# Why is 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?

• Does not imply what, where $T$ is what? Oct 7, 2014 at 2:39
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.