I need to prove the following theorem:
Let $X$ and $Y$ be two random variables. Then $$ Y = f(X) $$ for some measurable function $f$ if and only if $$ \sigma(Y)\subset\sigma(X). $$
The implication $\Rightarrow$ is straightforward. The other implication $\Leftarrow$ requires the construction of $f$. I have found the following proof:
Suppose that $\sigma(Y)\subset\sigma(X)$. We want to find $f$. For fixed $n$ consider the set $$ A_{n,m}=\left\{\omega\in\Omega\left| m\,\frac{1}{2^n}\leq Y(\omega)<(m+1)\,\frac{1}{2^n}\right.\right\},\quad m=0,\pm 1,\pm 2,... $$ Since $A_{n,m}\in\sigma(Y)$ then $A_{n,m}\in\sigma(X)$ and hence $$ A_{n,m}=\left\{\omega\in\Omega\left| X(\omega)\in B_{m,n}\right.\right\} $$ Consider the sequence of functions $$ f_n(x)=\sum_{m}m\,\frac{1}{2^n}\,I_{\left\{x\in B_{m,n}\right\}} $$
The proof then continues claiming that, for all $x$, the sequence $f_{n}(x)$ is monotonically increasing, and then has a limit. I cannot see this, any suggestion?