The Integral Is a Continuous Local Martingale

From Local Square Integrability to a Local Martingale

For an Ito integral to be a martingale by the direct square-integrability argument, its integrand must have finite expected energy on the time interval under consideration. A locally square-integrable integrand has a weaker property: its accumulated squared size is finite almost surely on each finite interval, but its expectation may be infinite. Stopping when that accumulated energy reaches a fixed level turns the integrand into a square-integrable one. The resulting stopped integrals are martingales, which is exactly the structure needed to establish the local martingale property.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) with respect to this filtration. In particular, Brownian increments after time \(s\) are independent of \(\mathcal{F}_s\). For a finite \(T\), an Ito integrand in \(\mathcal{L}^2\) is a progressively measurable process \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Its Ito integral has the continuous version established in the earlier material on Ito integrals.

The Square-Integrable Case

Proof. The Ito isometry gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds<\infty\), so each \(M_t\) is integrable. To check the conditional expectation, first take an adapted step-process integrand. Its integral over \((s,t]\) is a sum of terms of the form \(\xi(B_v-B_u)\), with \(s\le u<v\le t\) and \(\xi\) measurable at the beginning of the corresponding step. Brownian increments have conditional mean zero given the information at their starting time. Conditioning first at that time and then on \(\mathcal{F}_s\) shows that each term has conditional mean zero given \(\mathcal{F}_s\). Thus \(\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\) for step integrands. The Ito integral for \(H\) is obtained by approximation in the norm \(\big(\mathbb{E}\int_0^T H_s^2\,ds\big)^{1/2}\); the Ito isometry makes the corresponding integrals converge in \(L^2(\mathbb{P})\). Conditional expectation is a contraction in \(L^2\), so the conditional-mean-zero identity passes to the limit. Hence \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). \(\square\)

The proposition gives both ingredients of the martingale property: finite second moments and the conditional expectation identity. The second-moment bound alone would not establish the latter.

Stopping at an Accumulated-Energy Level

Now suppose \(H\) is progressively measurable and locally square-integrable, meaning that for every finite \(T\), \(\int_0^T H_s^2\,ds<\infty\) almost surely; its expected energy need not be finite. Define the accumulated energy \(A_t=\int_0^t H_s^2\,ds\). Almost surely, \(A_t\) is finite and continuous on every finite time interval. For each positive integer \(n\), stop at the first time that this energy reaches \(n\): \(\tau_n=\inf\{t\ge0:A_t\ge n\}\), with \(\inf\varnothing=\infty\).

Proof. Fix \(n\) and \(T<\infty\). The process \(H_s\mathbf{1}_{\{s\le\tau_n\}}\) is progressively measurable. Its energy up to \(T\) is \(\int_0^{T\wedge\tau_n}H_s^2\,ds=A_{T\wedge\tau_n}\le n\) almost surely: continuity of \(A\) means that it reaches \(n\) at the stopping time, if it reaches it at all. Thus the expected energy of this stopped integrand is at most \(n\), so it belongs to \(\mathcal{L}^2\) on \([0,T]\). The stopping identity for Ito integrals gives \(\int_0^t H_s\mathbf{1}_{\{s\le\tau_n\}}\,dB_s=M_{t\wedge\tau_n}\). Applying the square-integrable proposition proves that \(M_{t\wedge\tau_n}\) is a square-integrable martingale on \([0,T]\). Finally, \(A_T<\infty\) almost surely, so for each finite \(T\), all integers \(n>A_T\) satisfy \(\tau_n>T\). Hence \(\tau_n\) increases to infinity almost surely. The process \(M\) is adapted and continuous, and its stopped processes are martingales; therefore it is a continuous local martingale. \(\square\)

Here, the statement that the stopped processes are square-integrable is a statement on each finite interval \([0,T]\). The energy bound is pathwise, \(A_{T\wedge\tau_n}\le n\), and taking expectations preserves it. The conclusion does not require \(\mathbb{E}A_T<\infty\) for the original, unstopped integrand.

What Localization Does—and Does Not—Show

The proof works because stopping controls the quantity that enters the Ito isometry: the accumulated squared integrand. Each fixed energy threshold gives a square-integrable martingale, and the thresholds increase so that on any finite interval the stopping eventually has no effect. Continuity of the original integral is inherited from its continuous version; localization is used to prove the martingale property, not to replace continuity with a limit argument.

A common mistake is to infer a martingale identity for the unstopped process directly from the stopped identities. For example, one might write \(\mathbb{E}[M_t]=\mathbb{E}[M_{t\wedge\tau_n}]=0\) and then conclude \(\mathbb{E}[M_t]=0\). The displayed equality is not justified merely by knowing that \(M_{t\wedge\tau_n}\) is a martingale: it would require an additional argument connecting the stopped and unstopped expectations. In general, localization proves that \(M\) is a local martingale, not that it is a true martingale. Conditions that guarantee the stronger conclusion are a separate question.

Also keep the roles of the two coefficients in mind when an integral appears as part of an SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\): \(\mu\) is the drift coefficient, whereas \(\sigma\) is the diffusion coefficient and plays the role of the stochastic integrand when evaluated along \(X_t\). The result here applies to that stochastic integral when its integrand is progressively measurable and locally square-integrable; it makes no claim about the drift integral.