In Oksendal's book, in Theorem 3.2.1, the author states that $\int_S^T f dB_t$ is $\mathcal{F}_T$-measurable. In the proof he says that since it holds for elementary functions, when we take the limits we obtain this for all $f$ with some special properties.
There's a theorem stating that
the pointwise limit of a measurable function, is also measurable.
The problem is that the limit we're using in the definition of the Itô integral is with respect to the L2 norm, and convergence in Lp norms doesn't not imply pointwise convergence…
Therefore, how does one can simply take the limits and get a $\mathcal{F}_T$-measurable?
Best Answer
Convergence in $L^p$ norms implies pointwise convergence along a subsequence, though, so that gives the measurability you need. Suppose $(X_n) \rightarrow X$ in $L^1$, then we can take a subsequence such that $\sum_{k=1}^\infty \mathbb{E}[|X_{n_k}-X|] < \infty.$ By the monotone convergence theorem, we have $$\sum_{k=1}^\infty \mathbb{E}[|X_{n_k}-X|] = \mathbb{E}\left[\sum_{k=1}^\infty|X_{n_k}-X|\right] < \infty $$ so $\sum_{k=1}^\infty|X_{n_k}-X| < \infty$ almost surely, and hence $\lim_{k \rightarrow \infty} |X_{n_k}-X| = 0$ almost surely so we have $(X_{n_k}) \rightarrow X$ pointwise a.s. In particular, if each $X_n$ was measurable, this shows that $X$ is also measurable.
Edit: As pointed out in the comments, this relies on the fact that the $\mathcal F_t$ is complete. This is included in Definition 3.1.2 of Oksendal's book.