Absolutely Continuous Measure and the Radon-Nikodym Derivative

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.

First you have $$f \ge 0$$, so you do not need to worry about taking absolute values.
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 $$m(A)$$?
• 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. – K.Mor Jun 13 at 18:46
• 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 $A$. – gerw Jun 13 at 19:33