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Suppose that $nu$ is a $sigma$-finite signed measure and $mu$, $lambda$ are $sigma$-finite measures on $(X,M)$ such that $nu<<mu$ and $mu << lambda$.
a. If g $in L_1(mu)$ then
$int g dnu=int g frac{dnu}{dmu}dmu$

Folland's Proof:
By considering $nu+$ and $nu-$ separately we may assume $nu geq 0$. The equation $int g dnu=int g frac{dnu}{dmu}dmu$ is true when g=$X_E$ (indicator function on E), by definition of $dnu/dmu$.

I don't understand why this is true for g=$X_E$ by the definition of $dnu/dmu$? Could someone explain why this is true?

This is what I know:
the decomposition $nu=rho+lambda$ where $lambda$ is mutually singular to $mu$ and $rho << mu$ is called the Lebesgue decomposition of $nu$ with respect to $mu$.
In the case where $nu << mu$, we have $dnu=fdmu$ for some f, $nu(E)=int_E f dmu$ for all E and denote f by $dnu/dmu$. I guess thus we have $nu(E)=int_E frac{dnu}{dmu} dmu$, which I guess represents $dnu=frac{dnu}{dmu} dmu$.

All the notation is very confusing, so I'm very confused why when g=$X_E$ we can say $dnu=frac{dnu}{dmu}dmu$ in the integral in the proof.
Any help would be much appreciated. Thanks.

By definition, $dnu / dmu$ is a $mu$-locally integrable function for which

$$int_A frac{dnu}{dmu}, dmu = nu(A)$$

for all $mu$-measurable sets $A$. So if we take $g = chi_E$, then

$$int_X g frac{dnu}{dmu}, dmu = int_X chi_E frac{dnu}{dmu} , dmu = int_E frac{dnu}{dmu} , dmu = nu(E)$$

agreeing with the fact that $int_X g dnu = int_E dnu = nu(E)$.

Note that the singular part of the measure you refer to is zero in this context, because you're assuming that $nu$ is absolutely continuous with respect to $mu$.

Since $nu ll mu$, Radon-Nikodym states that

$$
nu(E) = int_E frac{dnu}{dmu}dmu tag{1}
$$

Now, note that for any measure $lambda$ we have that $lambda(E) = intchi_Edlambda$ and so

$$
intchi_E dnu = nu(E) = int_E frac{dnu}{dmu}dmu = intchi_Efrac{dnu}{dmu}dmu. tag{2}
$$

Now, just let $g = chi_E$ and replace it in $(2)$.