Let $\mu$ be a finite Borel measure on $R$. Show that $|\{x\in R : \sup_{r>0} \frac{1}{2r} \mu ([x-r,x+r]) \ge \lambda \}| \le \frac{C}{\lambda} \mu(R)$ for any $\lambda > 0$ and some absolute constant $C > 0$.
Get stuck with this one. Any idea is appreciated.
On
I assume $\mu$ is a non-negative measure. If $\mu = f(x) \, dx$ for some $f \in L^1$ (i.e. if $\mu$ is absolutely continuous with respect to Lebesgue measure), then we know it to be true by Hardy-Littlewood. Now replace the left hand side by $$ |\{x\in\mathbb R:r^{-1} \int \phi(r^{-1}x) \, d\mu(x)>\lambda\}| $$ where $\phi:\mathbb R \to [0,1]$ is such that $\phi(x) = 1$ if $|x|\le 1$, and $\phi(x) = 0$ if $|x|\ge 2$, and $\phi$ is continuous. Note that $$ |\{x\in\mathbb R:r^{-1} \int \phi(r^{-1}x) \, d\mu(x)>\lambda\}| \le 2C \mu(\mathbb R) $$ if $\mu$ is absolutely continuous with respect to Lebesgue measure.
Now if $\mu$ is any finite, Borel measure, take a sequence of measures $\mu_n$ that are absolutely continuous with respect to Lebesgue measure, such that $\mu_n$ converges weakly to $\mu$, remembering that the space of measures is the dual of $C_0(\mathbb R)$, and then maybe use the Lebesgue dominated convergence theorem on the left hand side.
Use Chebychev inequality (see here) and the fact that since $\mu$ is finite, it must be radon, and there exists a distribution function $F$ corresponding to $\mu$.