In Folland's Real Analysis, Second Edition, he proves the chain rule as Proposition 3.9:
''Suppose that $\nu$ is a $\sigma$-finite signed measure and $\mu, \lambda$ are $\sigma$-finite measures on $(X, \scr{M})$ such that $\nu < \!\! < \mu$ and $\mu < \!\! < \lambda$.
a. If $g \in L^1(\nu)$, then $g(d\nu / d\mu) \in L^1(\mu)$ and
\begin{equation} \int g d\nu = \int g \frac{d\nu}{d\mu}. \end{equation}
b. We have $\nu < \!\! < \mu$, and
\begin{equation} \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda} \;\;\;\; \mu-\text{a.e.''} \end{equation}
Assume that (a) has been proven. In Folland's proof of (b), he immediately asks the reader to set $g = \chi_E (d\nu / d\mu)$, in preparation to use (a), where $E \in \scr{M}$ is arbitrary.
My question is this: In the case that $|\nu|(E) = \infty$, how do we know that $g \in L^1(\mu)$, as the hypotheses of (a) demands. Of course, if it were the case that $|\nu|(E) < \infty$, then $\chi_E \in L^1(\nu)$ and so $g \in L^1(\mu)$ by applying (a) to $g' = \chi_E$. But in the case that $|\nu|(E) = \infty$, it seems that we must rely on the $\sigma$-finiteness of $\nu$ to show $g \in L^1(\mu)$.
I am fairly sure that this is a straightforward application of the definitions plus the MCT (or some variant of this idea), but I have not been able to put the pieces together correctly, it seems. Any clarification would be greatly appreciated.
Thanks.
Best Answer
The following is the original proof in Folland:
Note that in the beginning of the proof, we have the assumption that $\nu\geq 0$. In the case that $\nu(E)=\infty$ (with assumption $\nu\geq 0$), instead of showing $g=\chi_E(d\nu/d\mu)$ is in $L^1(\mu)$, one needs to use the $\sigma$-finiteness to reduce the proof to the finite case.