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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!

1 Answers 1

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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.

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    If you are willing to work over differential manifolds, however, then for diffeomorphisms (smooth invertible maps) the two concepts can be identified since now the measure/volume form can be both pushed forward and pulled back (using either the map or the inverse map) and seen to be identical. Smoothness will guarantee that the pushforward measure is absolutely continuous, has a RN derivative, and is given by the change of variables formula. But this amounts to throwing on a lot of extra structures to make the problem nice enough.2011-05-12