Show that $\Pr(G=1|Z=1, \eta)=1$ and $\mathbb{E}(\eta|Z=1)=0$ imply $\mathbb{E}(G\times \eta|Z=1)=0$ Take two binary random variables $G,Z$ and a continuous random variable $\eta$.
Assume
$$
\Pr(G=1|Z=1, \eta)=1 \text{ almost surely}
$$
and
$$
 \mathbb{E}(\eta|Z=1)=0
$$
Could you help me to show that this implies
$$
 \mathbb{E}(G\times \eta|Z=1)=0 \quad?
$$
If this implication is wrong, could you explain why?

Using the antecedent conditions in your post you have:
$$\begin{align}
\mathbb{E}(G \times \eta|Z=1) 
&= \int \eta \cdot p(\eta,G=1|Z=1) \ d\eta \\[6pt]
&= \int \eta \cdot \mathbb{P}(G=1|\eta,Z=1) \cdot p(\eta|Z=1) \ d\eta \\[6pt]
&= \int \eta \cdot p(\eta|Z=1) \ d\eta \\[10pt]
&= {E}(\eta|Z=1) \\[14pt]
&= 0. \\[6pt]
\end{align}$$
