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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.

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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}

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  • $\begingroup$ Thank you for always answering my questions, this is very helpful! $\endgroup$ Apr 18 at 23:35
  • $\begingroup$ I actually do have one follow up question: considering that $\nu(F) = m(F \cap B)$ why does it also equal $\int_{F \cap B} \chi_{B}d\mu$? $\endgroup$ Apr 19 at 1:04
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    $\begingroup$ @TaylorRendon See my edit $\endgroup$
    – angryavian
    Apr 19 at 1:41

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