Measures induced by integral and by measurable mapping Suppose $(X, \mathcal{F}, u)$ is a measure space. 
$f: X \rightarrow \mathbb{R}$ is a measurable mapping. Then $f$ induces a measure $v_f$ on $(X, \mathcal{F})$ as $v_f(A):=\int_A f du$, 
$g: X \rightarrow X$ is a measurable mapping. Then $g$ induces  a measure $w_g$ on $(X, \mathcal{F})$ as $w_g(A):=u(g^{-1}(A))$.
My question are:


*

*Are  there  some relations between
$V:=\{ v_f: \forall \text{
    measurable }f: X \rightarrow
    \mathbb{R} \}$ and $W:=\{ w_g:
    \forall \text{ measurable }g: X
    \rightarrow X \}$?

*One thing I noticed is that $v_f$
must be absolutely continuous wrt
$u$, while $w_g$ may not be required
as such. Is it true that $V
    \subseteq W$?


Thanks and regards!
 A: As stated, (2) is not true. Take $\mu$ be a measure so that $\mu(X) = 1$. Let $f$ be the function $f\equiv 2$. Then $v_f = 2\mu$, and so $v_f(X) = 2$. By the definition of $w_g$, for any self-map $g:X\to X$, $w_g(X) = \mu(g^{-1}(X)) \leq \mu(X) = 1$. Hence $V$ is not a subset of $W$. 
The reverse is also not true. Take $\mu$ to be an arbitrary atomless measure. Let $X_0\in \mathcal{F}$ be a set with positive measure, and let $\{x_0\}\in \mathcal{F}$ be a point. Then the map $g:X\to X$ such that $g|_{X\setminus X_0} = Id$, and that $g(y) = x_0$ for any $y\in X_0$ is measurable. But $\mu(\{x_0\}) = 0$, while $w_g(\{x_0\}) = \mu(X_0) > 0$, so $w_g$ is not absolutely continuous w.r.t $\mu$. 
In general I don't think there can be any relationships between those two sets you defined: one is the pushforward of $\mu$ under automorphisms of $X$, the other being essentially the class of all absolutely continuous measures w.r.t. $\mu$, it is not clear to me why one may expect the two to be connected, unless perhaps very stringent requirements on what $\mathcal{F}$ and $\mu$ one is allowed to take is made. 
