Define measures by $\nu(E) = \int_E f d\mu$ and $\eta(E) = \int_E g d\mu$. Is $\nu \ll \eta$? If yes, find the Radon-Nikodym derivative. Let $(X,\Sigma,\mu)$ be a finite measure space and $f(x),g(x)$ be $\mu$-measurable functions that are positive for all $x \in X$. Define measures by 
$$\nu(E) = \int_E f \, d\mu \ \text{ and  } \eta(E) = \int_E g \, d\mu.$$ Is $\nu \ll \eta$?
Here's what I've thought about to far:


*

*It's true that $\nu \ll \eta$ if $g$ is a simple function: suppose $g(x) = \sum_{i=1}^n c_i \chi(E_i)$ is a simple function with $E_i \cap Ej = \emptyset$ and  $c_i > 0$ for all $i,j=1,\ldots, n$. If $\eta(E) = 0$, then


\begin{align*}0  = \eta(E) & = \int g \, d\mu & \\
& = \int \sum_{i=1}^n c_i \chi(E_i) \, d\mu\\
& = \sum_{i=1}^n c_i \mu(E\cap E_i).
\end{align*}
Since $c_i >0$, this implies and that $\mu(E\cap E_i) = 0$ for all $i=1,\ldots,n$. Thus,
$$
\mu(E) = \mu(\cup_{i=1}^n E \cap E_i) = \sum_{i=1}^n \mu(E \cap E_i) = 0.
$$
So $E$ is a $\mu$-null set. Thus, $\nu(E) = \int_E f \,d\mu  = 0$. So $\nu \ll \eta$. 


*If it is true that $\nu \ll \eta$, then the Radon-Nikodym derivative is $d\nu/d\eta = f/g$. Suppose that $d\nu/d\eta = h$. Then for any $E\in \Sigma$
$$
\int_E f \, d \mu = \nu(E) = \int_E h  \, d\eta = \int_E hg \, d\mu.
$$
Hence, $h = f/g$ $\mu$-a.e. But also $\eta \ll \mu$ so $h = f/g$ $\eta$-a.e., which proves the claim. Edit: I just realized that by definition $f = d\nu/ d\mu$ and $g = d\eta/d\nu$. So by the chain rule $d\nu/ d\eta = (d\nu/d\mu \cdot d\mu / d\eta) = f/g$ $\eta$- a.e. 


*

*Does $\nu \ll \eta$ hold for any measurable function $g$?
Edit: I answered my own question below. 

*Is the Radon-Nikodym derivative correct and what is a general strategy for computing the Radon-Nikodym derivative?
 A: Any non-negative measurable function can be written as a sort of "infinite" simple function. I'll leave it to you to verify but there exist constants $a_n > 0$ and measurable sets $E_n$ with the property that
$$ g = \sum_n a_n \chi_{E_n}$$ everywhere on $X$. The monotone convergence theorem simplifies things:
$$\eta(E) = \int_E g \, d\mu = \int_E \sum_n a_n \chi_{E_n} \, d\mu = \sum_n a_n \int_E \chi_{E_n} \, d\mu = \sum_n a_n \mu(E \cap E_n).$$
Your proof should work from there.
A: Proof: Suppose that $\eta(E) = 0$. Then 
$$
\mu(E) = \int \chi_E d \mu  \leq \int g\chi_E d \mu = \int_E g \, d\mu = \eta(E) = 0.
$$
Thus, $E$ is a $\mu$-null set (in fact, this shows that $\mu \ll \eta,\nu$). It follows that 
$$
\nu(E) = \int_E f \, d\mu = 0.
$$
This proves that $\nu \ll \eta$ (and by symmetry $\eta \ll \nu$) . A simlar argument shows that $\nu \ll \eta$ as well. Since $\mu$ is finite, the measures $\eta,\nu$ are $\sigma$-finite (at most). Further, it is easily seen that $\eta,\nu \ll \mu$ and $\mu \ll \eta,\nu$. Thus, we can apply the chain rule, we have
$$
\frac{d\nu}{d\eta} = \frac{d\nu}{d\mu}\frac{d\mu}{d\eta} = f/g
$$
and similarly
$$
\frac{d\eta}{d\nu} = \frac{d\eta}{d\mu}\frac{d\mu}{d\nu} = g/f.
$$
