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?
 A: 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.
