Measure theoretic reasoning that $\mathbb E(Y\mid Z)=\mathbb E(Y)$ where $Y$ and $Z$ are independent 
Show that $\mathbb E(Y\mid Z)=\mathbb E(Y)$ where $Y$ and $Z$ are
independent.

I know how to show this without measure theory:
$$\mathbb E(Y\mid Z)=\int_{\mathcal Y} y f_{Y\mid Z}(y\mid z)\overset{\text{ind}}{=}\int_{\mathcal Y} y f_{Y}(y)=\mathbb E(Y)$$
where $\mathcal Y$ is the support of $Y$. However, I'm trying to give a measure theoretic justification of this. I'm new to conditional expectation as it relates to measure theory so please bear with me. I'm given the hint to integrate over the set $\Lambda\in\mathcal G$. Here I suppose that the hint is referring to $\Lambda$ as being an atom of a $\sigma$-algebra $\mathcal G$?
My initial thoughts were to consider the probability triple $(\Omega,\mathcal F, P)$. We have that $\mathbb E(Y\mid Z)=\mathbb E(Y\mid\sigma(Z))$. From here should I consider $\sigma(Z)$ to be $\mathcal G$? I'm just not sure how $\Lambda$ and $\mathcal G$ come into play here. I assume at some point I need to use $\mathbb P(Y\mid Z)=\mathbb P(Y)$ but I'm not sure how to use it.
I assume the proof is a one-liner but I would appreciate someone explaining it to me like I'm $5$.
 A: We can come back to the definition of a conditional expectation.
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $\mathcal{G} \subset \mathcal{F}$. We define a r.v. $X$  such that $\mathbb{E}[|X|] < \infty$.
The conditional expectation of $X$ knowing $\mathcal{G}$ is defined as any random variable $Y$ verifying the following two conditions:

*

*$Y$ is $\mathcal{G}$-measurable

*For all $A \in \mathcal{G}$, $\int_A Xd\mathbb{P} =\int_A Yd\mathbb{P} $ (i.e. $\mathbb{E}[X\mathbf{1}_A] = \mathbb{E}[Y\mathbf{1}_A]$  a.s.)

In fact, we define $Y = \mathbb{E}[X|\mathcal{G}]$ a.s.
Now using this definition, if $X$ is independent of $\mathcal{G}$ (i.e knowing $\mathcal{G}$ does not give any information on $X$) then we can show that $\mathbb{E}[X|\mathcal{G}] = \mathbb{E}[X]$. Let's verify the two points:

*

*$\mathbb{E}[X]$ is a degenerated r.v. integrable and $\mathcal{G}$-mesurable

*Let $A \in \mathcal{G}$, we have $\mathbb{E}[Y\mathbf{1}_A] = \mathbb{E}[X\mathbf{1}_A] = \mathbb{E}[\mathbf{1}_A]\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X]\mathbf{1}_A]$
The second equality comes from the independence.
