Understanding Folland 3.9 Proposition: Radon-Nikodym Derivative

Question

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 $\int g d\nu=\int g \frac{d\nu}{d\mu}d\mu$

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

I don't understand why this is true for g=$X_E$ by the definition of $d\nu/d\mu$? 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 $d\nu=fd\mu$ for some f, $\nu(E)=\int_E f d\mu$ for all E and denote f by $d\nu/d\mu$. I guess thus we have $\nu(E)=\int_E \frac{d\nu}{d\mu} d\mu$, which I guess represents $d\nu=\frac{d\nu}{d\mu} d\mu$.

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

Answer 1

By definition, $d\nu / d\mu$ is a $\mu$-locally integrable function for which

$$\int_A \frac{d\nu}{d\mu}\, d\mu = \nu(A)$$

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

$$\int_X g \frac{d\nu}{d\mu}\, d\mu = \int_X \chi_E \frac{d\nu}{d\mu} \, d\mu = \int_E \frac{d\nu}{d\mu} \, d\mu = \nu(E)$$

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

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 $\mu$.

Answer 2

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

$$ \nu(E) = \int_E \frac{d\nu}{d\mu}d\mu \tag{1} $$

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

$$ \int\chi_E d\nu = \nu(E) = \int_E \frac{d\nu}{d\mu}d\mu = \int\chi_E\frac{d\nu}{d\mu}d\mu. \tag{2} $$

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