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.

measure-theory

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:

  1. 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 $\nu \ll \eta$.

  1. 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?

Best Answer

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.

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