Let's consider $\Omega = [0,1]$ and Lebesgue measure. Also define $Y(x)=x(1-x)$. I want to prove that $$E[X\mid Y](x)=\frac{X(x)+X(1-x)}{2}$$
My first shot was to use $E[X\mid Y]=\frac{E[X1_Y]}{P(Y)}$ but I thought that it's not a good idea, because I cannot say anything about $P(Y)$ since $Y$ distribution is not known.Can you give me a hint how can I start proving this theorem ? I found it not so trivial due to generic form of variables (no distributions).
Definition of conditional expectation $\mathbb E[X|Y]$ is such a random variable $Z$ that is $\sigma(Y) := \{Y^{-1}[B] ; B \in \mathcal B(\mathbb R)\}$ measurable and for any $D \in \sigma(Y)$ we have $\mathbb E[Z1_D] = \mathbb E[X1_D]$, where $$1_D(x) = \begin{cases} 1 & x \in D \\ 0 & x \not \in D \end{cases}$$
Note that for any $B \in \mathcal B(\mathbb R)$ we get $Y^{-1}[B] = \{ x \in [0,1] : Y(x) \in B \} = \{ x \in [0,\frac{1}{2}] : Y(x) \in B \} \cup \{ x \in [\frac{1}{2},1] : Y(x) \in B\} $
But since for every $x \in [0,\frac{1}{2}]$ we have $Y(x) = Y(1-x)$, then such preimage is in form $A \cup (1-A)$, where $1-A := \{1-x \in [0,1] : x \in A \}$ and $A \subset [0,\frac{1}{2}]$ (and clearly we can get any $A \subset [0,\frac{1}{2}]$ measurable)
Note that function $Z(x)= \frac{X(x) + X(1-x)}{2}$ is $\sigma(Y)$ measurable, since $Z(x)= Z(1-x)$, (proceed analogously as above) such preimages $Z^{-1}[B]$ would be of form $C \cup (1-C)$, too, so in $\sigma(Y)$.
This establishes that $Z$ is $\sigma(Y)$ measurable. Now, we need to take any $D \in \sigma(Y)$. We know it is of form $A \cup (1-A)$ for some $A \subset [0,\frac{1}{2}]$. Hence:
$$ \mathbb E[Z1_D] = \frac{1}{2}\Big(\int_{A} X(x)+X(1-x) dx + \int_{1-A} X(x) + X(1-x)dx\Big) = $$ $$= \frac{1}{2}\Big(\int_{A \cup (1-A)} X(x)dx + \int_{A \cup (1-A)} X(1-x)dx \Big) $$
Now use substitution $y=1-x$ (note that under such transformation, set $D=A \cup (1-A)$ which is symetricc at $\frac{1}{2}$ would remain unchanged), you'll get:
$$ \mathbb E[Z1_D] = \frac{1}{2} \Big( \int_D X(x) dx + \int_D X(y)dy \Big) = \frac{1}{2}\cdot 2\int_D X(x)dx = \int_D X(x)dx = \mathbb E[X1_D]$$