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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!

Tim
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1 Answers1

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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.

Willie Wong
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  • +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
  • @Willie: The motivation of my questions is whether in integral, change of measures by Radon-Nikodym derivative and change of variables wrt a measure and its push-forward are related somehow. I can't see there is connection. Also see my previous question http://math.stackexchange.com/questions/31905/is-integration-by-substitution-a-special-case-of-radonnikodym-theorem, and comments on http://math.stackexchange.com/questions/38361/radonnikodym-derivative-and-normal-derivative/38364#38364 . – Tim May 12 '11 at 15:17
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    @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
  • @Willie: Oh, yeah, right, silly mistake. I'd have to think a little (and I'm honestly not that interested). – t.b. May 12 '11 at 15:37
  • @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
  • 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. – Willie Wong May 12 '11 at 15:59