For functions \(f \in\mathcal{V}\) we show how to define the Itô integral \[ \mathcal{I}[f](\omega)=\int_{S}^Tf(t,\omega)dB_{t}(\omega), \] where \(B_{t}\) is 1-dimensional Brownian motion. Our approach is to define the integral for elementary functions before extending results to all functions in \(\mathcal{V}\).
A function \(\phi \in\mathcal{V}\) is called elementary if it has the form \[ \phi(t,\omega)=\sum_{j}e_{j}(\omega)\cdot \mathbb{1}_{[t_{j}, t_{j+1})}(t). \]
This kind of function is also known as a step function. We note that since \(\phi \in \mathcal{V} \implies\) each \(e_{j}\) much be \(\mathcal{F}_{t_{j}}\)-measurable. For elementary functions \(\phi\) we define Itô’s integral by \[ \int_{S}^T \phi(t,\omega)dB_{t}(\omega) = \sum_{j \geq 0}e_{j}(\omega)[B_{t_{j+1}}-B_{t_{j}}](\omega). \]
If \(\phi(t,\omega)\) is bounded and elementary then \[ \mathbb{E}\left[ \left( \int_{S}^T \phi(t, \omega)dB_{t}(\omega) \right)^2 \right] = \mathbb{E}\left[ \int_{S}^T \phi(t, \omega)^2dt \right]. \]
Proof: We denote \(\Delta B_{j}=B_{t_{j+1}}-B_{t_{j}}\) and we note that \[ \mathbb{E}[e_{i}e_{j}\Delta B_{i}\Delta B_{j}]=\begin{cases}0 & i \neq j; \\\mathbb{E}[e_{j}^2]\cdot(t_{j+1}-t_{j}) & i = j,\end{cases} \] from the independent increments of Brownian motion. Thus we have \[ \mathbb{E}\left[ \left( \int_{S}^T \phi dB \right)^2 \right]=\sum_{i,j}\mathbb{E}[e_{i}e_{j}\Delta B_{i}\Delta B_{j}]=\sum_{j}\mathbb{E}[e_{j}^2]\cdot(t_{j+1}-t_{j})=\mathbb{E}\left[ \int_{S}^T\phi^2dt \right], \] giving the result. \(\square\)
We now proceed to extend the integral to be defined on all functions \(\mathcal{V}\) by applying three steps.
Step 1: Let \(g \in \mathcal{V}\) be bounded and \(g(\cdot, \omega)\) continuous for each \(\omega\). Then there exists elementary functions \(\phi_{n}\in\mathcal{V}\) such that \[ \mathbb{E}\left[ \int_{S}^T(g-\phi_{n})^2dt \right]\to 0~~\text{as}~~n \to \infty. \]
Proof: We define \(\phi_{n}(t, \omega)=\sum_{j}g(t_{j}, \omega)\cdot \mathbb{1}_{[t_{j},t_{j+1})}\) so that \(\phi_{n}\) is elementary since \(g\in\mathcal{V}\) and further for each \(\omega\) \[ \int_{S}^T (g-\phi_{n})^2 dt \to 0~~\text{as}~~n \to \infty, \] since \(g(\cdot, \omega)\) is continuous for each \(\omega\). Hence \(\mathbb{E}\left[ \int_{S}^T (g-\phi_{n})^2dt \right]\to 0\) as \(n \to \infty\) by bounded convergence. \(\square\)
Step 2: Let \(h \in\mathcal{V}\) be bounded. Then there exist bounded functions \(g_{n} \in\mathcal{V}\) such that \(g_{n}(\cdot, \omega)\) is continuous for all \(\omega\) and \(n\), and \[ \mathbb{E}\left[ \int_{S}^T(h-g_{n})^2dt \right]\to 0. \] Proof: