# Finding the Radon-Nikodym Derivative for Given Measures $\mu$ and $\nu$

I am currently reading about the Radon-Nikodym derivative and came across a problem in my textbook the author attempts to work through. The problem is as follows:

Given $$(\Omega, \mathcal{F})$$, Let $$\Omega = [0,1]$$ with Lebesgue measure $$m$$ and consider measures $$\mu, \nu$$ given by densites $$\chi_{A}$$, $$\chi_{B}$$ respectively. Find a condition on the sets $$A,B$$ so that $$\mu$$ dominates $$\nu$$ and find the Radon-Nikodym derivative $$\frac{d\nu}{d\mu}.$$

While this question is pretty straightforward, I have a few questions about his work:

First assume that $$m(A) \neq 0$$. Then $$B \subset A$$ clearly implies that $$\mu$$ dominates $$\nu$$. Now consider the partition $$\mathcal{P} := \{B, A \setminus B, \Omega \setminus A\}$$ of $$\Omega$$. Therefore for a set $$F \in \mathcal{F}$$, $$\nu(F)=m(F \cap B) =\int_{F \cap B}\chi_{B}dm=\int_{F \cap B}\chi_{B}d\mu.$$

My Questions:

1. In the first line: I don't think I understand why we are employing the Lebesgue measure to show that $$\mu$$ dominates $$\nu$$, Could someone explain this in more detail since it is not as clear to me as the author makes it out to be? My definition of a measure dominating another is the following: $$\mu$$ dominates $$\nu$$ $$\iff$$ $$0\leq \nu(F) \leq \mu(F)$$ $$\forall F \in \mathcal{F}$$.

2. I understand why we chose the partition $$\mathcal{P} := \{B, A \setminus B, \Omega \setminus A\}$$, however, i'm not following why $$\nu(F) = m(F \cap B)$$ and how we conclude this equals $$\int_{F \cap B}\chi_{B}d\mu$$? Does this maybe have something to do with the Lebesgue decomposition? I am pretty new to this information, so i'm sure I am just overlooking something.

The measures are themselves defined using the Lebesgue measure via $$\mu(F) := \int_F \chi_A \, dm = m(F \cap A).$$ This is what it means when they say "consider measure $$\mu$$ given by density $$\chi_A$$ w.r.t. Lebesgue measure." (In fact, $$\chi_A$$ is actually the Radon-Nikodym derivative of $$\mu$$ with respect to $$m$$. Using the suggestive notation $$\chi_A = \frac{d\mu}{dm}$$ we have $$\mu(F) = \int_F \, d\mu = \int_F \frac{d\mu}{dm} \, dm = \int_F \chi_A \, dm$$.)
If $$B \subseteq A$$ then for any measurable subset $$F$$ we have $$\nu(F) = m(F \cap B) \leq m(F \cap A) = \mu(F)$$ which shows the dominance.
\begin{align} m(F \cap B) &= \int_{F \cap B} \, dm & \text{definition of m} \\ &= \int_{F \cap B} \chi_B \chi_A \, dm & \text{\chi_A(x) \chi_B(x) = 1 for x \in F \cap B} \\ &= \int_{F \cap B} \chi_B \, d\mu & \text{\chi_A \, dm = d\mu by definition of \mu} \end{align}
• I actually do have one follow up question: considering that $\nu(F) = m(F \cap B)$ why does it also equal $\int_{F \cap B} \chi_{B}d\mu$? Apr 19 at 1:04