Why the Choice of Localizer Must Not Matter
When an adapted integrand is not square-integrable in expectation over a whole time interval, its Ito integral can still be constructed by stopping it where its accumulated size is controlled. This construction uses a sequence of stopping times. But there may be many sequences that make the stopped integrands square-integrable. The resulting integral should describe the integrand, not the particular sequence used to define it.
We prove that independence by comparing two constructions on intervals where both are valid. The comparison has three ingredients: a common smaller stopping time, the stopping identity for Ito integrals, and uniqueness of the L2 Ito integral. We work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration.
For each finite horizon \(T\), the L2 Ito integral is unique up to almost-sure equality at each time. Its isometry says that the expected squared difference of two such integrals is the expected integral of the squared difference of their integrands. We also use the stopping identity: stopping an Ito integral at a stopping time \(\rho\) gives the integral with its integrand stopped at \(\rho\). These are standard results from the earlier material on Ito integrals. Continuous versions of the L2 integral processes are used when we compare whole paths.
The Common-Refinement Argument
Proof. Since \(\rho_{n,m}\le\tau_n\), we have \(\mathbf{1}_{\{s\le\rho_{n,m}\}}\le\mathbf{1}_{\{s\le\tau_n\}}\). Therefore $$ \mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\le\rho_{n,m}\}}\,ds \le \mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\le\tau_n\}}\,ds<\infty. $$ The stopped integrand is progressively measurable, so its L2 Ito integral is defined. Write \(J^n_t=\int_0^t H_s\mathbf{1}_{\{s\le\tau_n\}}\,dB_s\) and \(K^m_t=\int_0^t H_s\mathbf{1}_{\{s\le\sigma_m\}}\,dB_s\). By the stopping identity and \(\rho_{n,m}\le\tau_n,\sigma_m\), $$ J^n_{t\wedge\rho_{n,m}} = \int_0^t H_s\mathbf{1}_{\{s\le\rho_{n,m}\}}\,dB_s = K^m_{t\wedge\rho_{n,m}}. $$ The equalities hold at each time by uniqueness of the L2 Ito integral. Taking continuous versions, they hold simultaneously for all times: they hold almost surely at every nonnegative rational time, and two continuous paths agreeing on the rationals agree everywhere. \(\square\)
Proof. A patched process is defined by using \(J^n_t\) whenever \(t\le\tau_n\); the localizing property ensures that for each fixed \(t\), such an \(n\) exists almost surely. The lemma gives a consistent definition on overlaps: if \(t\le\tau_n\wedge\tau_k\), applying the lemma with the two stopping times \(\tau_n,\tau_k\) shows \(J^n_t=J^k_t\) almost surely. The same consistency holds for the sequence \((\sigma_m)\).
For each pair \(n,m\), the lemma gives equality of the two stopped integral processes up to \(\rho_{n,m}\), outside a null event. There are only countably many pairs, so we can exclude the union of these null events and retain a single full-probability event on which all these equalities hold. Also restrict to the full-probability event where both localizing sequences tend to infinity and all the stopped L2 integral versions have continuous paths.
Fix an outcome in this event and a finite time \(t\). Since \(\tau_n\to\infty\) and \(\sigma_m\to\infty\), choose \(n,m\) so that \(t\le\tau_n\) and \(t\le\sigma_m\) for this outcome. Then \(t\le\rho_{n,m}\), and the lemma gives \(J^n_t=K^m_t\). By consistency, these are exactly the respective patched values at \(t\). The argument applies to every \(t\) on the same event, so the patched processes are indistinguishable. \(\square\)
The proof uses only local comparisons. It does not require either full, unstopped integrand to have finite expected squared integral. The common refinement keeps the integrand inside both regions where the L2 construction is available.
Worked Examples of the Comparison
Worked Example: The Constant Integrand
Take \(H_s=1\), with localizing sequences \(\tau_n=n\) and \(\sigma_m=2m\). For each finite \(T\), both stopped integrands are progressively measurable and square-integrable in expectation: $$ \mathbb{E}\int_0^T \mathbf{1}_{\{s\le n\}}\,ds=T\wedge n, \qquad \mathbb{E}\int_0^T \mathbf{1}_{\{s\le 2m\}}\,ds=T\wedge(2m). $$ The corresponding integrals are \(J^n_t=B_{t\wedge n}\) and \(K^m_t=B_{t\wedge 2m}\), because the Ito integral of the indicator of an interval from zero to a stopping time is the stopped Brownian motion.
The common refinement is \(\rho_{n,m}=n\wedge 2m\), so both stopped processes equal \(B_{t\wedge\rho_{n,m}}\). For example, with \(n=2\), \(m=1\), and \(t=3\), the common stopping time is \(2\), and both values are \(B_{3\wedge2}=B_2\). As \(n\) and \(m\) tend to infinity, each patched process at any fixed finite \(t\) is \(B_t\). Check: if \(t\le n\wedge2m\), both formulas reduce to \(B_t\); if \(t>n\wedge2m\), the comparison is made only after stopping both at \(n\wedge2m\), where both equal \(B_{n\wedge2m}\).
Worked Example: A Deterministic Time-Dependent Integrand
Let \(H_s=2s\), and use \(\tau_n=n\) and \(\sigma_m=3m\). The L2 conditions on \([0,T]\) follow from direct integration: $$ \mathbb{E}\int_0^T (2s)^2\mathbf{1}_{\{s\le n\}}\,ds =4\int_0^{T\wedge n}s^2\,ds =\frac{4(T\wedge n)^3}{3}<\infty. $$ The same calculation with \(3m\) in place of \(n\) gives \(4(T\wedge3m)^3/3<\infty\).
For a deterministic cutoff \(c\), Ito's formula applied to \(f(t,x)=tx\) gives \(f_t=x\), \(f_x=t\), \(f_{xx}=0\), and hence \(d(tB_t)=B_t\,dt+t\,dB_t+\tfrac12(0)\,dt\). Rearranging and integrating, $$ \int_0^c 2s\,dB_s=2cB_c-2\int_0^c B_s\,ds. $$ The variance, checked independently by the Ito isometry, is \(4\int_0^c s^2\,ds=4c^3/3\). For the common cutoff \(c=1\), the integral is \(2B_1-2\int_0^1B_s\,ds\) and its variance is \(4/3\). Both sequences give this same expression once their cutoffs exceed \(1\). Check: the boundary term is zero at \(c=0\), and the variance formula also gives zero there, as required.
Worked Example: A Brownian-Dependent Integrand
Let \(H_s=B_s\). One possible localizing sequence is \(\tau_n=n\wedge\inf\{t\ge0:|B_t|\ge n\}\); another is \(\sigma_m=2m\wedge\inf\{t\ge0:|B_t|\ge2m\}\). Both increase to infinity almost surely: on each finite time interval a continuous Brownian path is bounded, so sufficiently high levels are not reached there, and the deterministic caps tend to infinity.
Before \(\tau_n\), \(|B_s|\le n\), so for every finite \(T\), $$ \mathbb{E}\int_0^T B_s^2\mathbf{1}_{\{s\le\tau_n\}}\,ds\le n^2T<\infty. $$ Before \(\sigma_m\), the corresponding bound is \((2m)^2T<\infty\). Thus both stopped integrands are in the L2 class. Set \(\rho=\tau_n\wedge\sigma_m\). Ito's formula for \(f(x)=x^2\) has \(f'(x)=2x\), \(f''(x)=2\), and correction \(\tfrac12 f''(x)\,dt=dt\). It yields $$ \int_0^{t\wedge\rho} B_s\,dB_s =\frac12\left(B_{t\wedge\rho}^2-(t\wedge\rho)\right). $$ Both localizing sequences therefore give this same stopped value on their common interval. Check: at \(t=0\), the right side is \(\tfrac12(B_0^2-0)=0\), matching the integral from \(0\) to \(0\).
Why This Independence Matters
The theorem makes the phrase “the local Ito integral” unambiguous. A localizing sequence is a tool for constructing the object, not extra data that become part of the answer. This matters whenever one later proves a property by choosing a convenient sequence—for instance, one that stops when an energy or a process level crosses a threshold. A different admissible sequence can be used for another argument without changing the integral.
There is an important limit to what the comparison says. Each stopped integral is an L2 Ito integral, but the patched integral need not have finite second moment on an entire fixed time interval. Thus the comparison establishes equality of the constructed processes; it does not assert that the unstopped integral is itself an L2 integral. Nor does it by itself show that the patched process is a true martingale. Those properties require their own hypotheses and arguments.
Common mistake. It is not valid to compare \(J^n_t\) and \(K^m_t\) at a time \(t\) after one of the localizers has stopped and then claim that the unstopped formulas must agree. For instance, in the constant-integrand example, \(J^2_3=B_2\) while \(K^2_3=B_3\); these are not generally equal. The correction is to compare only up to the common stopping time: here \(\rho_{2,2}=2\), and \(J^2_{3\wedge2}=K^2_{3\wedge2}=B_2\). Once both sequences have reached a given time, the patched values agree there.
Check Your Understanding
Use the common-refinement proof to answer the following questions.
- For two localizing sequences, why is the minimum of one stopping time from each sequence still sufficient to make the stopped integrand square-integrable?
- State the stopping identity used to compare the two stopped Ito integrals.
- Why does equality at each fixed time not, by itself, establish indistinguishability of two processes?
- For \(H_s=2s\), compute the variance of \(\int_0^c 2s\,dB_s\) from the Ito isometry.
- In the constant-integrand example, why is it incorrect to compare \(B_{t\wedge n}\) and \(B_{t\wedge2m}\) after their common stopping time?