Show that if $Y$ is $\sigma (X)-$measurable, there is a measurable function $f:\mathbb R\to \mathbb R$ s.t. $Y=f(X)$.
I really have problem to show that. It's clear that if $Y=f(X)$, then $Y$ is $\sigma (X)$ measurable. But the converse looks complicate.
Attempt
Step 1 : Let $Y=\boldsymbol 1_A$ for $A\in \sigma (X)$. In particular, there is a $B\in \mathcal B(\mathbb R):=\{borel\ set\ of\ \mathbb R\}$ s.t. $$X^{-1}(B)=A. $$ If I set, $$f(x)=\boldsymbol 1_B(x),$$
then $$Y(\omega )=\boldsymbol 1_A(\omega )=\boldsymbol 1_{X^{-1}(B)}(\omega )=\boldsymbol 1_{B}(X(\omega ))=f(X(\omega )).$$
Step 2 : If $Y$ is simple, the same argument work.
Step 3 : If $Y$ is positive, there is a sequence $Y_n$ of simple function s.t. $Y_n\nearrow Y$. Let $(f_n)_n$ s.t. $$Y_n=f_n(X).$$
Since $Y_n$ are a.s. increasing, $(f_n(X(\omega ))_n$ is a.s. increasing. But how can I prove that is converge to $f(X(\omega ))$ for some measurable $f$ ? I just have that $(f_n(X(\omega )))_n$ converge to $f(X(\omega ))=Y(\omega )$, but it doesn't tell me that the function $f:\mathbb R\longrightarrow \mathbb R$ obtained is Lebesgue measurable.
Best Answer
Let $\{Y_n\}_n$ be a sequence of $\sigma(X)$-measurable simple functions converging pointwise to $Y$. You know that the result holds for simple functions so for each $n$ let $f_n$ be the corresponding Borel function such that $Y_n=f_n(X)$. That $\{Y_n\}_n$ converges pointwise this tells us that $\{f_n\}_n$ converges pointwise on $X(\Omega)$. Now let $A\subset \mathbb{R}$ denote the Borel measurable set of the points $s\in \mathbb{R}$ for which $\{f_n(s)\}_n$ converges pointwise (we just showed that $X(\Omega)\subset A)$. Then we can define the Borel function $f:=\lim\limits_{n \rightarrow \infty}\mathcal{X}_Af_n$, and by simply taking pointwise limits on both sides of the equality $Y_n=f_n(X)$ we see that $Y=f(X)$.