Stochastic integration with non-caglad integrand

Question

In Oksendal's book on Malliavin calculus. The $\sigma$-algebra $\mathcal{F}_G$ is defined as the completed one generated by all random variables of the form $$ F = \int_0^T\chi_A(t) \, dW(t), $$ for all $A\subseteq G$, where $G$ can by any Borel set in $[0,T]$

Even the process $\chi_A(t)$ is deterministic, how does one justify such integration? As I know, integrands for stochastic integration is required to be caglad, i.e. is left continuous and has left limit. But $\chi_A(t)$ can be non-caglad, for example when $A=\{t\}$.

Answer 1

It's not true that stochastic integrals are only defined for caglad integrands. In order to make sense of the stochastic integral, we need that the integrand satisfies a certain "good" measurability property. In case of Brownian motion, the integrand has to be progressively measurable, see e.g. Section 15.5 in the book Brownian Motion by Schilling & Partzsch for details. If you are dealing with deterministic integrands, then this progressive measurability boils down to Borel measurability.

Lemma: Let $f:[0,\infty) \to \mathbb{R}$ be a (deterministic) Borel measurable function such that $\int_0^T f(s)^2 \, ds < \infty$ for all $T>0$. Then the stochastic integral $$\int_0^T f(s) \, dW_s$$ exists for all $T>0$.