I am working on the book Measure, Integration and Real Analysis by Sheldon Axler. I am stuck on Problem 9 of Section 4A. For $h: \mathbb{R} \to \mathbb{R}$ Lebesgue measurable, $h^*$ is defined as follows:
$$ h^*(x)=\sup_{r>0}\frac{1}{|B(x,r)|} \int_{B(x,r)} |h(y)|dy.$$
Suppose $h: \mathbb{R} \to \mathbb{R}$ is Lebesgue measurable. Prove that $$\left\{ b \in \mathbb{R} : h^{\ast}(b) > c\right\}$$ is an open subset of $\mathbb{R}$ for every $c \in \mathbb{R}$.
My Idea is like this: Let $A=\left\{ b \in \mathbb{R} : h^{\ast}(b) > c\right\}$. Pick $b \in A.$ I want to show that there is a ball $B(b,\epsilon)$ contained inside of $A$. But I don't know how to get to this. Any help?
Fix $c$. Let $b$ be such that $\sup_r \frac1{|B(b,r)|}\int_{B(b,r)}|h|>c$. Then there exists $r>0$ and $\alpha>1$ such that $$\frac1{|B(b,r)|}\int_{B(b,r)}|h|> \alpha c.$$ Let $\epsilon>0$ and $b'\in B(b,\epsilon)$. Then $$ B(b',r+\epsilon)\supset B(b,r)$$ and $$c< \frac1{\alpha|B(b,r)|}\int_{B(b,r)}|h| \le \frac1{\alpha|B(b,r)|}\int_{B(b',r+\epsilon)}|h| = \frac{|B(b',r+\epsilon)|}{\alpha |B(b,r)|}\frac1{|B(b',r+\epsilon)|}\int_{B(b',r+\epsilon)}|h|$$ since (independently of $b$ and $b'$) $$ \frac{|B(b',r+\epsilon)|}{\alpha |B(b,r)|}\xrightarrow[\epsilon\to 0]{} \frac1\alpha < 1$$ there exists $\epsilon>0$ sufficiently small such that for all $b'\in B(b,\epsilon)$, $$c< \frac1{|B(b',r+\epsilon)|}\int_{B(b',r+\epsilon)}|h|$$ and hence $c < h^*(b')$ as needed.