Why $\{x: Mf(x) > \lambda\} $ is open for any $\lambda \in \mathbb{R}$?

Question

For a measurable function $f$, define the (centered) Hardy-Littlewood maximal function by \begin{equation} Mf(x) = \sup_{r > 0} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)| dy \end{equation} where $|B(x,r)|$ denotes the Lebesgue measure of $B(x,r) := \{y \in \mathbb{R}^n: |x-y| < r\}$.

I want to prove that if $f \in L^1_{\rm loc}(\mathbb{R}^n)$, then for any $\lambda \in \mathbb{R}$, $$\{x: Mf(x) > \lambda\} $$ is open. But I don't know how to prove it.

Answer 1

Hints: $\{x:M(f,x)>\lambda\}=\cup_r \{x: \int_{B(x,r)} |f(y)|dy >\lambda|B(0,r)|\}$ because $|B(x,r)|$ is independent of $x$. Since $f$ is locally integrable you only need to use the fact that $|B(x_n,r)\Delta B(x,r)| \to 0$ as $x_n \to x$ [For, then, $\{x: \int_{B(x,r)} |f(y)|dy >\lambda|B(0,r)|\}$ becomes the inverse image of an open interval under a continuous map]. I will let you finish.