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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 ?

smiley06
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    The definition of Lebesgue measurability that I've seen is weaker than Borel measurability, and thus every Lebesgue measurable function is Borel measurable. – SBF Mar 19 '13 at 21:22
  • @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. – smiley06 Mar 20 '13 at 05:23
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    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 – SBF Mar 20 '13 at 07:26

2 Answers2

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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.

icurays1
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  • I didn't get your point... – user39843 May 05 '13 at 16:32
  • Where are you stuck? – icurays1 May 05 '13 at 21:46
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    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 $x\in B$? – Nirav Feb 20 '14 at 03:46
  • @Nirav this is implicit - the supremum is over all balls containing $x$, though not necessarily centered at $x$. – icurays1 Feb 20 '14 at 03:50
  • @icurays1 There is an exercise which asked to prove the same given the balls are centered at the point where we are evaluating the function...in this case how to prove {Mf> alpha} is open? – mathgirl Sep 08 '20 at 17:58
  • One should note that $Mf$ is a function with codomain $\overline{\mathbb{R}}$, although your argument still works. – Simon SMN Dec 15 '22 at 16:12
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It seems you confused with the definition of Lebesgue measurable.

Certainly, a function is said to be Lebesgue measurable, if preimage of every Borel (not Lebesgue!) set is Lebesgue measurable. The preimage of a Lebesgue measurable set may not be Lebesgue measuable, even when $f$ is continuous. See this post for the most classical example.

icurays' answer is right if one consider the ball is not necessarily centered at $x$. Here is a proof which is also work for the case of centered balls.

Assume $$Mf(x):=\sup_{x\in B}\frac{1}{|B|}\int_B |f|,$$ where the supremum is over all balls centered at $x$. Using the absolute continuity of integral, the supremum can be replaced by considering all balls centered at $x$ with rational radicals. Given $r\in \mathbb{Q}$, put $$F_r(x):=\frac{1}{|B(x,r)|}\int_{B(x,r)}|f|.$$ Using the absolute continuity of integral again, we conclude that $F_r$ is continuous. $Mf$ is the point supremum of $\{F_r\}_{r\in \mathbb{Q}}$. So $Mf$ is Borel measurable, hence Lebesgue measurable.

Landau
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