Let $(Y,S)$ a measurable space and $\phi :X\to Y$ any function where $X\neq \emptyset$.
Suppose that $f:X\to\mathbb{R}$ is measurable over $(X,S')$, where $S'=\phi^{-1}(S)$. I want to prove that there exists $g:Y\to\mathbb{R}$ measurable such that $f=g\circ\phi$.
How can we define such $g$?
Thanks!
This is a result due to Doob and Dynkin.
When $f$ is the characteristic function of an element of $\mathcal S'$, it is clear.
By linearity, we can deal with the case where $f$ is a linear combination of characteristic functions of measurable sets (simple function).
If $f$ is measurable and non-negative, there is a sequence $(f_n)_{n\geqslant 1}$ a sequence of simple functions such that for each $x\in X$, $f_n(x)\uparrow f(x)$. Consider the $g_n$ obtained in 2. and notice that the sequence $(g_n)_{n\geqslant 1}$ is pointwise non-decreasing. The limit will be the wanted $g$.
To solve the general case, write $f$ as a difference of two non-negative measurable functions.
