I have a proof for the following statement
Let $h:\Bbb R \to \Bbb R $ Lebesgue measurable and set the Hardy-Littlewood maximal function $h^*:\Bbb R \to [0,\infty ]$ by $$h^*(b):=\sup_{t>0}\frac1{2t}\int \chi _{(b-t,b+t)}|h|$$ Show that $\{x\in \Bbb R :h^*(x)>c\}$ is open for each $c>0$.
I had read in MSE that the proof of this statement is "easy", however my proof is a bit long so I guess that probably there is a shorter one. Can you show me a shorter one? Moreover, if I had a mistake in the proof below, can you show me it? Thank you.
My proof:
We set $$ g(b,t):=\frac1{2t}\int\chi _{(b-t,b+t)}|h|\tag1 $$ Then for arbitrary $\delta >0$ we can see that $$ g(b\pm\delta ,t \pm\delta )=\frac{\int_{b-t }^{b+t }|h|+r}{2t +2\delta }\geqslant g(b,t )\left(\frac{t }{t +\delta }\right)\tag2 $$ for some $r\geqslant 0$.
Now WLOG assume that $\int|h|>0$, then its easy to check that $h^*>0$, therefore for any chosen $b\in \Bbb R $ there are $t>0$ such that $g(b,t)>0$ and from $\rm (2)$ we also can see that for any chosen $\epsilon_t \in (0,g(b,t))$ there is $\delta_{\epsilon,t} > 0$ such that $$ g(b,t )\left(\frac{t }{t +\delta_{\epsilon,t} }\right)\geqslant g(b,t)-\epsilon_t \implies g(b\pm\delta_{\epsilon,t} ,t\pm\delta_{\epsilon,t} )+\epsilon_t\geqslant g(b,t) \tag3 $$
Case 1: $h^*(b)<\infty $
Suppose that $h^*(b)<\infty $ for some fixed $b$, and because $h^*(b)=\sup_{t>0}g(b,t)$, then for any chosen $\gamma >0$ there is some $t_\gamma >0$ such that $h^*(b)<\gamma+g(b,t_\gamma )$. Then from $\rm (2)$ and $\rm (3)$ we find that $$ \forall s\in (0,\delta _{\epsilon_{t_\gamma },t_\gamma }):h^*(b)<\gamma+\epsilon _{t_\gamma }+g(b\pm s,t_\gamma \pm s)\tag4 $$ So we can conclude that if $h^*(b)<\infty $ then $$ \forall \epsilon >0,\exists t\in \Bbb R ,\exists \delta >0,\forall s\in (0,\delta ):h^*(b)<\epsilon +g(b\pm s,t\pm s)\tag5 $$ what imply that $$ \begin{align*} \forall \epsilon >0,\exists \delta >0:|s|<\delta &\implies h^*(b)\leqslant \epsilon +h^*(b+ s)\\ &\implies (h^*(b+s)\geqslant h^*(b))\,\lor\,(|h^*(b+s)-h^*(s)|\leqslant \epsilon ) \end{align*} \tag6 $$ so is clear that $b$ is an interior point of $\{x\in \Bbb R :h^*(x)>c\}$ for any $c<h^*(b)$.
Case 2: $h^*(b)=\infty $
If $h^*(b)=\infty $ then for any chosen $n \in\Bbb N$ there is some $t_n>0$ such that $g(b,t_n)\geqslant n$ and by $\rm (2)$ we can see that there is a $\delta _n>0$ such that $g(b+s,t_n+s)\geqslant n\left(\frac{t _n}{t_n+\delta_n}\right)$ for any $|s|<\delta _n$. Then setting $\delta _n:=t_n$ we can see that $g(b+s,t_n+s)\geqslant n/2$, and taking the supremum we find that $h^*(b+s)\geqslant n/2$ for all $|s|<t_n$.
Thus $(b-t_n,b+t_n)\subset \{x\in \Bbb R :h^*(x)>n/3\}$ holds for all $n\in\Bbb N$, thus by the Archimedean property of the real numbers we finally find that $b$ is an interior point of $\{x\in \Bbb R :h^*(x)>c\}$ for any $c>0$.$\Box$