# Image measure of a $\sigma$-finite measure is not $\sigma$-finite

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 $$(a,b],\ a to the real line. I denote it by $$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.

• 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 $0\in A$. Nov 9, 2019 at 13:32
• @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? Nov 9, 2019 at 13:39

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