shows why that inference fails: each stopped energy is \(\log n\), while the full energy diverges. Another error is to assert equality at \(T\) merely because the integral has continuous paths on each stopped interval. The proof must first establish finite total energy, here by identifying it with the finite terminal quadratic variation of the given continuous local martingale.
Check Your Understanding
Use the localization theorem and the distinction between local and global energy to answer these questions.
- Why does stopping a continuous local martingale when its absolute value first reaches \(n\) produce a bounded stopped process?
- For \(m\ge n\), which equality of stopped processes leads to equality of \(H^m\) and \(H^n\) on \([0,\tau_n]\)?
- What condition on the filtration is used to represent each stopped martingale?
- Why do local expected-energy bounds alone not justify an Ito integral at the terminal time?
- In the theorem's endpoint argument, which identity gives finite total energy for the pasted integrand?