Why a Representing Integrand Is Unique Only Almost Everywhere
A Brownian stochastic integral records an integrand through its accumulated fluctuations, but it cannot detect changes on sets of zero time-probability measure. The Itô isometry makes this precise: the mean square of an integral is exactly the integral of the squared integrand. Consequently, two integrands give the same terminal integral precisely when their difference vanishes almost everywhere with respect to time and probability. Continuity of the integral processes then gives the corresponding statement for equality of whole processes.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual conditions: the filtration is complete and right-continuous. Assume \(B_t\) is a Brownian motion with respect to this filtration. In particular, \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\) whenever \(0\le s<t\). This condition is stronger than adaptedness alone and is needed for the Itô integral construction and its martingale properties.
For \(H\in\mathcal{L}^2\), the Itô integral \(\int_0^t H_s\,dB_s\) is a continuous square-integrable martingale. The Itô isometry, established in the earlier material on square-integrable predictable integrands and the Itô isometry, states that $$ \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right] =\mathbb{E}\int_0^T H_s^2\,ds. $$ The same formula applies to \(H-K\), since \(H,K\in\mathcal{L}^2\) implies \(H-K\in\mathcal{L}^2\).
The Equivalence Theorem
- \(\int_0^T H_s\,dB_s=\int_0^T K_s\,dB_s\) almost surely;
- \(H=K\) \(dt\otimes d\mathbb{P}\)-almost everywhere on \([0,T]\);
- the continuous integral processes \(\left(\int_0^t H_s\,dB_s\right)_{0\le t\le T}\) and \(\left(\int_0^t K_s\,dB_s\right)_{0\le t\le T}\) are indistinguishable.
Proof. Set \(D=H-K\), which belongs to \(\mathcal{L}^2\) because \(D_s^2\le2H_s^2+2K_s^2\). First suppose the terminal integrals agree almost surely. Then \(\int_0^T D_s\,dB_s=0\) almost surely, so the Itô isometry gives $$ \mathbb{E}\int_0^T D_s^2\,ds =\mathbb{E}\left[\left(\int_0^T D_s\,dB_s\right)^2\right]=0. $$ The random variable \(\int_0^T D_s^2\,ds\) is nonnegative. A nonnegative random variable with expectation zero is zero almost surely; hence \(D=0\) \(dt\otimes d\mathbb{P}\)-almost everywhere. This proves the first implication.
Conversely, suppose \(D=0\) \(dt\otimes d\mathbb{P}\)-almost everywhere. Then \(\mathbb{E}\int_0^T D_s^2\,ds=0\), and the isometry gives \(\mathbb{E}[(\int_0^T D_s\,dB_s)^2]=0\). Thus the terminal integrals agree almost surely. The same argument applies on \([0,t]\) for each fixed \(t\le T\), so the two integral processes agree almost surely at each fixed time. To obtain indistinguishability, take the intersection of the probability-one events of equality over the countable set of rational times in \([0,T]\), including \(T\). On that event, the processes agree at all those times. Both have continuous paths, so they agree at every time by continuity.
Finally, indistinguishability implies equality at \(T\), so it implies the first condition. All three conditions are therefore equivalent. If \(T=0\), every integral is zero and every integrand is equal almost everywhere on the zero-length time interval; the conclusion is then trivial. \(\square\)
What the Isometry Says About the Difference
The theorem is a direct consequence of measuring the difference between two candidates. For every \(t\in[0,T]\), the isometry gives the quantitative identity $$ \mathbb{E}\left[\left(\int_0^t H_s\,dB_s-\int_0^t K_s\,dB_s\right)^2\right] =\mathbb{E}\int_0^t (H_s-K_s)^2\,ds. $$ Thus a zero left-hand side forces the integrands to agree on the whole interval up to \(t\), in the product-measure sense. In the other direction, a zero right-hand side forces the difference of the integrals to vanish at that time.
The martingale property used in this framework also depends on the filtration condition stated above. For an elementary adapted integrand, the integral is a finite sum of terms of the form \(\xi_i(B_{t_{i+1}}-B_{t_i})\), with \(\xi_i\) measurable at time \(t_i\). Independence and mean zero of Brownian increments imply that these sums have conditional mean zero over future time intervals. General progressively measurable square-integrable integrands are obtained by approximation in the norm \(\big(\mathbb{E}\int_0^T H_s^2\,ds\big)^{1/2}\). The isometry gives convergence of the corresponding integrals in \(L^2(\mathbb{P})\); conditional expectation is a contraction in \(L^2\), so the conditional-mean identities pass to the limit. The integral is square-integrable by the isometry, and it is a true martingale, not merely a local martingale.
Worked Example: A Difference That Cannot Integrate to Zero
On \([0,2]\), compare \(H_s=s\) and \(K_s=0\). Both are progressively measurable and square-integrable. Compute the mean square of the difference of their terminal integrals.
Apply the isometry to \(H-K=s\): $$ \mathbb{E}\left[\left(\int_0^2 s\,dB_s-\int_0^2 0\,dB_s\right)^2\right] =\mathbb{E}\int_0^2 s^2\,ds =\int_0^2 s^2\,ds =\left[\frac{s^3}{3}\right]_0^2 =\frac{8}{3}. $$ The mean square is positive, so the two terminal integrals cannot be equal almost surely.
Check: the integrands differ on the entire interval except at \(s=0\). Their squared difference has integral \(8/3\), exactly as the isometry predicts. Also, the stochastic integral has mean zero and variance \(8/3\), because it is square-integrable and its second moment is given by the isometry.
Worked Example: Changing an Integrand at One Time
On \([0,1]\), let \(H_s=0\) for every \(s\), and let \(K_s=\mathbf{1}_{\{1/2\}}(s)\). These deterministic processes are progressively measurable and square-integrable. They differ at exactly one time. Show that their integral processes agree.
The squared difference has zero time integral: $$ \mathbb{E}\int_0^1 (H_s-K_s)^2\,ds =\int_0^1\mathbf{1}_{\{1/2\}}(s)\,ds =0, $$ because a singleton has Lebesgue measure zero. For each \(t\in[0,1]\), the isometry gives $$ \mathbb{E}\left[\left(\int_0^t (H_s-K_s)\,dB_s\right)^2\right] =\int_0^t\mathbf{1}_{\{1/2\}}(s)\,ds =0. $$ Therefore the integrals agree almost surely at each fixed \(t\). Since their integral versions have continuous paths, they are indistinguishable.
Check: the integral of \(H\) is identically zero. The theorem shows that the integral of \(K\) is also the zero process, even though \(K_{1/2}=1\). A pointwise value at one time carries no weight in the time integral that defines the Itô isometry.
Using Uniqueness in a Representation
A Brownian martingale representation writes a random variable or martingale using a stochastic integral. For uniqueness, the key issue is equality of the represented quantities, not how the representations were found. If two candidate integrands are in \(\mathcal{L}^2\) and their terminal integrals represent the same random variable almost surely, the theorem forces them to agree \(dt\otimes d\mathbb{P}\)-almost everywhere. This is uniqueness in the natural equivalence class of integrands, not pointwise uniqueness.
Proof. Subtract the two process identities. Indistinguishability gives \(\int_0^T(H_s-K_s)\,dB_s=0\) almost surely. Applying the Itô isometry as in the theorem yields \(\mathbb{E}\int_0^T(H_s-K_s)^2\,ds=0\), which is exactly equality of the integrands \(dt\otimes d\mathbb{P}\)-almost everywhere. \(\square\)
Worked Example: The Integrand in a Centered Brownian-Square Representation
For fixed \(T>0\), identify the integrand in a representation of \(B_T^2-T\), and use the theorem to test the proposed alternative integrand \(K_s=2B_s+1\).
Apply Itô's formula to \(f(x)=x^2\). Its derivatives are \(f'(x)=2x\) and \(f''(x)=2\), so the correction term is \(\tfrac12 f''(B_s)\,ds=ds\). Since \(dB_s\,dB_s=ds\), Itô's formula gives $$ d(B_s^2)=2B_s\,dB_s+\frac12(2)\,ds =2B_s\,dB_s+ds. $$ Integrating from \(0\) to \(T\), using \(B_0=0\), gives $$ B_T^2-T=\int_0^T 2B_s\,dB_s. $$ The integrand \(H_s=2B_s\) is in \(\mathcal{L}^2\), since $$ \mathbb{E}\int_0^T (2B_s)^2\,ds =4\int_0^T\mathbb{E}[B_s^2]\,ds =4\int_0^T s\,ds =2T^2<\infty. $$
The proposed integrand differs from \(H\) by \(1\) at every time. The isometry therefore yields $$ \mathbb{E}\left[\left(\int_0^T(2B_s+1)\,dB_s-\int_0^T2B_s\,dB_s\right)^2\right] =\mathbb{E}\int_0^T 1^2\,ds=T. $$ Since \(T>0\), the mean square difference is positive, so \(K\) cannot represent the same terminal variable.
Independent check: the difference of the two integrals is \(\int_0^T1\,dB_s=B_T\), whose second moment is \(\mathbb{E}[B_T^2]=T\), in agreement with the isometry. Also, \(B_T^2-T\) has mean zero and second moment \(2T^2\): using \(\mathbb{E}[B_T^4]=3T^2\), its second moment is \(3T^2-2T^2+T^2=2T^2\). The isometry for its representation gives \(4\int_0^T s\,ds=2T^2\), the same value.
Common Pitfalls
Do not infer pointwise equality. The isometry proves equality almost everywhere for the product measure \(dt\otimes d\mathbb{P}\); it does not prove \(H_t(\omega)=K_t(\omega)\) for every \(t\) and every outcome. The one-time modification above is a concrete counterexample to that stronger claim.
Do not confuse terminal equality with an unproved pathwise assertion. Equality of terminal integrals almost surely is enough to infer equality of integrands almost everywhere. Conversely, equality of integrands almost everywhere gives equality of integral processes indistinguishably, but this last step uses continuity and a countable dense set of times; equality separately at each time is not itself the definition of indistinguishability.
Do not use adaptedness as a substitute for Brownian motion relative to the filtration. The conditional mean-zero increment argument requires future Brownian increments to be independent of the current sigma-field. If the filtration contains information about future Brownian values, adaptedness alone does not provide this property or justify the stated Itô-integral framework.
| Claim being compared | What it implies | Measure of equality |
|---|---|---|
| Equal terminal integrals almost surely | Integrands agree almost everywhere | \(dt\otimes d\mathbb{P}\) |
| Integrands agree almost everywhere | Integral processes are indistinguishable | One probability-one event for all times |
| Integrands agree at every time and outcome | More than uniqueness requires | Pointwise equality |
Check Your Understanding
Use the Itô isometry and the distinction between pointwise and almost-everywhere equality to answer these questions.
- If two square-integrable integrands have equal terminal integrals almost surely, what quantity does the Itô isometry show is zero?
- Why does equality \(dt\otimes d\mathbb{P}\)-almost everywhere imply indistinguishability of the continuous integral processes?
- On \([0,3]\), compute the mean square of \(\int_0^3 2\,dB_s\) using the isometry.
- Can integrands that differ at one deterministic time have different Itô integrals? Explain using the time integral in the isometry.
- Why is Brownian motion relative to the filtration a necessary assumption for the martingale argument?