Density of Simple Predictable Integrands

Why Density Is the Key Step

The Ito integral is first defined for simple predictable processes, where it is a finite sum of adapted coefficients multiplied by Brownian increments. To pass from these finite sums to general square-integrable predictable integrands, one needs an approximation result: every such integrand must be close, in mean square over time and probability, to a simple predictable process. This tutorial establishes that density statement and identifies precisely which measurability and integrability conditions it requires.

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. Fix $T>0$. Let $\mathcal{P}$ denote the predictable sigma-field on $\Omega\times[0,T]$. We use its standard generating description: it is generated by $\{0\}\times A$, $A\in\mathcal{F}_0$, and rectangles $(s,t]\times A$, where $0\le s<t\le T$ and $A\in\mathcal{F}_s$. A predictable process is a process measurable with respect to $\mathcal{P}$. The time-zero slice has zero measure for the product measure used below.

Definition: Write $\mu=\mathbb{P}\times dt$ on $\Omega\times[0,T]$. The space $L^2_{\mathrm{pred}}([0,T])$ consists of predictable processes $H$, identified when they agree $\mu$-almost everywhere, such that $$ \|H\|_2^2=\mathbb{E}\int_0^T H_t^2\,dt<\infty. $$ A simple predictable process is a finite sum of processes $\xi_i\mathbf{1}_{(t_i,t_{i+1}]}$, where $0\le t_0<\cdots<t_n\le T$, each $\xi_i$ is $\mathcal{F}_{t_i}$-measurable, and $\mathbb{E}[\xi_i^2]<\infty$. Such a process is in $L^2_{\mathrm{pred}}([0,T])$, since $\mathbb{E}\int_{t_i}^{t_{i+1}}\xi_i^2\,dt=(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]$.

The admissible class here is the class of progressively measurable, adapted integrands with $\mathbb{E}\int_0^T H_t^2\,dt<\infty$; the density result concerns its predictable members. Predictability is the measurability that matches the simple processes just defined: on an interval $(s,t]$, the coefficient is known at time $s$, not chosen using information arriving later.

Approximating Predictable Measurable Sets

The central measure-theoretic step is to approximate predictable sets by finite combinations of the generating rectangles. Let $\mathcal{A}$ be the algebra generated by those rectangles and the time-zero sets. On $\Omega\times[0,T]$, $\mu$ is finite, with total mass $T$. This finite-measure property is important: it lets us approximate countable unions by sufficiently large finite unions.

Lemma: For every $E\in\mathcal{P}$ and every $\varepsilon>0$, there is $A\in\mathcal{A}$ such that $\mu(E\triangle A)<\varepsilon$, where $E\triangle A$ is the symmetric difference.

Proof. Let $\mathcal{C}$ be the collection of all predictable sets $E$ for which the stated approximation property holds. If $E\in\mathcal{C}$, then $\mu(E^c\triangle A^c)=\mu(E\triangle A)$, and $A^c\in\mathcal{A}$. Thus $E^c\in\mathcal{C}$. Now let $E_j\in\mathcal{C}$ for $j\ge1$, and put $E=\bigcup_{j\ge1}E_j$. Since $\mu$ is finite and $\bigcup_{j=1}^mE_j$ increases to $E$, continuity from below gives an $m$ such that $\mu(E\setminus\bigcup_{j=1}^mE_j)<\varepsilon/2$. For each $j\le m$, choose $A_j\in\mathcal{A}$ with $\mu(E_j\triangle A_j)<\varepsilon/(2m)$. The finite union $A=\bigcup_{j=1}^m A_j$ belongs to $\mathcal{A}$, and $$ \mu(E\triangle A)\le \mu\left(E\setminus\bigcup_{j=1}^mE_j\right)+\sum_{j=1}^m\mu(E_j\triangle A_j)<\varepsilon. $$ Therefore $\mathcal{C}$ is a sigma-field. It contains the generating rectangles and time-zero sets, so it contains $\mathcal{P}$. $\square$

For sets in $\mathcal{A}$, their indicators are finite combinations of predictable rectangle indicators. Indeed, the intersection of two generating rectangles is either empty or another rectangle of the same type: the later left endpoint determines the time interval, and the intersection of the two events is measurable at that endpoint. Boolean combinations can therefore be written as finite sums of rectangle indicators, after refining the finitely many time endpoints. A rectangle indicator $\mathbf{1}_{(s,t]\times A}$ is the simple process $\mathbf{1}_A\mathbf{1}_{(s,t]}$; its coefficient is $\mathcal{F}_s$-measurable and square-integrable. Time-zero indicators have zero $L^2(\mu)$ norm.

The Density Theorem

Theorem: The simple predictable processes are dense in $L^2_{\mathrm{pred}}([0,T])$. Precisely, for each predictable $H$ satisfying $\mathbb{E}\int_0^T H_t^2\,dt<\infty$, there is a sequence of simple predictable processes $H^n$ such that $$ \mathbb{E}\int_0^T |H_t-H^n_t|^2\,dt\longrightarrow0. $$

Proof. Fix $H\in L^2_{\mathrm{pred}}([0,T])$ and $\varepsilon>0$. First truncate $H$: define $H^{(M)}=(-M)\vee(H\wedge M)$. Then $$ \|H-H^{(M)}\|_2^2 \le \mathbb{E}\int_0^T H_t^2\mathbf{1}_{\{|H_t|>M\}}\,dt\longrightarrow0 $$ as $M\to\infty$, by dominated convergence, since $H^2$ is integrable. Choose $M$ so that this norm is less than $\varepsilon/3$. Quantize the bounded predictable process $H^{(M)}$ into a finite-valued predictable process $q=\sum_{j=1}^r c_j\mathbf{1}_{E_j}$, choosing the quantization mesh small enough that $\|H^{(M)}-q\|_2<\varepsilon/3$. This is possible because pointwise error at most $\delta$ gives $L^2(\mu)$ error at most $\delta\sqrt{T}$.

For each $j$, the lemma supplies $A_j\in\mathcal{A}$ such that $\mu(E_j\triangle A_j)$ is as small as desired. Since $$ \left\|\sum_{j=1}^r c_j(\mathbf{1}_{E_j}-\mathbf{1}_{A_j})\right\|_2 \le \sum_{j=1}^r |c_j|\sqrt{\mu(E_j\triangle A_j)}, $$ choose the approximations so that this bound is less than $\varepsilon/3$. The process $S=\sum_{j=1}^r c_j\mathbf{1}_{A_j}$ is a finite combination of predictable rectangle indicators, hence is simple predictable after refining the finitely many time endpoints. The triangle inequality gives $\|H-S\|_2<\varepsilon$. As this holds for every $\varepsilon>0$, choosing errors tending to zero gives the required sequence. $\square$

The theorem is a statement about the product-space norm, not pointwise convergence of paths. In particular, its conclusion controls the average squared error over both probability and time. It requires both predictability and finite $L^2$ norm; without the latter, the norm to be approximated need not be finite.

Worked Approximations

Worked Example: A Deterministic Integrand

On $[0,1]$, approximate $H_t=t$ by $H^n_t=k/n$ on each interval $(k/n,(k+1)/n]$, for $k=0,\ldots,n-1$. These deterministic coefficients are measurable at time $k/n$, so $H^n$ is simple predictable.

On the $k$-th interval, the error is $t-k/n$. Therefore $$ \mathbb{E}\int_0^1|H_t-H^n_t|^2\,dt =\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}(t-k/n)^2\,dt =n\int_0^{1/n}u^2\,du =\frac{1}{3n^2}. $$

Check: for $n=2$, the two interval errors are each $\int_0^{1/2}u^2\,du=1/24$, so the total is $1/12$. The formula gives $1/(3\cdot 2^2)=1/12$. It tends to zero as $n\to\infty$.

Worked Example: Approximating Brownian Motion

On $[0,1]$, let $H_t=B_t$ and use $H^n_t=B_{k/n}$ on $(k/n,(k+1)/n]$. The coefficient $B_{k/n}$ is $\mathcal{F}_{k/n}$-measurable. Also $\mathbb{E}[B_{k/n}^2]=k/n<\infty$, and $\mathbb{E}\int_0^1 B_t^2\,dt=\int_0^1t\,dt=1/2$. Thus both the integrand and its approximants meet the square-integrability requirement.

For $t\in[k/n,(k+1)/n]$, Brownian increments have second moment $t-k/n$. Hence $$ \mathbb{E}\int_0^1|B_t-H^n_t|^2\,dt =\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}(t-k/n)\,dt =n\int_0^{1/n}u\,du =\frac{1}{2n}. $$

Check: when $n=4$, each of the four intervals contributes $\int_0^{1/4}u\,du=1/32$; the sum is $4/32=1/8$, equal to $1/(2\cdot4)$. The mean-square error vanishes as the partition is refined.

Worked Example: A Square-Integrable Nonlinear Integrand

On $[0,1]$, take $H_t=B_t^2$ and $H^n_t=B_{k/n}^2$ on $(k/n,(k+1)/n]$. Since a centered normal variable with variance $t$ has fourth moment $3t^2$, $$ \mathbb{E}\int_0^1 H_t^2\,dt=\int_0^1\mathbb{E}[B_t^4]\,dt =\int_0^1 3t^2\,dt=1. $$ Also $\mathbb{E}[(B_{k/n}^2)^2]=3(k/n)^2<\infty$, so every coefficient is square-integrable.

Set $s=k/n$ and $u=t-s$. Write $B_t=B_s+Z$, where $Z=B_t-B_s$ is independent of $B_s$, has mean zero, variance $u$, and fourth moment $3u^2$. Then $B_t^2-B_s^2=2B_sZ+Z^2$. Expanding the square, independence and the zero odd moments give $$ \mathbb{E}\big[(B_t^2-B_s^2)^2\big] =4\mathbb{E}[B_s^2]\mathbb{E}[Z^2]+\mathbb{E}[Z^4] =4su+3u^2. $$ Consequently, with $h=1/n$, $$ \mathbb{E}\int_0^1|H_t-H^n_t|^2\,dt =\sum_{k=0}^{n-1}\int_0^h(4(k/n)u+3u^2)\,du =\sum_{k=0}^{n-1}\left(\frac{2k}{n}h^2+h^3\right) =\frac{n-1}{n^2}+\frac{1}{n^2} =\frac1n. $$

Check: for $n=1$, the coefficient is $B_0^2=0$, so the error is $\mathbb{E}\int_0^1B_t^4\,dt=1$, matching $1/n=1$. The exact error tends to zero as $n$ increases.

Common Pitfall: Adapted Is Not the Same as Simple Predictable

A tempting but invalid approximation is to put the right endpoint value $B_{(k+1)/n}$ on $(k/n,(k+1)/n]$. That coefficient is generally measurable only at time $(k+1)/n$, not at the left endpoint $k/n$. The resulting process is not of the required simple predictable form: it uses information from the end of the interval throughout that interval. For the Brownian approximation above, the valid coefficient is $B_{k/n}$. Its squared error is computed from the increment after $k/n$, and that error is exactly $1/(2n)$ on $[0,1]$.

The density theorem matters because it reduces questions about all square-integrable predictable integrands to finite adapted step processes. Its proof depends on approximating in the product $L^2$ norm, preserving predictability, and checking the integrability of the random coefficients. Merely approximating sample paths, or using coefficients that depend on future information, does not establish the required density.

Check Your Understanding

Use the product-space $L^2$ norm and the left-endpoint measurability condition in your answers.

Why does the time-zero slice have no effect on the $L^2(\mathbb{P}\times dt)$ norm?
For $H_t=t$ on $[0,1]$, what is the squared approximation error using $k/n$ on each interval $(k/n,(k+1)/n]$?
Why is $B_{k/n}^2$ an admissible coefficient for the simple approximation to $B_t^2$?
Which finite-measure property is used to approximate a countable union by a finite union in the lemma?
Explain why $B_{(k+1)/n}$ is not the appropriate coefficient for a simple predictable process on $(k/n,(k+1)/n]$.