Accumulated Energy as the Admissibility Test
For an Ito integral on a fixed time interval, the familiar square-integrability condition controls the expected accumulated squared size of its integrand. A broader class becomes relevant when that expectation is infinite, even though the accumulated squared size is finite along almost every sample path. This distinction is the meaning of “locally square-integrable” here: the word “local” refers to finite accumulated energy almost surely on each finite horizon, not to a bound on its expectation.
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. In particular, \(B\) has continuous paths and its increments are independent of the past. An integrand \(H\) in this tutorial is real-valued and predictable; predictability means measurability with respect to the sigma-field generated by adapted left-continuous processes. In particular, predictable processes are progressively measurable, the measurability needed for the time integrals below.
For \(H\in\mathcal{L}^2\) on \([0,T]\), the Ito integral \(\int_0^T H_s\,dB_s\) is defined as an \(L^2(\mathbb{P})\) limit, with the Ito isometry relating its second moment to \(\mathbb{E}A_T(H)\). Local square integrability does not assert that this expectation is finite. It identifies integrands whose energy is finite path by path almost surely, which allows the Ito integral to be treated through localization. The local condition alone does not claim that an integral has the unconditional expectation or martingale properties enjoyed by an \(L^2\) integral.
Testing the Condition on All Finite Horizons
Proof. If \(H\) is locally square-integrable, its defining condition applies to each finite integer horizon \(n\), so \(A_n(H)<\infty\) almost surely. Conversely, suppose this holds for every positive integer \(n\). Fix any finite \(T\ge0\), and choose a positive integer \(n\ge T\). Since the integrand is squared, its energy is nondecreasing in the horizon: $$ 0\le A_T(H)=\int_0^T H_s^2\,ds\le\int_0^n H_s^2\,ds=A_n(H). $$ Thus \(A_T(H)<\infty\) almost surely. The choice of \(T\) was arbitrary, proving the condition for every finite horizon. \(\square\)
The proposition is a countable-horizon test: it is enough to check the integer horizons, rather than intersecting uncountably many probability-one events. The energy need not have a common deterministic upper bound across paths. It only has to be finite almost surely at each finite horizon.
Worked Example: A Deterministic Singularity at the Initial Time
Let \(H_0=0\) and \(H_t=t^{-1/4}\) for \(t>0\). This deterministic Borel process is predictable. Compute its accumulated energy on \([0,T]\), for finite \(T\ge0\), and determine whether it is locally square-integrable and in \(\mathcal{L}^2\).
For \(T=0\), the integral is zero. For \(T>0\), \(H_t^2=t^{-1/2}\) for \(t>0\), so $$ A_T(H)=\int_0^T t^{-1/2}\,dt =\left[2\sqrt{t}\right]_{0}^{T} =2\sqrt{T}. $$ This is finite and deterministic for every finite \(T\), hence finite almost surely. Since its expectation is the same deterministic value, \(\mathbb{E}A_T(H)=2\sqrt{T}<\infty\); therefore \(H\) is also in \(\mathcal{L}^2\) on every finite horizon.
Check: at \(T=4\), the formula gives \(A_4(H)=2\sqrt{4}=4\). Integrating directly gives \(\int_0^4t^{-1/2}\,dt=2(2)-0=4\), in agreement. The value assigned at \(t=0\) does not affect this Lebesgue time integral.
Local Square Integrability Is Weaker Than \(\mathcal{L}^2\)
If \(H\in\mathcal{L}^2\) on \([0,T]\), then \(A_T(H)\) is finite almost surely. Indeed, it is a nonnegative random variable with finite expectation, and a nonnegative random variable that equals infinity on an event of positive probability has infinite expectation. The converse fails: a random variable can be finite almost surely while having infinite expectation. Accumulated energy can have exactly this behavior.
Proof. For the first assertion, let \(A_T(H)=\int_0^T H_s^2\,ds\) and suppose \(\mathbb{E}A_T(H)<\infty\). If \(\mathbb{P}(A_T(H)=\infty)>0\), then for every positive integer \(m\), \(A_T(H)\ge m\mathbf{1}_{\{A_T(H)=\infty\}}\), giving \(\mathbb{E}A_T(H)\ge m\mathbb{P}(A_T(H)=\infty)\). Letting \(m\) increase contradicts finite expectation. Hence \(A_T(H)<\infty\) almost surely. The failure of the converse is demonstrated by the next example, which satisfies local square integrability but has infinite expected energy. \(\square\)
Worked Example: A Finite Random Amplitude with Infinite Second Moment
Let \(U\) be uniform on \((0,1)\), independent of \(B\). Use a filtration that includes \(U\) at time zero and the Brownian history up to each time \(t\), completed in the usual way. Independence ensures that \(B\) remains Brownian relative to this filtration. Set \(A=U^{-1/2}\) and \(H_t=A\mathbf{1}_{[0,1]}(t)\). Since \(A\) is \(\mathcal{F}_0\)-measurable, this process is predictable. Determine its energy on any finite horizon and compare the two conditions.
For \(T\ge0\), the integrand is constant with value \(A\) from time \(0\) through time \(1\), and zero afterwards. Consequently, $$ A_T(H)=\int_0^T A^2\mathbf{1}_{[0,1]}(s)\,ds =A^2\min(T,1) =\frac{\min(T,1)}{U}. $$ Because \(U>0\) almost surely, this energy is finite almost surely for every finite \(T\). Thus \(H\) is locally square-integrable. If \(T>0\), however, then $$ \mathbb{E}A_T(H)=\min(T,1)\int_0^1\frac{1}{u}\,du=\infty. $$ The integral diverges since \(\int_\varepsilon^1u^{-1}\,du=-\log(\varepsilon)\) tends to infinity as \(\varepsilon\) decreases to zero. Hence \(H\) is not in \(\mathcal{L}^2\) on any horizon \(T>0\). At \(T=0\), its energy is zero.
Check: at \(T=2\), \(A_2(H)=1/U\), which is finite for every outcome with \(U>0\), an event of probability one. Its expectation is \(\int_0^1u^{-1}\,du=\infty\). This verifies directly that almost-sure finiteness and finite expected energy are different requirements.
Basic Closure Properties
Local square integrability is stable under finite linear combinations and multiplication by a bounded predictable process. The bound is pathwise; no moment estimate on the energy is needed.
Proof. Fix a finite \(T\). For real numbers \(x,y\), \((x+y)^2\le2x^2+2y^2\), since \(0\le(x-y)^2\). Apply this inequality with \(x=aH_s\) and \(y=bK_s\), then integrate: $$ \int_0^T(aH_s+bK_s)^2\,ds \le 2a^2\int_0^T H_s^2\,ds +2b^2\int_0^T K_s^2\,ds. $$ Both energies on the right are finite almost surely. Their finite linear combination is therefore finite almost surely, proving the first assertion. For the second, the pointwise bound \((G_sH_s)^2\le C^2H_s^2\) gives \(\int_0^T(G_sH_s)^2\,ds\le C^2A_T(H)<\infty\) almost surely. Predictability of the resulting processes follows from measurability of products and sums of predictable processes. Since \(T\) was arbitrary, both conclusions hold on every finite horizon. \(\square\)
Worked Example: Combining Two Locally Square-Integrable Integrands
On a Brownian filtration, take \(H_t=B_t\) and \(K_t=1\). Verify their local square integrability and use the proposition to check \(J_t=2B_t-3\). Then compare \(H\)'s energy expectation with its pathwise finiteness.
Brownian paths are continuous, so on every finite interval \([0,T]\), \(B_t^2\) is continuous and bounded along each path. Thus \(\int_0^T B_s^2\,ds<\infty\) almost surely. Also \(A_T(K)=\int_0^T1\,ds=T<\infty\). The closure proposition gives local square integrability of \(J\). Directly, using \((x+y)^2\le2x^2+2y^2\) with \(x=2B_s\) and \(y=-3\), $$ \int_0^T(2B_s-3)^2\,ds \le 8\int_0^T B_s^2\,ds+18T, $$ which is finite almost surely.
For \(H_t=B_t\), nonnegativity permits Tonelli's theorem. Since \(B_s\) has mean zero and variance \(s\), \(\mathbb{E}[B_s^2]=s\), so $$ \mathbb{E}\int_0^T B_s^2\,ds =\int_0^T\mathbb{E}[B_s^2]\,ds =\int_0^T s\,ds =\frac{T^2}{2}. $$ This is finite for every finite \(T\); \(H\) is therefore in \(\mathcal{L}^2\) on each such horizon as well.
Check: at \(T=2\), the expected energy is \(2^2/2=2\). Integrating the variance directly gives \(\int_0^2s\,ds=[s^2/2]_0^2=2\), confirming the calculation. This example also shows that the energy itself is random, even though its expectation is finite.
Common Mistake: Replacing “Almost Surely Finite” with “Finite Mean”
A common error is to conclude that a locally square-integrable integrand must belong to \(\mathcal{L}^2\), because its energy is finite on almost every path. That conclusion silently replaces pathwise finiteness with finite expectation. In the random-amplitude example, the mistaken calculation would be: “\(U>0\) almost surely, so \(1/U\) is finite almost surely; therefore \(\mathbb{E}[1/U]<\infty\).” The final implication is false. In fact, \(\mathbb{E}[1/U]=\int_0^1u^{-1}\,du=\infty\). The correction is to keep the two tests separate: local square integrability asks whether \(A_T(H)<\infty\) almost surely; \(\mathcal{L}^2\) asks whether \(\mathbb{E}A_T(H)<\infty\).
Likewise, a singularity at a single endpoint should not be judged by its pointwise value alone. What matters is the time integral of \(H^2\). For instance, \(t^{-1/2}\) has finite values for \(t>0\), but its squared energy near zero is \(\int_0^T t^{-1}\,dt=\infty\) for every \(T>0\). It is therefore not locally square-integrable on horizons starting at zero. By contrast, the preceding \(t^{-1/4}\) example has squared energy \(t^{-1/2}\), which is integrable at zero.
The local condition is useful because it admits integrands with pathwise finite energy even when expected energy is infinite. The stronger \(\mathcal{L}^2\) condition remains essential when an argument needs the Ito isometry or an unconditional second-moment bound. Keeping those roles distinct prevents an almost-sure statement from being used as an expectation estimate.
Check Your Understanding
For each question, distinguish finiteness almost surely from finiteness of the expected energy.
- If \(H\in\mathcal{L}^2\) on \([0,T]\), why must \(A_T(H)<\infty\) almost surely?
- For \(H_t=t^{-1/4}\) when \(t>0\), what is \(A_T(H)\), and is \(H\) in \(\mathcal{L}^2\) for finite \(T\)?
- For the random-amplitude example, compute \(A_T(H)\) separately for \(0<T\le1\) and \(T>1\). Which condition fails?
- Why is it enough to test \(A_n(H)<\infty\) almost surely at positive integer horizons?
- If \(H\) and \(K\) are locally square-integrable, which inequality verifies local square integrability of \(aH+bK\)?