Construction of Ito Intergration
1. Idea
Just like the construction of Lebesgue integral, we first define the integral for elementary (simple functions), then extend it to the class of where the integral is well-defined.
Elementary Functions
For a given partition $P = {t_0, t_1, \cdots, t_n}$ of $[0, T]$, we define the elementary function $\phi$ as
$$
\phi = \sum_{i=0}^{n-1} e_i(\omega) \chi_{[t_{i}, t_{i+1})}(\omega)
$$
where $e_i$ is a $\mathcal{F}{t{i}}$-measurable random variable.
2. Ito Isometry
For $\phi$ is an elementary function, we define the Ito integral of $\phi$ as
$$
\int_{S}^{T} \phi(t, \omega) \mathrm{d}B_t = \sum_{i=0}^{n-1} e_i(\omega) (B_{t_{i+1}} - B_{t_{i}})
$$
The Ito isometry states that
$$
\mathbb{E} \left[ \left( \int_{S}^{T} \phi(t, \omega) \mathrm{d}B_t \right)^2 \right] = \mathbb{E} \left[ \int_{S}^{T} \phi^2(t, \omega) \mathrm{d}t \right]
$$
3. Ito Integral
$\mathcal{V}$
Denote the class we want to define the integral as $\mathcal{V}$. Let $\mathcal{V} = \mathcal{V}(S, T)$ be the class of all functions $f(t, \omega): [0, \infty) \times \Omega \rightarrow \mathbb{R}$ such that
$f(t, \omega)$ is $\mathcal{B} \times \mathcal{F}$-measurable, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $[0, \infty)$.
$f(t, \omega): \omega \mapsto f(t, \omega)$ is $\mathcal{F}_t$-measurable for each $t \geq 0$.
$\mathbb{E} \left[ \int_{0}^{\infty} f^2(t, \omega) \mathrm{d}t \right] < \infty$.
Step 1
Handle the bounded and continuous case.
Bounded and continuous functions in $\mathcal{V}$ can be approximated by elementary functions.
Lemma 1. If $g \in \mathcal{V}$ is bounded and continuous, then there exists a sequence of elementary functions $\phi_n \in \mathcal{V}$ such that
$$
\mathbb{E} \left[ \int_{S}^{T} (g(t, \omega) - \phi_n(t, \omega))^2 \mathrm{d}t \right] \rightarrow 0
$$
as $n \rightarrow \infty$.
Here bounded convergence theorem is used.
Theorem 1. (Bounded Convergence Theorem) Let $f_n$ be a sequence of bounded measurable functions that are supported on a set $E$ of finite measure. If $f_n \rightarrow f$ pointwise a.e. on $E$, then
$$
\lim_{n \rightarrow \infty} \int_{E} |f_n - f| \mathrm{d}\mu = 0
$$
Consequently,
$$
\lim_{n \rightarrow \infty} \int_{E} f_n \mathrm{d}\mu = \int_{E} f d\mu
$$
Proof of Lemma 1.
Define $\phi_n$ as
$$
\phi_n(t, \omega) = \sum_{i=0}^{n-1} g(t_i, \omega) \chi_{[t_{i}, t_{i+1})}(t)
$$
thus $\phi_n$ is an elementary function.
$$
\int_S^T (g-\phi_n)^2 \mathrm{d}t \rightarrow 0， \text{as } n \rightarrow \infty \tag{1}
$$
since $g$ is continuous and $[S, T]$ is compact.
Proof of (1)
Given $\omega$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that
$\forall t, t’ \in [S, T]$ with $|t - t’| < \delta$, we have $|g(t, \omega) - g(t’, \omega)| < \sqrt{\frac{\epsilon}{T-S}}$. Choose $n$ such that $\frac{1}{n} < \frac{\delta}{2}$, then
$$
|g(t, \omega) - \phi_n(t, \omega)|^2 = |g(t, \omega) - g(t_i, \omega)|^2 < \frac{\epsilon}{T-S}
$$
since $|t - t_i| < \frac{2}{n} < \delta$.
Thus we have
$$
\int_S^T (g-\phi_n)^2 \mathrm{d}t < \frac{\epsilon}{T-S} \int_S^T \mathrm{d}t = \epsilon
$$
Denote $I_n(\omega) = \int_S^T (g-\phi_n)^2 \mathrm{d}t$, Eq.(1) suggests that $I_n(\omega) \rightarrow 0$ pointwisely respect to $\omega$.
Here we just need to verify that $I_n(\omega)$ is bounded and measurable, then we can apply the bounded convergence theorem.
- Boundedness: $I_n(\omega) \leq \int_S^T g^2 \mathrm{d}t < M^2 (T-S)$, where $M$ is the bound of $g$.
- Measurability: $I_n(\omega)$ is measurable since $g$ is measurable according to Fubini’s theorem.
Step 2
Handle the bounded (but not necessarily continuous) case.
Bounded functions in $\mathcal{V}$ can be approximated by bounded and continuous functions.
Lemma 2. If $h \in \mathcal{V}$ is bounded, then $\exists$ bounded and continuous functions $g_n$ such that
$$
\mathbb{E} \left[ \int_{S}^{T} (h - g_n)^2 \mathrm{d}t \right] \rightarrow 0, \text{as } n \rightarrow \infty
$$
Proof of Lemma 2.
Suppose $h$ is bounded by $M$, first $\phi_n : \mathbb{R} \rightarrow \mathbb{R}$ is constructed as follows:
- $\phi_n(x) =0 $ for $x \le -\cfrac{1}{n}$ and $x \ge 0$.
- $\int_{-\infty}^{\infty} \phi_n(x) \mathrm{d}x = 1$.
Then define $g_n$ as
$$
g_n(t, \omega) = \int_{-\infty}^{\infty} \phi_n(t-s) h(s, \omega) \mathrm{d}s
$$
Then $g_n$ is bounded and continuous for each $\omega$. And $g_n$ is $\mathcal{F}_t$-measurable for each $t$ since $h$ is $\mathcal{F}t$-measurable. Also $\phi_n$ and $g_n$ are seleted for
$$
\int{S}^{T} (h - g_n)^2 \mathrm{d}t \rightarrow 0, \text{as } n \rightarrow \infty
$$
- $\phi_n$:
- $h$:
- $g_n$ ($n=5$):
Step 3
Handle the general case.
Every function in $\mathcal{V}$ can be approximated by bounded functions.
Lemma 3. If $f \in \mathcal{V}$, then there exists a sequence of bounded functions $h_n \in \mathcal{V}$ such that
$$
\mathbb{E} \left[ \int_{S}^{T} (f - h_n)^2 \mathrm{d}t \right] \rightarrow 0, \text{as } n \rightarrow \infty
$$
Proof of Lemma 3.
For each $n$, let $h_n$ satisfy
$$
h_n(t, \omega) = \begin{cases}
f(t, \omega), & |f(t, \omega)| \le n \
n, & f(t, \omega) > n \
-n, & f(t, \omega) < -n
\end{cases}
$$
By Dominated Convergence Theorem, we have the desired result.
Definition of Ito Integral
Definition 1. If $f \in \mathcal{V}$, one can find a sequence of elementary functions $\phi_n \in \mathcal{V}$ such that
$$
\mathbb{E} \left[ \int_{S}^{T} (f - \phi_n)^2 \mathrm{d}t \right] \rightarrow 0, \text{as } n \rightarrow \infty
$$
Then we define the Ito integral of $f$ as
$$
\int_{S}^{T} f(t, \omega) \mathrm{d}B_t = \lim_{n \rightarrow \infty} \int_{S}^{T} \phi_n(t, \omega) \mathrm{d}B_t
$$
The limit exists in $L^2(\Omega)$ since $L^2(\Omega)$ is complete and
Denote $J_n(\omega) = \int_{S}^{T} \phi_n(t, \omega) \mathrm{d}B_t$, then
$$
J_{n+k}(\omega) - J_n(\omega) = \int_{S}^{T} (\phi_{n+k} - \phi_n) \mathrm{d}B_t
$$