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I constructed an example in which the pushforward of a $\sigma$-finite measure is itself not $\sigma$-finite.

Let $\lambda$ be the Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. This measure is $\sigma$-finite.

I know that there exists an homeomorphism from the intervals $(a,b],\ a<b$ to the real line. I denote it by $f'$. Now since $f'$ is continuous because it is a homeomorphism, it is also a Borel measurable map.

Then I slightly modify $f'$ to be $f\colon \mathcal{B}(\mathbb{R})\rightarrow \mathcal{B}(\mathbb{R})), (a,b]\mapsto (a,\infty)$

Since, $f$ is still a homeomorphism, I just picked the image measure $f^{-1}_*\lambda=\lambda\circ f$ which is obviously not $\sigma$-finite.

This construction seems to trivial, so I assume there must be some mistake I don't see.

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    $\begingroup$ Consider a constant function, like $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=0$. Then $f_*(\lambda)(A)=\lambda(f^{-1}(A))$ is $0$ if $0\notin A$ and $\infty$ if $0\in A$. $\endgroup$ Nov 9, 2019 at 13:32
  • $\begingroup$ @conditionalMethod This also came to my mind, as also does the projection. But I wanted to force the example in my post to work. Do you have any idea if this can be done? $\endgroup$ Nov 9, 2019 at 13:39

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The half-open, half-closed interval $(a,b]$ is not homeomorphic to the real line.

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  • $\begingroup$ The rest doesn't really make sense, too. I have only defined the map on the intervals. $\endgroup$ Nov 9, 2019 at 13:31

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