TITLE: Characterization of $\sigma(Y)$ measurable random variable
QUESTION [4 upvotes]: I cannot solve the following question. Please provide me some helps for dealing with this problem.
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $Y$ be a mesurable random variable with respect to $\mathcal{F}$. Then define $$\mathcal{G} = \sigma(Y) = Y^{-1}(\mathcal{B}_\mathbb{R}) = \{Y^{-1}(A) : A \in \mathcal{B}_\mathbb{R}\}$$ where $\mathcal{B}_\mathbb{R}$ is a Borel sigma algebra on $\mathbb{R}$. If $X$ is $\mathcal{G}$-measurable, then $X$ must be a function of $Y$, i.e., there exists a measurable function $f$ such that $$X = f(Y).$$

REPLY [2 votes]: First assume that $X$ is the characteristic function of an element of $\mathcal G$: which function $f$ would you choose?
Then assume that $X$ is a linear combination (with non-negative coefficients) of characteristic function of measurable sets (simple function). Since it is assumed to be $\mathcal G$-measurable, these sets are in $\mathcal G$. Which function $f$ would you choose?
Now, if $X$ is non-negative and $\mathcal G$-measurable, we can find a sequence of non-negative simple functions $(X_n)_{n\geqslant 1}$ such that $X_n(\omega)\uparrow X(\omega)$ for each $\omega\in \Omega$. We can write $X_n=f_n(Y)$. It remains to show that the sequence $\left(f_n(x)\right)_{n\geqslant 1}$ is convergent for $x\in Y(\Omega)$.
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