I had a problem when I came across a proposition in Oksendal's book on Malliavin calculus. In the book, it claims
$$ D_t\mathbb{E}[F|\mathcal{F}_G] = \mathbb{E}[D_tF|\mathcal{F}_G]\chi_G(t) $$ where G can be any Borel set in [0,T], $\chi$ is the indicator function.
However, when choosing G_1 = [0,t] and G_2 = [0,t), by the above theorem we have $$ D_t\mathbb{E}[F|\mathcal{F}_t] = \mathbb{E}[D_tF|\mathcal{F}_t] $$ while $$ D_t\mathbb{E}[F|\mathcal{F}_{t-}] = 0 $$
Clearly the RHS are different in general, but the LHS is the same if the filtration is continuous, for now $\mathbb{E}[F|\mathcal{F}_{G_1}] = \mathbb{E}[F|\mathcal{F}_{G_2}]$ and so is the Malliavin derivative. This leads to a contradiction. Can anyone point where I had gone wrong?
The answer to this question is that the malliavin derivative is not defined pathwise. We use the optional version of the process.