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Suppose $\lambda,\mu$ are $\sigma$-infinite measures on the measurable space $(X,\mathcal{X}),$ with $\lambda\ll\mu$ So, I need to prove that the Radon-Nikodym derivative $f$ can be taken to be finite-valued in $X$. This is a question on Bartle's book.

I have a few questions:

1) What exactly means $\sigma$-infinite measure? It is just not a $\sigma$-finite measure?

2) This is not what the Radon-Nikodym prove?

3) If 2) is incorrect, how can I prove this?

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I believe your theorem should be revised as the following:

Suppose $\lambda, \mu$ are $\sigma$-finite measures on the measurable space $(X,\mathcal{X})$, with $\lambda \ll \mu$. Show that the Radon-Nikodym derivative $f$ can be taken to be finite-valued in $X$.

Note I simply changed your term of $\sigma$-infinite to $\sigma$-finite. Like @copper.hat, I have never heard of $\sigma$-infinite.

The theorem has some value since the Radon-Nikodym does not directly state the derivative can be taken to be finite-valued.

Proof:

Assume, on the contrary, $B\equiv f^{-1}(\{\infty\}) \subset X$ has $\mu(B) > 0$. Since $\lambda$ is $\sigma$-finite, we can find a sequence of sets $A_n$ so that $A_n \uparrow X$, and $0 < \lambda(A_n) < \infty$. It follows that $\mu(B\cap A_n) > 0$ for sufficiently large $n$. By Randon-Nikodym theorem, $\lambda(A_n) = \int_{A_n}fd\mu \ge \int_{B\cap A_n}fd\mu=\infty$. This contradicts the condition that $\lambda$ is $\sigma$-finite. Since $\mu\left(f^{-1}(\{\infty\})\right) = 0$, we can always take the Random-Nikodym derivative to be finite-valued by zeroing out those data points with infinite values.

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  • $\begingroup$ Just to be sure, what do you call $\Omega$ is the whole set $X$, right? And the $U$ set is actually the $B$ set? $\endgroup$ – Mateus Rocha Oct 27 '19 at 22:21
  • $\begingroup$ @MateusRocha, yes, corrected. Thanks for noticing. $\endgroup$ – Xiaohai Zhang Oct 28 '19 at 2:12

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