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What is a example that shows that $\mu$ $\sigma$ -finite does not imply $\mu \cdot T^{-1}$. I have a basic understanding of what $\sigma$-finite means but if $\mu$ is finite implies $\mu \cdot T^{-1}$ is finite, why isn't this also true for sigma-finite? Thanks.

edit: T is a mapping T: $\Omega$ -> $\Omega$'

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  • $\begingroup$ Does not imply what, where $T$ is what? $\endgroup$ Oct 7, 2014 at 2:39
  • $\begingroup$ sorry about that $\endgroup$
    – klib
    Oct 7, 2014 at 2:49

2 Answers 2

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So $\mu$ is a $\sigma$-finite measure on set $\Omega$ with $\sigma$-algebra $\mathcal B$, and $T$ is a measurable mapping from $(\Omega, \mathcal B)$ into $(\Omega', \mathcal B')$ where $\mathcal B'$ is a $\sigma$-algebra of subsets of $\Omega'$. Then $\mu \circ T^{-1}$ is a measure on $(\Omega',\mathcal B')$.

For example, take $\Omega = \Omega' = \mathbb Z$ with $\sigma$-algebra $\mathcal B$ consisting of all subsets of $\mathbb Z$, $T$ the identity map, and $\mu$ counting measure, but let $\mathcal B'$ be the trivial $\sigma$-algebra $\{\emptyset, \Omega'\}$. $\mu \circ T^{-1}$ is not $\sigma$-finite because no nonempty members of $\mathcal B'$ have finite measure.

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  • $\begingroup$ Thanks this really helps. I would vote this up but I do not have the required reputation. $\endgroup$
    – klib
    Oct 7, 2014 at 14:59
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Here is a simple counter-example. Consider the measure space $(\mathbb{N},2^{\mathbb{N}},\#)$ where $\#$ is the counting measure. The function $T\equiv1$ is $2^\mathbb{N}-\{\emptyset,\mathbb{N}\}$ measureble. However, $\# T^{-1}$ is not $\sigma$--finite since it takes the values $0$ and $\infty$ on $\{\emptyset,\mathbb{N}\}$.

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