This was talked about in the notes (see for instance D. Chua's notes for the course taught by M. Lis, which was based on the same class), and I don't understand why it is true.
Basically, suppose $(X, W)$ are Gaussian random variables, and our goal (quoting D. Chua's notes) is to compute the conditional expectation $E(X|W)$. Suppose we random variable $Y$ such that $EX = EY$, $X - Y$ is independent of $W$, and $Y$ is $W$-measurable, then $Y = E(X|W)$, since $E(X - Y)1_A = 0$ for all $\sigma(W)$-measurable $A$.
My question is: I know we are supposed to use $X - Y$ being independent of $W$ together with $EX = EY$ over entire probability space $\Omega$ to show $E(X - Y) 1_A = 0$ for all $\sigma(W)$-measurable $A$, where $1_A$ is the indicator function, but I cannot figure out how to do it.
For your question the answer is $I_A$ is independent of $X-W$ so $E(X-W)I_A=E(X-W)P(A)=0$.
However, you don't have to go through the definition of conditional expectation to find $E(X|W)$. Choose $a$ such that $E[(X-aW)W]-E(X-aW)EW=0$. Then $E(X|W)=E(X-aW)|W)+E(aW|W)=EX-aEW+aW$ because $X-aW$ and $W$ are independent. [ If jointly normal random variables have covariance $0$ then they are independent].