$\mathbb{E}(X|Y)$ and $\mathbb{P}(X \in A | Y)$ for r.v.s $X = x$ and $Y=\sin(\pi x)$ on $([0,1], \mathcal{B}_{[0,1]}, dx)$ Suppose we have r.v.s $X = x$ and $Y=\sin(\pi x)$ on $([0,1], \mathcal{B}_{[0,1]}, dx)$.

*

*what is $\sigma(Y)$?

*calculate $\mathbb{E}(X|Y)$

*calculate $\mathbb{P}(X \in A | Y)$ for some borel set A from $\mathcal{B}_{[0,1]}$

*calculate $\mathbb{P}(X \in A | \sigma(Y))$ for some borel set A from $\mathcal{B}_{[0,1]}$
Now, by definition $\sigma(Y) = \{Y^{-1}(B), B \in \mathcal{B}_{[0,1]}\}$. Does that mean $\sigma(Y) = \mathcal{P}([0,1])$? Taking arbitrary set from [0,1] and plugging it into $\sin(\pi x)$ gives sets from $[0,1]$ too. Is that correct?
How can I calculate the rest? I see that $\mathbb{P}(X \in A | Y) = \mathbb{E}(\mathbb{1}_{A}|Y)$ but I usually had densities available to calculate $f_{X|Y}(x,y)$. Here this seems different.
 A: As Kavi Rama Murthy suggests, let us conjecture that
$$\sigma(Y) = \{B\in\mathcal{B}_{[0,1]} \: : \: B \text{ is symmetric around }\frac12\}$$
For the inclusion $"\subseteq"$, note that $\sin(\pi(\frac12+h))=\sin(\pi(\frac12-h))$ for all $h\in[0,\frac12]$ and for the other inclusion $"\supseteq"$ consider a set $B\in \mathcal{B}_{[0,1]}$ symmetric around $\frac12$ and note that $B=Y^{-1}(Y(B))\in \sigma(Y)$.
Now in order to prove $\mathbb{E}[X \:| \: Y] = \frac12$, we would first have to argue that $\frac12$ is $\sigma(Y)$ measurable, which is obvious, and secondly we would need to prove that $\mathbb{E}[X1_B]=\mathbb{E}[\frac12 1_B]$ forall $B\in \sigma(Y)$. Let's verify this
\begin{align*}
\mathbb{E}[X1_B] &= \int_B x \: dx \\
&= \int_{B-\frac12}x \: dx \: + \frac12 \int_{B-\frac12}dx \\
&= 0 + \frac12 \mathbb{E}[1_B] = \frac12\mathbb{P}(B)
,\end{align*}
where we have used that $B-\frac12$ is symmetric around $0$ for $B \in \sigma(Y)$ and also we used the translation invariance of the lebesgue measure.
To find $\mathbb{P}(X\in A \: | \: Y)$ (which is the same as $\mathbb{P}(X\in A \: | \: \sigma(Y))$), we need to find a variable $Z$, such that $Z$ is $\sigma(Y)$ measurable and such that $\mathbb{P}(A \cap B)=\mathbb{E}[Z 1_B]$ for all $B \in \sigma(Y)$. A reasonable guess is
$$Z = \frac12 1_{A} + \frac12 1_{1-A},$$
where $1-A$ is the reflection of $A$ around $\frac12$, which also makes $Z$ symmetric around $\frac12$ and hence $\sigma(Y)$ measurable. Now for $B\in \sigma(Y)$ we compute
\begin{align*}
\mathbb{E}[Z1_B]&=\frac12 \int_{A\cap B} dx + \frac12 \int_{(1-A)\cap B}dx \\
&= \int_{A\cap B}dx = \mathbb{P}(A \cap B),
\end{align*}
here we used that the reflection of $(1-A)\cap B$ around $\frac12$ is $A \cap B$, since $B$ is it's own reflection. And we may conclude that $Z=\mathbb{P}(X\in A \: | \: Y)$.
