Why Localization Builds the Integral in Stages
The Ito integral is first constructed for progressively measurable integrands whose expected squared energy is finite. A locally square-integrable integrand need not satisfy that condition on the whole time interval: its energy may become unbounded as time approaches a terminal point. Localization handles this by stopping before the problematic region. Each stopped integrand has finite expected energy, so its ordinary \(L^2\) Ito integral exists. The construction then patches these integrals together, using their agreement on the time intervals where both are defined.
Fix a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix \(T>0\). All stochastic integrals below are Ito integrals. An integrand \(H\) is assumed progressively measurable. For an ordinary \(L^2\) Ito integral on \([0,T]\), we require \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); the class of such integrands is denoted \(\mathcal{L}^2\). We use the Ito isometry and the stopping identity for stochastic integrals established in the earlier material.
For each \(n\), the stopped integrand \(H_s\mathbf{1}_{\{s\le\tau_n\}}\) belongs to \(\mathcal{L}^2\) on \([0,T]\), since its expected squared energy is \(\mathbb{E}\int_0^{\tau_n}H_s^2\,ds\). Define its ordinary Ito integral process by $$ I^{(n)}_t=\int_0^t H_s\mathbf{1}_{\{s\le\tau_n\}}\,dB_s,\qquad 0\le t\le T. $$ Equivalently, by the stopping identity, \(I^{(n)}_t=\int_0^{t\wedge\tau_n}H_s\,dB_s\). The choice of whether to include the single time \(s=\tau_n\) does not affect the integral.
Agreement on Overlapping Stopped Intervals
Proof. Since \(\tau_n\le\tau_m\), stopping the level-\(m\) integral at \(\tau_n\) gives, by the stopping identity, $$ I^{(m)}_{t\wedge\tau_n} =\int_0^{t\wedge\tau_n}H_s\mathbf{1}_{\{s\le\tau_m\}}\,dB_s. $$ Up to the stopping time \(\tau_n\), the indicator \(\mathbf{1}_{\{s\le\tau_m\}}\) equals one, except possibly at an endpoint that does not change the integral. The right side is therefore \(\int_0^{t\wedge\tau_n}H_s\,dB_s=I^{(n)}_t\), again by the stopping identity. On \(\{t\le\tau_n\}\), \(t\wedge\tau_n=t\), giving the stated agreement there. \(\square\)
The Patched Integral Before the Limiting Time
Proof. Fix \(t<T\). Since \(\tau_n\uparrow T\) almost surely, for almost every \(\omega\) the sequence \(\tau_n(\omega)\) eventually exceeds \(t\): its limit \(T\) is strictly greater than \(t\). Thus the set of indices \(n\) with \(t\le\tau_n(\omega)\) is nonempty almost surely. Define \(N(\omega)\) to be its first index on this event, and set \(M_t=I^{(N)}_t\) there; set \(M_t=0\) on the null event where no such index exists. If two indices \(n\le m\) both satisfy \(t\le\tau_n\), the lemma gives \(I^{(m)}_t=I^{(n)}_t\) almost surely. Hence the selected value is unchanged if a later covering index is used. Measurability also follows from the countable patch: the events \(\{t\le\tau_n\}\) are in \(\mathcal{F}_t\), and each \(I^{(n)}_t\) is \(\mathcal{F}_t\)-measurable. This defines the patched value for every fixed \(t<T\). The argument does not cover \(t=T\), because \(\tau_n\uparrow T\) need not imply \(\tau_n=T\) for any \(n\). \(\square\)
This theorem constructs the integral from the ordinary \(L^2\) integrals, rather than taking a limit of their terminal values. At a fixed time \(t<T\), eventually the stopped intervals reach \(t\), and the lemma ensures that all later choices give the same value. No assertion that the resulting process has continuous paths is needed for this pointwise construction.
Worked Example: A Deterministic Singularity and a Missing Endpoint
On \([0,1]\), take \(H_s=(1-s)^{-1/2}\) for \(s<1\), and choose any value for \(H_1\). Let \(\tau_n=1-e^{-n}\), for integers \(n\ge1\). These are deterministic stopping times increasing to \(1\), and every \(\tau_n<1\).
The expected stopped energy is deterministic, so direct integration gives $$ \mathbb{E}\int_0^{\tau_n}H_s^2\,ds =\int_0^{1-e^{-n}}\frac{1}{1-s}\,ds =-\log(1-(1-e^{-n}))=n. $$ Thus each stopped integral is an ordinary \(L^2\) Ito integral. For each \(t<1\), some \(\tau_n\) reaches \(t\), and the patched value is the deterministic Wiener integral \(\int_0^t(1-s)^{-1/2}\,dB_s\). Its variance, by the Ito isometry, is \(\int_0^t(1-s)^{-1}\,ds=-\log(1-t)\).
At \(t=1\), however, no \(\tau_n\) reaches the endpoint. The stopped terminal values \(I^{(n)}_1\) are centered normal random variables with variance \(n\), because their deterministic integrands have squared integral \(n\). For any fixed \(K>0\), $$ \mathbb{P}(|I^{(n)}_1|\le K) =\Phi(K/\sqrt n)-\Phi(-K/\sqrt n)\longrightarrow0. $$ Consequently these terminal values cannot converge in probability to a finite random variable: convergence in probability to a finite random variable would make the sequence bounded in probability, contrary to the displayed limit. Check: substituting \(n=1\) into the energy gives \(\int_0^{1-e^{-1}}(1-s)^{-1}ds=1\), as required by the general value \(n\). The example shows both why the construction works for every fixed \(t<1\) and why it cannot be extended to \(t=1\) by the same argument.
Worked Example: A Random Amplitude with Infinite Total Energy
Let \(T=1\). Suppose \(\xi\) is \(\mathcal{F}_0\)-measurable, independent of \(B\), and has probabilities \(\mathbb{P}(\xi=k)=c/k^3\) for integers \(k\ge1\), where \(c=1/\sum_{j=1}^{\infty}j^{-3}\). Set \(H_s=\xi\), and define \(\tau_n=\mathbf{1}_{\{\xi\le n\}}\), taking values \(0\) and \(1\). These are stopping times because they are \(\mathcal{F}_0\)-measurable, they increase, and \(\tau_n\uparrow1\) almost surely since \(\xi\) is finite almost surely.
The stopped energy is $$ \mathbb{E}\int_0^{\tau_n}\xi^2\,ds =\mathbb{E}\big[\xi^2\mathbf{1}_{\{\xi\le n\}}\big] =c\sum_{k=1}^{n}\frac{k^2}{k^3} =c\sum_{k=1}^{n}\frac1k<\infty. $$ In contrast, the full expected energy is \(c\sum_{k=1}^{\infty}1/k=\infty\). The stopped integral is \(I^{(n)}_t=\xi\mathbf{1}_{\{\xi\le n\}}B_t\): on \(\{\xi\le n\}\) the stopping time is \(1\), and otherwise it is \(0\). For each fixed \(t\le1\), the event \(\{\xi\le n\}\) holds for all sufficiently large \(n\), so the patched value is \(\xi B_t\). This includes \(t=1\), because in this example the localizing sequence actually reaches \(1\) eventually on every outcome with finite \(\xi\).
Check: for \(n=2\), the stopped energy is \(c(1+\tfrac12)=\tfrac32c\), obtained from the two terms \(1/1\) and \(1/2\). It is finite, while the unrestricted harmonic series diverges. This example illustrates that finite expected energy is needed for each stopped integral, not necessarily for the unstopped integrand.
Worked Example: Patching a Step Integrand
On \([0,1]\), let \(H_s=2\) for \(s\le\tfrac12\) and \(H_s=-1\) for \(s>\tfrac12\). Use \(\tau_1=\tfrac12\) and \(\tau_n=1\) for \(n\ge2\). The sequence is nondecreasing and converges to \(1\). The first stopped energy is \(4\cdot\tfrac12=2\); the full energy is \(4\cdot\tfrac12+1\cdot\tfrac12=\tfrac52\), so both stopped integrals are in \(\mathcal{L}^2\).
For \(t\le\tfrac12\), the first integral is \(I^{(1)}_t=2B_t\). For \(t>\tfrac12\), the second integral is $$ I^{(2)}_t=2B_{1/2}-(B_t-B_{1/2}). $$ At the overlap endpoint, \(I^{(2)}_{1/2}=2B_{1/2}=I^{(1)}_{1/2}\). For instance, at \(t=\tfrac14\) the patch uses \(I^{(1)}_{1/4}=2B_{1/4}\), while at \(t=\tfrac34\) it uses \(I^{(2)}_{3/4}=2B_{1/2}-(B_{3/4}-B_{1/2})\). Check: substituting \(t=\tfrac12\) in the second formula makes its increment term \(B_{1/2}-B_{1/2}=0\), so the two definitions agree exactly where the stopped intervals meet.
What Localization Does Not Guarantee
Common mistake: treating a limiting stopping time as though it were attained. A wrong endpoint argument says, “Since \(\tau_n\uparrow T\), choose \(n\) with \(T\le\tau_n\), then define \(M_T=I^{(n)}_T\).” The choice need not exist. In the singular-integrand example, \(\tau_n=1-e^{-n}<1=T\) for every \(n\), so the event \(\{T\le\tau_n\text{ for some }n\}\) is empty. The correction is to use the patching argument only at fixed \(t<T\), where eventual coverage does follow, and to state an additional condition for the endpoint.
One sufficient endpoint condition is that \(\tau_n=T\) eventually almost surely. Then the same patching argument defines \(M_T\), since at least one stopped integral covers \(T\) on almost every outcome. Another possibility is to prove separately that the patched values have a finite terminal limit in an appropriate sense and use that limit to define \(M_T\). Neither condition follows just from \(\tau_n\uparrow T\).
The construction is useful because it separates two questions: whether the integrand has finite energy on each stopped interval, and whether it has finite energy over the entire horizon. The first question is enough to define consistent integral values before the limiting time. It does not, by itself, imply that the unstopped integral is square-integrable, that terminal values converge, or that expectations can be passed through a limit. Those require additional hypotheses and arguments.
Check Your Understanding
For each question, distinguish carefully between times strictly before the limiting time and the endpoint.
- Why does \(\tau_n\uparrow T\) ensure that some \(\tau_n\) reaches a fixed \(t<T\), but not necessarily \(T\)?
- For the integrand \(H_s=(1-s)^{-1/2}\) on \([0,1)\), compute the expected energy stopped at \(1-e^{-3}\).
- In the random-amplitude example, why is each stopped expected energy finite even though the full expected energy is infinite?
- State the stopping-identity fact that ensures two stopped integrals agree on their common interval.
- Give a sufficient additional condition on a localizing sequence that allows the same patching argument to define the integral at \(T\).