# Proving that the Radon-Nikodym derivative is finite

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?

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.

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