Why is a a pushforward of a $\sigma$-finite measure not necessarily $\sigma$-finite? What is an example that shows that if $\mu$ is a $\sigma$-finite measure on $\Omega$ and $T:\Omega\to\Omega'$ is measurable then $\mu \cdot T^{-1}$ is not necessarily $\sigma$-finite? I have a basic understanding of what $\sigma$-finite means but if $\mu$ is finite that implies $\mu \cdot T^{-1}$ is finite, so why isn't this also true for $\sigma$-finite?
 A: 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.
A: 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 $\{\emptyset,\mathbb{N}\}$.
