From Stopped Integrals to One Process
For an integrand that is only locally square-integrable, an Ito integral is first constructed after stopping the integrand at increasing energy levels. Each stopped integral is well behaved on its own: it is square-integrable and has a continuous adapted version. The construction still needs a pathwise step. We must choose versions that agree with one another, then define a single process whose paths are continuous on every bounded time interval.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Assume the filtration satisfies the usual conditions. Let \(H_t\) be progressively measurable and locally square-integrable, meaning that, almost surely, \(\int_0^T H_s^2\,ds<\infty\) for every finite \(T\). For a fixed \(T\), the class \(\mathcal{L}^2\) consists of progressively measurable integrands \(G\) satisfying \(\mathbb{E}\int_0^T G_s^2\,ds<\infty\). The Ito integral on \([0,T]\) is defined for integrands in this class.
The stopping times are nondecreasing, and \(\tau_n\to\infty\) almost surely. Indeed, on any path with finite \(A_T\) for every finite \(T\), a fixed finite \(T\) cannot satisfy \(A_T\ge n\) for arbitrarily large \(n\). Also, continuity of \(A\) gives \(\int_0^{t\wedge\tau_n}H_s^2\,ds\le n\). Consequently the stopped integrand belongs to \(\mathcal{L}^2\) on every bounded interval, and in fact its expected total energy is at most \(n\).
Why the Stopped Versions Are Compatible
For \(m\ge n\), the stopped integrands agree up to \(\tau_n\): stopping the \(m\)th one again at \(\tau_n\) gives the \(n\)th stopped integrand, up to an endpoint that has zero Lebesgue measure. The stopping identity for Ito integrals therefore gives \(I^m_{t\wedge\tau_n}=I^n_{t\wedge\tau_n}\) for all \(t\), up to indistinguishability. Here indistinguishability means equality for every time outside one null event, rather than equality almost surely separately at each fixed time.
Each stopped integral has an adapted continuous version because its integrand is in \(\mathcal{L}^2\) on bounded intervals. The Ito isometry gives finite second moments, and the martingale property follows first for adapted step integrands from the zero conditional mean of Brownian increments, then for \(\mathcal{L}^2\) integrands by approximation in \(L^2\). The continuous-version result for \(L^2\) Ito integrals supplies the continuous paths.
Proof. For each \(n\), choose a continuous version of \(I^n\); the event that this version has continuous paths has probability one. For every pair \(m\ge n\), the stopping identity gives indistinguishable equality of \(I^m_{t\wedge\tau_n}\) and \(I^n_{t\wedge\tau_n}\), so this equality also holds on an event of probability one. There are only countably many indices and pairs. Intersecting these probability-one events with the probability-one event on which \(\tau_n\to\infty\) still gives an event of probability one. On that intersection, every stated property holds at once. \(\square\)
Defining the Integral at Each Time
Compatibility lets us use any stopped integral whose stopping time has not yet occurred. To make this choice precise even on exceptional paths, define \(N_t\) to be the least positive integer \(n\) for which \(\tau_n\ge t\), if such an integer exists. Set \(I_t=I^{N_t}_t\) when \(N_t\) exists, and \(I_t=0\) otherwise. Since \(\tau_n\to\infty\) almost surely, the fallback value is needed only on a null event for each fixed \(t\).
Proof. First fix \(t\). The event \(\{N_t=n\}\) is \(\{\tau_n\ge t\}\cap\bigcap_{j<n}\{\tau_j<t\}\), which belongs to \(\mathcal{F}_t\), since \(\tau_j\) are stopping times. Each \(I^n_t\) is \(\mathcal{F}_t\)-measurable. Thus \(I_t\), which equals \(I^n_t\) on \(\{N_t=n\}\) and equals zero if there is no such \(n\), is \(\mathcal{F}_t\)-measurable. This proves adaptedness.
Now take a path on the probability-one event from the Proposition. For any \(T<\infty\), choose \(n\) with \(\tau_n>T\); such an \(n\) exists because \(\tau_n\to\infty\). For each \(t\in[0,T]\), both the index selected in the definition of \(I_t\) and \(n\) have stopping times at least \(t\). Compatibility implies that the corresponding stopped integral values at \(t\) agree. Hence \(I_t=I^n_t\) for every \(t\in[0,T]\) on this path. Since \(I^n\) is continuous, \(I\) is continuous on \([0,T]\). The same probability-one event works for every finite \(T\); no new exceptional event is chosen for each interval. Finally, the same compatibility argument gives \(I_t=I^n_t\) whenever \(t\le\tau_n\). \(\square\)
Worked Examples of the Construction
Worked Example: Constant Integrand
Let \(H_t=1\). Its accumulated energy is \(A_t=\int_0^t1\,ds=t\), so the stopping time at level \(n\) is \(\tau_n=n\). The stopped integral is \(I^n_t=B_{t\wedge n}\), since the Ito integral of the constant integrand over \([0,t\wedge n]\) is the Brownian increment \(B_{t\wedge n}-B_0\), and \(B_0=0\).
At a fixed time \(t\), any \(n\ge t\) gives \(I^n_t=B_t\). Thus the assembled process is \(I_t=B_t\). Check: for \(t\le n\), \(B_{t\wedge n}=B_t\), and Brownian paths are continuous, so the assembled process has the required continuous paths.
Worked Example: A Time-Dependent Integrand
Let \(H_t=t\). Then \(A_t=\int_0^t s^2\,ds=t^3/3\), and solving \(t^3/3=n\) gives \(\tau_n=(3n)^{1/3}\). The stopped integrand has total energy \(\int_0^{\tau_n}s^2\,ds=n\), so its expected energy is exactly \(n\). The stopped Ito integral is \(I^n_t=\int_0^{t\wedge(3n)^{1/3}}s\,dB_s\).
For any fixed \(T\), choose an integer \(n>T^3/3\). Then \(\tau_n>T\), so the assembled process on \([0,T]\) is the continuous version of \(\int_0^t s\,dB_s\) there. Check: if \(T=1\), then \(A_1=1/3\); choosing \(n=1\) gives \(\tau_1=3^{1/3}>1\), so the first stopped integral already agrees with the assembled process throughout \([0,1]\).
Worked Example: Finite Pathwise Energy Without Finite Expected Energy
Let \(H_t=e^{B_t^2}\). Brownian paths are continuous, so \(H_t\) is continuous and bounded on every compact time interval along each path. It follows that \(\int_0^T H_s^2\,ds<\infty\) almost surely for each finite \(T\), and the energy stopping times tend to infinity almost surely.
However, for \(s>0\), the normal density of \(B_s\) gives \(\mathbb{E}[e^{2B_s^2}]=(2\pi s)^{-1/2}\int_{\mathbb{R}}\exp((2-1/(2s))x^2)\,dx\). This integral is infinite when \(s\ge1/4\), because its exponent has a nonnegative quadratic coefficient and the integrand does not decay at infinity. Therefore for \(T>1/4\), \(\mathbb{E}\int_0^T H_s^2\,ds=\infty\), by Tonelli's theorem. Check: for \(T=1/8\), every \(s\in(0,T]\) has \(2-1/(2s)<0\), so the Gaussian integral is finite; for \(T=1/2\), it is infinite for every \(s\in[1/4,1/2]\), an interval of positive length.
This example separates pathwise local square-integrability from membership in \(\mathcal{L}^2\) on a long interval. The localized construction applies even though the unstopped integrand need not have finite expected energy on that interval.
Martingale Status and a Common Pitfall
For each \(n\), the stopped process \(I^n\) is a square-integrable martingale: its integrand has expected total energy at most \(n\), and for \(s\le t\) the conditional expectation of the integral over \((s,t]\) given \(\mathcal{F}_s\) is zero. The construction also gives \(I_{t\wedge\tau_n}=I^n_{t\wedge\tau_n}\), and \(\tau_n\to\infty\) almost surely. Thus the constructed process is a local martingale, by the definition using a sequence of stopping times whose stopped processes are martingales.
This does not prove that \(I\) is a true martingale. The argument gives square-integrability for each stopped process, but it does not give a global bound \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) for the unstopped integrand on every interval under consideration. Without such a bound, the direct \(L^2\) argument for the unstopped integral and its conditional expectations is unavailable. A local martingale need not be a true martingale; the localization construction alone does not justify removing the stopping in an expectation identity.
Common mistake. It is incorrect to say, “Each \(I^n\) is a martingale, therefore their pointwise limit \(I\) is a martingale.” The construction does not establish that \(I^n_t\) converges to \(I_t\) in \(L^1\) for a fixed \(t\), or that the family is uniformly integrable. Instead, it establishes the local martingale property through the stopped processes \(I_{t\wedge\tau_n}\). This distinction preserves exactly what the stopping argument proves, without asserting an unverified true-martingale property.
The essential point is simultaneous consistency: one countable intersection gives a single event on which the stopped versions fit together at all times. On each bounded interval, some stopped version remains unstopped throughout that interval, so its continuous paths supply the paths of the assembled process there.
Check Your Understanding
For each question, distinguish statements that hold pathwise from those that require an expectation bound.
- Why does \(\tau_n\to\infty\) almost surely when \(\int_0^T H_s^2\,ds<\infty\) almost surely for every finite \(T\)?
- For \(H_t=t\), compute \(A_T\) and determine a condition on \(n\) that ensures \(\tau_n>T\).
- Why is the selected value \(I_t\) measurable with respect to \(\mathcal{F}_t\), even though the selected index depends on the stopping times?
- Explain why continuity on every bounded interval follows on one probability-one event rather than requiring a separate event for each \(T\).
- Which additional integrability property is missing if one wants to use the direct \(L^2\) argument to establish that the unstopped integral is a true martingale on \([0,T]\)?