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Let $(\Omega, \mathcal{F}_1, \mu)$ be a $\sigma$-finite measure space, Let T:$\Omega \to \mathbb R$ be $\langle\mathcal{F}, B(R)\rangle$-measurable. Could anyone think about an example to show $\mu$'s induced measure $\mu T^{-1}$ needs not to be $\sigma$-finite?

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    $\begingroup$ Take $(\Omega, \mathcal{F}_1, \mu)$ to be $(\mathbb R, B(R), \lambda)$, where $B(R)$ is the Borel $\sigma$-algebra and $\lambda$ is the usual Lebesgue measure. We know that $(\mathbb R, B(R), \lambda)$ is a $\sigma$-finite measure space. Let $T$:$\mathbb R \to \mathbb R$ be defined by $T(x)=0$ for all $x \in \mathbb R$. Then $T$ is $\langle B(R), B(R)\rangle$-measurable. But for all $A\in B(R)$, $\mu T^{-1}(A)$ is either $0$ or $+\infty$. So $\mu T^{-1}$ is NOT $\sigma$-finite. $\endgroup$
    – Ramiro
    Sep 27, 2015 at 17:51
  • $\begingroup$ @Ramiro Thanks for the help. I really learn something. $\endgroup$
    – user268866
    Sep 28, 2015 at 16:22

1 Answer 1

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Consider the case where $\mu(\Omega) = \infty$ and $T(\omega) = 0$ for all $\omega\in \Omega$. Then $\mu T^{-1} (A)$ is either $0$ or $\infty$, so we can't express $\mathbb{R}$ as a union of finite measure sets.

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  • $\begingroup$ is measure $\mu()$ sigma -finite? $\endgroup$
    – user268866
    Sep 27, 2015 at 15:40
  • $\begingroup$ There's no reason why it can't be. It just needs to be such that $\mu(\Omega) = \infty$. That is consistent with being $\sigma$-finite. $\endgroup$
    – user24142
    Sep 27, 2015 at 18:01
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    $\begingroup$ @meng Just take $(\Omega, \mathcal{F}_1, \mu)$ to be any $\sigma$-finite measure space, such that $\mu(\Omega)=+\infty$. $\endgroup$
    – Ramiro
    Sep 27, 2015 at 18:02

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