Characterization of elements of $X^*$ via the Radon-Nikodym theorem

I am reading Lindenstrauss' Classical Banach Spaces II and I am having trouble with the following characterization of integrals. First a couple of preliminary definitions:

Let $(\Omega, \Sigma, \mu)$ a complete $\sigma$-finite space. A Banach space $X$ consisting of equivalence classes, modulo equality almost everywhere, of locally integrable real valued functions on $\Sigma$ is called a Köhte function space if the following conditions hold:

1. If $\vert f (\omega) \vert \le \vert g(\omega) \vert$ a.e. on $\Omega$ with $f$ measurable and $g \in X$, then $f \in X$ and $\Vert > f \Vert \le \Vert g \Vert$.
2. For every $\sigma \in \Sigma$ with $\mu ( \sigma ) < \infty$ the characteristic function $\chi_\sigma$ of $\sigma$ belongs to X.

Also we have that:

Every measurable function $g$ on $\Omega$ such that $gf \in L_1(\mu)$ for every $f \in X$ defines an element $x_g^* \in X^*$ by $x_g^*(f) = \int_\Omega > f(\omega) g(\omega) d\mu$. Any functional of the form $x_g^*$ is called an integral.

Then the book states (this is the part that I haven't been able to work around):

It is an immediate consequence of the Radon-Nikodym theorem that a functional $x^* \in X^*$ is an integral is and only if, for every sequence $\{ f_n \} \in X$ with $f_n ( \omega) \downarrow 0$ a.e. we have that $\vert x^* \vert (f_n) \to 0$.

I don't understand how exactly this follows from the Radon-Nikodym theorem. Could anyone clarify it to me?

• When you say $gf\in L^1$ you mean that $gf\in L^1$ for all $f\in X$, right? – David C. Ullrich Apr 28 '16 at 15:08
• @DavidC.Ullrich Yes, I forgot to add that. I have edited the post to reflect it. – A. A. Apr 28 '16 at 15:09
• Also what does $|x^*|(f)$ mean? Is that $|x^*(f)|$ or something else? – David C. Ullrich Apr 28 '16 at 15:09
• The reason I suspect it may be something else is that if $\nu$ is a complex measure then $|\nu|(E)\ne|\nu(E)|$. Also if it just meant $|x^*(f)|$ it's hard to see why he'd include the absolute value sign when he says $|x^*|(f_n)\to0$. So: What is $|x^*|(f)$? – David C. Ullrich Apr 28 '16 at 15:21

Somehow we reduce to the case $\mu(\Omega)<\infty$. Define a complex measure $\nu$ by $$\nu(E)=x^*(\chi_E).$$Detail: Something shows somehow that $\nu$ is in fact countably additive. Now $$x^*\phi=\int\phi\,d\nu$$for integrable simple functions $\phi$. Detail: Somehow, maybe simple functions are dense, this extends to $\phi\in X$.
And now somehow that last condition shows somehow that $\nu<<\mu$.