At page 291 of Nonlinear Option Pricing by Julien Guyon and Pierre Henry-Labordère, the Clark-Ocone's Formula is applied to $\mathbb{1}_{S_t > K}$.
I do not get how to get from the second to the third step of the following:
$$ \mathbb{E}^{\mathbb{Q}^{t} }[(r_t-r_t^0)\mathbb{1}_{S_t > K}]= $$ $$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\mathbb{E}^{\mathbb{Q}^{t}}[\mathbb{1}_{S_t > K}] + \int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{B^t}\mathbb{1}_{S_t > K}]dB_s^t + \int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg) \bigg]= $$
$$ \mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg( \int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{B^t}\mathbb{1}_{S_t > K}]dB_s^t \bigg) \bigg]$$
I tried to use the Malliavin integration by parts formula and to manipulate the second integral $\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t $ but did not manage to get to the third step.
Is it the case that: $\mathbb{E}^{\mathbb{Q}^{t}}[\mathbb{1}_{S_t > K}] = -\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t $ and why?
You don't have the right argument, the real reason why you can get the third line is because first $B$ and $Z$ are orthogonal so that $$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg)\bigg] =0$$ and because : $$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\mathbb{E}^{\mathbb{Q}^{t}}[\mathbb{1}_{S_t > K}]\bigg)\bigg]=\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\int_0^t\sigma_r^t(s)dB_s^t \bigg]\cdot\mathbb{E}^{\mathbb{Q}^{t}}\bigg[\mathbb{1}_{S_t > K}\bigg] $$ and on the right hand side the first term is null because $\int_0^t\sigma_r^t(s)dB_s^t$ is a martingale.
Best regards