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Consider the problem mentioned in the question as follows.

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

This question has been given a detailed solution in the above-mentioned link. But notice that, by Example 4.1.7 in Durrett's "Probability: Theory and Examples" (fifth edition), we have $$ \mathbb{E}[ \varphi(X, Y) | Y] = \psi(Y). $$ A natural question is: is there a theorem in the book talking about the following: $$ \mathbb{E}[ \varphi(X, Y) | \mathcal{G}] = \mathbb{E}[ \varphi(X, Y) | Y]? $$ This question is also mentioned as equation (*) in link.

Greenhand
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  • Just approximate $\varphi(x,y)$ with functions of the form $(x,y)\mapsto f_1(x)f_2(y)$, and then use what you already know about conditional expectation. – Andrew Jan 15 '25 at 07:14
  • Hi @Andrew. Thanks for your comments. It is an interesting idea but I think it will not be trivial since it hides difficulties in finding the sequence of approximations. It would be better to have a theorem in the book talking about this. – Greenhand Jan 15 '25 at 07:40
  • It's a fact from analysis that such functions are dense. See e.g. https://math.stackexchange.com/questions/4737770/tensor-product-of-lp-spaces-is-dense-in-the-product-lp-space – Andrew Jan 15 '25 at 07:59
  • @Andrew I totally agree with the correctness of this approach. – Greenhand Jan 15 '25 at 08:20

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