This question maybe embarrassingly simple, but still I wish to ask whether the Hardy Littlewood maximal function is lebesgue measurable. I know it is Borel measurable as it is lower semi continuous if the function is locally integrable. Is there any shorthand proof of Lebesgue measurablity ?
[Math] Lebesgue measurablity of Hardy Littlewood maximal function
calculusmeasure-theoryreal-analysis
Best Answer
Recall that $f:\Bbb{R}^d\rightarrow \Bbb{R}$ is Lebesgue measurable if $\{f>\alpha\}$ is open for every real number $\alpha$ (this follows from the standard definition that $f$ is measurable if $f^{-1}([-\infty,\alpha))$ is measurable).
Then, let the maximal function be defined as usual $$ Mf(x)=\sup_{B\ni x}\frac{1}{\vert B\vert}\int_B\vert f(y)\vert dy $$
Now, $\{Mf>\alpha\}$ is open since if $y\in \{Mf>\alpha\}$, there is a ball $B$ such that $y\in B$ and
$$ \frac{1}{\vert B\vert}\int_B\vert f\vert >\alpha $$ And, for any other $x\in B$, we have
$$ Mf(x)\geq\frac{1}{\vert B\vert}\int_B\vert f\vert>\alpha $$and hence $x\in \{Mf>\alpha\}$ as well.