# 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:

1. Are there some relations between $V:=\{ v_f: \forall \text{ measurable }f: X \rightarrow \mathbb{R} \}$ and $W:=\{ w_g: \forall \text{ measurable }g: X \rightarrow X \}$?
2. 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!

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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.
+1: Nice examples. Maybe a better way of proceeding is to restrict the functions $f$ and $g$ one is allowed to take. E.g. integrable for $f$ and essentially surjective for $g$ –  t.b. May 12 '11 at 15:13
@Theo You need stronger than integrable. You need $f$ to effectively have "mass at most 1". Or, a more interesting question could be, for every $f$ taking values in $\mathbb{R}_+$ (or maybe its closure), can we find a positive $\lambda$ such that $v_{\lambda f} \in W$. And I don't know the answer to that. –  Willie Wong May 12 '11 at 15:29
@Tim: the two answers/comments you linked to are actually very strongly tied to the translation and scaling invariance of the Lebesgue measure on $\mathbb{R}^n$. For starters, on a general measure space you don't have the structure of a differentiable manifold, and there really is no appropriate notion of Jacobian determinant. –  Willie Wong May 12 '11 at 15:43