For two random variables, if $\sigma(Y) \subset \sigma (X) $ then $Y $ is a function of $X $? I found the statement in the title above in a book on statistics by Mark J. Schervish (but only put in words so I'm a litle bit unsure if I understood it correctly) and I would like to know (1) if it means that $Y=g(X) $ for some $\sigma(X)/\sigma (Y)$-measurable function $g $, and (2) how could I prove the statement?
Thanks in advance!
 A: This is sort of a classical case of abstract nonsense. The statement is that if $Y$ is real-valued and $\sigma(X)$-measurable, then there exists a Borel measurable $T$ such that $Y=T(X)$.
To see this, note that $Y=\lim_{n\to\infty}\sum_{k=-\infty}^{\infty} k2^{-n}1_{(k2^{-n}\leq Y<(k+1)2^{-n})}$. Now, if $Y$is $\sigma(X)$-measurable, then, for every $k$ and $n$, we have
$
(k2^{-n}\leq Y<(k+1)2^{-n})\in\sigma(X), 
$ and every element of this $\sigma$-algebra has the form $(X\in B)$ for some appropriate Borel set $B$. 
Accordingly, pick $B_{n,k}$ such that $(k2^{-n}\leq Y<(k+1)2^{-n})=(X\in B_{n,k})$. Then, we have now argued that
$$
Y=\lim_{n\to\infty}\sum_{k=-\infty}^{\infty} k1_{(X\in B_{n,k})},
$$
so, fixing $T(x)=\limsup_{n\to\infty}\sum_{k=-\infty}^{\infty} k1_{(x\in B_{n,k})}$ works, since limit superiors of Borel maps are again Borel (we use $\limsup$ instead of $\sup$ to avoid problems with convergence for defining the map).
A: Trivial, if that happens:$\quad\quad Y=E[Y|X]=f(X)$
(The correct thing would be to put: $Y=E[Y|\sigma(X)]=E[Y|X]\circ X=f(X)$)
What follows is a little hint for one of the comments.:
$$E[I_B(Y)X]=\int_B dP_Y \int xdP_{X/Y}(y)=\int_{Y\in B}dP_{\sigma(Y)} \int xdP_{X/Y}\circ Y(\omega)$$
also
$$E[I_B(Y)X]=\int_{Y\in B}dP_{\sigma(Y)} \int xdP_{X/\sigma(Y)}(\omega)$$
from where:
$$E[X|\sigma(Y)](\omega)= E[X|Y]\circ Y(\omega)$$
without the need for the Doob-Dynkin-Lemma.
