A question about measurable functions 
This is an exercise problem I am stuck at. I think I have to construct the function f based on X and T, starting from X being a nonnegative simple function However, I can't find a reasonable way to construct such f.
Could anyone help me with this?
 A: First we will prove the statement when $X$ is a simple function which is $\sigma(T)$-measurable. Notice that any set in $\sigma(T)$ is of the form $T^{-1}(B)$, for some $B \in \mathcal{A'}$. Notice that $1_B(T)=1_{T^{-1}(B)}$, which is easy to show directly. Now, any simple function $X$ which is $\sigma(T)$-measurable has the form $X=\sum_{i=1}^n a_i 1_{T^{-1}(B_i)}$, for some $B_i \in \mathcal{A'}$. In this case define $f=\sum_{i=1}^n a_i 1_{B_i}$. Then $f(T)=\sum_i a_i1_{B_i}(T)=\sum_i a_i1_{T^{-1}(B_i)} = X$. This proves the statement for simple functions.
Now, for any nonnegative $\sigma(T)$-measurable function $X$, we can find a sequence $X_n$ of simple functions such that $X_n \uparrow X$. For each $n$, since $X_n$ is a simple function, we can use what we just proved to find some $\mathcal{A'}$-measurable function $f_n$ such that $X_n=f_n(T)$. Define $f=\sup_n f_n$. Then $f_n(T)=X_n \uparrow X$, so it follows that $X=\sup_n f_n(T)=f(T)$. This proves the statement when $X$ is nonnegative.
Now, if $X$ is any $\sigma(T)$-measurable function, we write $X=X^+-X^-$, where $X^+=\max\{X,0\}$ and $X^-=-\min\{X,0\}$. Then $X^+$ and $X^-$ are nonnegative, so from what we just proved, we can find $\mathcal{A'}$-measurable functions $f_+$ and $f_-$ such that $X^+=f_+(T)$ and $X^-=f_-(T)$. Then $X=(f_+-f_-)(T)=:f(T)$.
