# Probability and Measure: Sigma-finite

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: $\Omega$ -> $\Omega$'

• Does not imply what, where $T$ is what? – Robert Israel Oct 7 '14 at 2:39
• sorry about that – klib Oct 7 '14 at 2:49

## 2 Answers

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.

• Thanks this really helps. I would vote this up but I do not have the required reputation. – klib Oct 7 '14 at 14:59

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 $\{\emptyset,\mathbb{N}\}$.