Construction of Ito integral

Question

This is in regards to constructing the Ito integral, specifically the first step of approximating bounded, continuous functions by elementary functions.

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $V = V(S,T)$ be the class of functions $f: [0,\infty) \times \Omega \to \mathbb{R}$ such that

1. $(t,\omega) \mapsto f(t,\omega)$ is $\mathcal{B} \otimes \mathcal{F}$ - measurable, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $[0,\infty)$,
2. $f$ is $\mathcal{F}_t$-adapted,
3. $E[ \int_S^T f(t,\omega)^2 dt ] < \infty$.

Oksendal's 5th ed. of "Stochastic Differential Equations," on page 31, states:

Let $g \in V$ be bounded and $g(\cdot,\omega)$ continuous for each $\omega$. Then there exist elementary functions $\phi_n \in V$ such that $$ \lim_{n \to \infty} E\left[ \int_S^T (g - \phi_n)^2 dt\right] = 0. $$

In his proof, for each $n \in \mathbb{N}$ he sets $t_j = j \cdot 2^{-n}$ and defines the elementary functions $\phi_n(t,\omega) = \sum_{j=1}^\infty g(t_j, \omega) \mathbb{I}_{[t_j, t_{j+1})}(t)$. He then states that, for each $\omega$, $$ \lim_{n \to \infty} \int_S^T (g - \phi_n)^2 dt = 0 \qquad (1) $$ since $g(\cdot,\omega)$ is continuous for each $\omega$.

I'm having some trouble verifying this for myself. I don't think he uses boundedness of $g$ to reach (1). First, I guess for each $\omega$, $\phi_n \to g$ for all $t \in [0,\infty)$, perhaps uniformly? (I'm having trouble showing this, actually). If the convergence is uniform, then since $[S,T]$ has finite Lebesgue measure, $\phi_n \to g$ in $L^1[S,T]$; i.e., $$ \lim_{n \to \infty} \int_S^T (g - \phi_n) dt = 0. $$

I'm not sure why this would imply the $L^2[S,T]$ convergence of (1), though, and I'm not sure how to use continuity of $g$, either.

Answer 1

First, recall the following elementary result:

Lemma: Let $f:[0,\infty) \to \mathbb{R}$ be a continuous function and set $$f_n(t) := \sum_{j=0}^{\infty} f(t_j) 1_{[t_j,t_{j+1})}(t)$$ where $t_j := j 2^{-n}$. Then $f_n(t) \to f(t)$ as $n \to \infty$ for all $t \geq 0$.

Applying this result for each fixed $\omega$, we get that $\phi_n(t,\omega) \to g(t,\omega)$ as $n \to \infty$. Moreover,

$$|\phi_n(t,\omega)-g(\omega)| \leq |\phi_n(t,\omega)| + |g(\omega)| \leq 2 \|g\|_{\infty}<\infty.$$

Now it follows from the dominated convergence theorem that

$$\mathbb{E}\left( \int_S^T (g-\phi_n)^2 \, dt \right) \to 0$$

as $n \to \infty$.