Vrftsjtryk

Suppose $h: Bbb R to : Bbb R$; Hardy Littlewood maximal function $h^*: Bbb R to [0, infty]$ defined by $h^*(x)=sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h|$.

Now to prove $|h(x)|leq|h^*(x)|$ for almost every $x in Bbb R$

I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $fin L^{1}Bbb R$. Then

$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b in Bbb R$.

My attempt: $||h(x)|-|h^*(x)||leq |h(x)-h^*(x)|=|h(x)-sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(y)|dy|leq sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|leq h^*(x)$ for allmost every $x$.

Is my attempt coorect or is there any other way to approach? Please help

This version of Lebesgue differentiation, which correctly stated is
$$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
$$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
Therefore, for almost every $x$,
$$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$