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Let $\mu$ be a regular Borel Measure on $\mathbb{R}$ and suppose that there exists a $C > 0$ such that for every $x \in \mathbb{R}$ and $r > 0$ \begin{align*} \mu((x - r,x+r)) \leq Cr. \end{align*} Let m be the Lebesgue measure on $\mathbb{R}$. I have shown that $\mu << m$ ($\mu$ is absolutely continuous wrt m). I now want to show that the Radon-Nikodym derivative of $\mu$ wrt m is in $L^{\infty}(\mathbb{R},m)$.

I know that $\mu$ and m are two $\sigma$-finite measures and $\mu << m$. So, by the Radon-Nikdoym Theorem, there exists a positive non-negative measure function f so that \begin{align*} \mu(E) = \int_E f dm \end{align*} for every measurable set E. So, I need to show that $f \in L^{\infty}(\mathbb{R},m)$. So I need to show that for every $N \in \Sigma$ with $m(N) = 0$ there exists $C > 0$ so that \begin{align*} \left\vert{f(s)}\right\vert \leq C \end{align*} for every $s \in N^c$.

I am not sure how to proceed/how to relate this to the above integral or the first equation. Any ideas or recommendations would be greatly appreciated. Thank you in advance.

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First you have $f \ge 0$, so you do not need to worry about taking absolute values.

Hints:

  1. Can you show $\mu( A ) \le d \, m(A)$ for all $A$ and some constant $d > 0$?
  2. Assume that $f \ge c$ on some (measurable) set $A$. What does this tell you about $\mu(A)$ and $m(A)$?
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  • $\begingroup$ 1. Yes I showed this. 2. So we assume $f \not\in L^{\infty}(\mathbb{R},m)$. Then for every $M > 0$ there exists a measurable set N with $m(N) = 0$ so that $f(s) > M$ for some $s \in N^c$. In particular we have a set N with $m = 0$ so that for some $s \in A = N^c$, $f(s) > C$. So $\mu(A) = \int_A f(s) dm > \int_A C dm = Cm(N^c)$. However, this contradicts the first fact. $\endgroup$ – K.Mor Jun 13 at 18:46
  • $\begingroup$ Your solution to 2 is not entirely correct. If $f \not\in L^\infty(m)$, then for every $M > 0$ there is a set $A$ of positive measure with $f > M$ on $A$. $\endgroup$ – gerw Jun 13 at 19:33

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