Lebesgue measurablity of Hardy Littlewood maximal function 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 ?  
 A: 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.  
A: 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.
