$\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$ What's the name for this proposition? Or alternatively what's happening?

If there isn't a name for this like Doob-Dynkin-Brown-Markov Tower Lemma / Theorem, then at least what's going on here so that I can describe this proposition in words?

(I guess the ff is in probability space $(\Omega, \mathcal F, \mathbb P)$.)

Let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}, X, Y$ be two random variables such that $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable, and let $\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a Borel-measurable function such that $\mathbb{E}(|\varphi(X, Y)|)<$$+\infty$. Then$$\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \psi(Y) \quad \text { a.s., } \quad \text { where } \psi(y)=\mathbb{E}(\varphi(X, y)).$$

From here:

1. If $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable, then $\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$

2. Why is Borel measurability needed here?

3. Example 4.1.7 of Durrett PTE

Also I do recall it is in my class notes from before:

Edit 1 for searchability: It's called freezing lemma. I like to call this independence-stability generalisation.

Edit 2: Btw I'm not sure the remark (2) is right unless $X$& $Y$ are independent, in w/c case obviously you choose $\mathcal G = \sigma(Y)$.