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?
 A: Yes, it is correct.
To put the discussion under the measure-theoretic conditional expectation/probability framework, first let's clarify the exact meaning of the notation $P(G = 1 | Z = 1, \eta)$.  As $P(G = 1 | Z, \eta)$ is $\sigma(Z, \eta)$-measurable, we can denote it by $f(Z, \eta)$, where $f$ is a Borel function from $(\mathbb{R}^2, \mathscr{R}^2)$ to $(\mathbb{R}^1, \mathscr{R}^1)$.  $P(G = 1 | Z = 1, \eta) = 1$ almost surely then means $f(Z(\omega), \eta(\omega)) = 1$ for $\omega \in A := \{\omega \in \Omega: Z(\omega) = 1\}$ with probability $1$.  Accordingly, $P(G = 1 | Z, \eta) = E(G | Z, \eta)$ can be written as
\begin{align}
E(G | Z, \eta) = P(G = 1|Z, \eta)I_A + P(G = 1|Z, \eta)I_{A^c} = I_A + I_{A^c}f(Z, \eta).
\end{align}
Now applying the property of conditional expectation: $E[X|\mathscr{F}_1] = E[E[X|\mathscr{F_2}]|\mathscr{F}_1]$ if $\mathscr{F}_1 \subset \mathscr{F}_2$ yields
\begin{align}
 & E[G\eta|Z] = E[E[G\eta|Z, \eta]|Z] \\
=& E[\eta(I_A + I_{A^c}f(Z, \eta))|Z] = I_AE[\eta|Z] + I_{A^c}E[\eta f(Z, \eta)|Z]. \tag{1}
\end{align}
Again, by noticing that $E[G\eta|Z = 1]$ means the (common) value of $E[G\eta|Z]$ on set $A$, it follows by $(1)$ and $E[\eta|Z = 1] = 0$ that for all $\omega \in A$ but a null-set that
\begin{align}
E[G\eta|Z](\omega) = 1 \times E[\eta|Z](\omega) + 
0 \times E[\eta f(Z, \eta)|Z](\omega) = E[\eta | Z](\omega) = 0.  
\end{align}
This shows $E[G\eta | Z = 1] = 0$.
A: Using the antecedent conditions in your post you have:
$$\begin{align}
\mathbb{E}(G \times \eta|Z=1) 
&= \int \eta \cdot p(\eta,G=1|Z=1) \ d\eta \\[6pt]
&= \int \eta \cdot \mathbb{P}(G=1|\eta,Z=1) \cdot p(\eta|Z=1) \ d\eta \\[6pt]
&= \int \eta \cdot p(\eta|Z=1) \ d\eta \\[10pt]
&= {E}(\eta|Z=1) \\[14pt]
&= 0. \\[6pt]
\end{align}$$
