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 ?

  • $\begingroup$ The definition of Lebesgue measurability that I've seen is weaker than Borel measurability, and thus every Lebesgue measurable function is Borel measurable. $\endgroup$ – Ilya Mar 19 '13 at 21:22
  • $\begingroup$ @Ilya Well, Borel measurable implies inverse image of Borel sets are Borel, but we require inverse image of Lebesgue measurable sets to be Lebesgue measurable. All Borel sets are Lebesgue measurable but converse is not true. $\endgroup$ – smiley06 Mar 20 '13 at 5:23
  • $\begingroup$ I see, then I confused it with the situation when the preimages of Borel sets are Lebesgue, like the 2nd paragraph here. I would be thus interested if you can provide a source for the definition you meant $\endgroup$ – Ilya Mar 20 '13 at 7:26

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.

  • $\begingroup$ I didn't get your point... $\endgroup$ – user39843 May 5 '13 at 16:32
  • $\begingroup$ Where are you stuck? $\endgroup$ – icurays1 May 5 '13 at 21:46
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    $\begingroup$ If $B=B(x, r)$ then shouldn't the supremum in the definition of the Hardy Littlewood maximal function be over all $r>0$ and not $x\in B$? $\endgroup$ – Nirav Feb 20 '14 at 3:46
  • $\begingroup$ @Nirav this is implicit - the supremum is over all balls containing $x$, though not necessarily centered at $x$. $\endgroup$ – icurays1 Feb 20 '14 at 3:50

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