Proof for factorization lemma I am confused by the proof of Factorization lemma.
Specifically the part "f is a step function" confuses me.
$f$ has a representation where all $A_i$ are disjunctive. I prefer to work with this representation further on.
The proof constructs $g$ and claims that this function is exactly what we need: 


*

*$g$ is $\sigma(T)-\mathcal{B}(\overline{\mathbb{R}})$ measurable
(and it is clear because $\forall i$ $A'_i \in \mathcal{A}'$),

*$f=g \circ T$ (and that I do not see).


Indeed, take $\omega \in A_i$, then it follows $f(w)=\alpha_i$, also follows that $T(w) \in A'_i$, but there might be $A'_j$ such that $T(w) \in A'_j$ because sets $A'_i$ must not be disjunctive as we assumed for sets $A_i$. Thus $g \circ T(w)=\alpha_i+\alpha_j \ne f(w)$.  
 A: I guess I see where you problem lies. Assume you have a step function 
$$
f=\sum_{i=1}^n  \alpha_i 1_{A_i} 
$$
where $A_i \in \sigma(T)$ and for $i\ne j$ we have $A_i\cap A_j=\emptyset$. We may choose such a representation as you pointed out. Since $T$ is measurable (I am referencing to the Wikipedia link) by construction, we have 
$$
\forall i \text{ there exists }A^{'}_i\in \mathcal{A^{'}} \text{ such that } A_i=T^{-1}(A^{'}_i) 
$$ 
But if we assume that the $(A_i)_{i\in I}$ are disjoint, so are the $(A^{'}_i)_{i\in I}$. And therefore no contradiction occurs.
Why is that:
Let's say $A_i=T^{-1}(A^{'}_i)$ and $A_j= T^{-1}(A^{'}_j)$ with $A_i\cap A_j=\emptyset$. Now assume  $A^{'}_j\cap A^{'}_i = C \neq \emptyset$ and take $x\subset C$, then we have 
$$
T^{-1}(x)\subset A_i \text{ and } T^{-1}(x)\subset A_j \Rightarrow A_i\cap A_j\neq\emptyset
$$
which would be of course a contradiction.
bests
A: The sets $A'_i$ have not to be disjunctive. So it is true that if $x\in A'_i\cap A'_j$ (assuming that $x\notin A_k, k\neq i,j$), then $g(x)=\alpha_i+\alpha_j$. But this is not a problem. When we compose $g\circ T$, the function $g$ will take only values in the image of $T$, then things like $(g \circ T)(w)=\alpha_i+\alpha_j$ can't happen. To see this and conclude that $g\circ T = f$, let $x\in \Omega$. So $T(x)\in\Omega'$.
First suppose that $x\in\bigcup_{i=1}^n A_i$. Then $x\in A_i=T^{-1}(A_i')$ for some $i\in\{1,2,...,n\}$, that is, $T(x)\in A_i'$ and $T(x)\notin A_j'$ for all $j\neq i$ in $\{1,2,...,n\}$. In fact, if $T(x)\in A_j'$, then $x\in T^{-1}(A_j')=A_j$, what implies $x\in A_i\cap A_j$, that is, $i=j$. Nice. So $g(T(x))=\alpha_i=f(x)$ for all $x\in\bigcup_{i=1}^n A_i$.
Now, take $x\in\Omega\backslash(\bigcup_{i=1}^n A_i)$. Then $T(x)\notin A_i'$ for all $i\in\{1,2,...,n\}$. In fact, if $T(x)\in A_i'$ for some $i\in\{1,2,...,n\}$, then $x\in T^{-1}(A_i')=A_i$, what is an absurd $(x\notin\bigcup_{i=1}^n A_i)$. Therefore $g(T(x))=0=f(x)$ for all $x\in\Omega\backslash(\bigcup_{i=1}^n A_i)$.
Thus $g\circ T = f$, as we wanted.
You can find a nice proof of this theorem in the book Measure and Integration Theory by Heinz Bauer. Consider ask a new question to clarify new misunderstandings about the proof.
