> [!theorem] > > Let $f \in \loci$ be a [[Locally Integrable|locally integrable]] function, and define the **Hardy-Littlewood maximal function** $Hf$ by > $ > Hf(x) = \sup_{r > 0}A_r\abs{f}(x) = \sup_{r > 0}\frac{1}{m(B(r, x))}\int_{B(r, x)}\abs{f(y)}dy > $ > then $Hf$ is [[Measurable Function|measurable]]. > > *Proof*. Let $a \in \real$, then $(Hf)^{-1}((a, \infty)) = \bigcup_{r > 0}(A_r\abs{f})^{-1}((a, \infty))$. Since each $A_r\abs{f}$ is jointly continuous, $(Hf)^{-1}$ is measurable. > [!theorem] The Maximal Theorem > > There exists $C > 0$ such that for all $f \in L^1$ and $\alpha > 0$, > $ > m(\bracs{x: Hf(x) > \alpha}) \le \frac{C}{\alpha} \int \abs{f(x)}dx > $ > *Proof*. Let $E_\alpha = \bracs{x: Hf(x) > \alpha}$, then for each $x \in E_\alpha$, there exists $r_x > 0$ such that $A_{r_x}\abs{f}(x) > \alpha$. Hence the balls $B(r_x, x)$ cover $E_\alpha$. Let $c < m(E_\alpha)$, then there exists $\bracs{x_j}_1^k \subset E_\alpha$ such that $\bracs{B(x_j, r_{x_j})}_1^k$ are disjoint with $\sum_{j \in [k]}m(B(x_j, r_{x_j})) > 3^{-n}c$. Since $A_{r_x}\abs{f}(x) > \alpha$, > $ > \frac{1}{\alpha }\int_{B(x_j. r_x)} \abs{f} > m(B(x_j, r_{x_j})) > $ > so as the balls are disjoint > $ > c < 3^n\sum_{j \in [k]}m(B(x_j, r_{x_j})) < \frac{3^n}{\alpha}\int_{B(x_j, r_{x_j})}\abs{f} \le \frac{3^n}{\alpha}\int \abs{f} > $ > as this holds for all $c < m(E_\alpha)$, we have the desired inequality.