The L2 Space of Ito Integrands

Measuring an Integrand Across Time and Probability

For an Ito integrand, the relevant size is not its value at one time or its size on a typical sample path. It is the average of its squared magnitude over both time and outcomes. This gives a norm that controls the second moment of a stochastic integral: for simple predictable integrands, the Ito isometry identifies the second moment of the integral with this squared norm.

Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Write \(dt\otimes\mathbb{P}\) for the product measure on \([0,T]\times\Omega\), and write \(\mathcal{P}_T\) for the predictable sigma algebra on that time interval. A process is predictable when it is measurable with respect to \(\mathcal{P}_T\); equivalently, this sigma algebra is generated by adapted left-continuous processes. In this tutorial we consider predictable processes \(H\) that also satisfy the usual progressive-measurability condition for Ito integrands.

The expectation and time integral can be read as one integral over the product space: \(\|H\|_2^2=\int_{[0,T]\times\Omega}|H_s(\omega)|^2\,(dt\otimes\mathbb{P})(d s,d\omega)\). Thus a large value on a set of product measure zero does not affect the norm. Nor does a process need to be bounded to have finite norm; it needs only to have a finite average squared magnitude.

Why Processes Are Identified Almost Everywhere

Equality in this space is not pointwise equality of paths. Two processes can differ at a time of zero Lebesgue measure, or on an event of probability zero, and still have zero distance in the norm. This is the appropriate identification for integration: changing an integrand on a \(dt\otimes\mathbb{P}\)-null set does not change its norm. The following proposition makes the zero-distance criterion precise.

Proof. By the definition of the norm, \(\|H-K\|_2^2=\int|H-K|^2\,d(dt\otimes\mathbb{P})\). If \(H=K\) almost everywhere, this nonnegative function is zero almost everywhere, so the integral is zero. Conversely, suppose the integral is zero. For each positive integer \(n\), the set \(A_n=\{|H-K|^2\ge 1/n\}\) must have product measure zero: otherwise the integral would be at least \((dt\otimes\mathbb{P})(A_n)/n>0\). The set where \(H\ne K\) is \(\bigcup_{n=1}^{\infty}A_n\), so it too has product measure zero. \(\square\)

The integral \(\mathbb{E}\int_0^T HK\,ds\) defines an inner product on the equivalence classes. It is finite by Cauchy–Schwarz, and changing either representative on a product-null set leaves it unchanged. The associated norm is exactly \(\|\cdot\|_2\).

The Space Is Complete

Completeness means that every sequence whose terms become arbitrarily close in this norm converges in the norm to an element of the space. This property matters because the Ito integral for general square-integrable integrands is built using limits in this norm. The result below treats the predictable restriction as well as the usual square-integrability requirement.

Proof. The inner product and induced norm are well-defined on equivalence classes by the preceding discussion, and the inner product is positive definite there by the proposition. It remains to prove completeness. Regard predictable processes as measurable functions on the measure space \(([0,T]\times\Omega,\mathcal{P}_T,dt\otimes\mathbb{P})\), where the total measure is \(T<\infty\). Let \((H_n)\) be a Cauchy sequence in the norm. Choose a subsequence \((H_{n_k})\) such that \(\|H_{n_{k+1}}-H_{n_k}\|_2\le 2^{-k}\) for every \(k\ge1\). Cauchy–Schwarz on this finite measure space gives \(\|H_{n_{k+1}}-H_{n_k}\|_1\le \sqrt{T}\,2^{-k}\). Therefore the integral of \(\sum_{k\ge1}|H_{n_{k+1}}-H_{n_k}|\) is finite, by summing these \(L^1\) bounds. The sum is finite almost everywhere, so the subsequence converges pointwise almost everywhere to a limit \(H\). Defining \(H=0\) on the exceptional measurable set gives a predictable version, since a pointwise limit of measurable functions is measurable. For each fixed \(k\), Fatou's lemma gives \(\|H-H_{n_k}\|_2^2\le\liminf_{j\to\infty}\|H_{n_j}-H_{n_k}\|_2^2\). The right side tends to zero as \(k\to\infty\), because the original sequence is Cauchy. Hence the subsequence converges to \(H\) in norm. In particular \(H\) has finite norm, and the Cauchy property then implies the full sequence converges to \(H\) in norm. Thus the space is complete and is a Hilbert space. \(\square\)

If \(T=0\), every time integral is zero and the quotient space consists only of the zero element. The theorem's positive-horizon assumption excludes this degenerate case.

Worked Examples: What the Norm Includes

A Common Misreading of the Condition

Pitfall: confusing pathwise behavior with the product-space norm. A wrong calculation might argue that \(H_t=t^{-1/2}\) is acceptable because it is finite for every \(t>0\), then use \(\int_0^1t^{-1/2}\,dt=2\) as its squared norm. The correction is that the norm squares the integrand first: for \(H_t=t^{-1/2}\), \(|H_t|^2=1/t\), and \(\int_0^1dt/t=\infty\). This process is not in the space. In contrast, \(t^{-1/4}\) has square \(t^{-1/2}\) and finite squared norm \(2\). Pointwise finiteness alone is not the membership test.

For a process with randomness, the same principle applies: the calculation must include both expectation and time integration. For example, the Brownian integrand has finite norm because \(\int_0^2\mathbb{E}[B_t^2]\,dt=2\), even though it has no finite essential bound. The norm is an average-square criterion, not a requirement of boundedness on every path.