maximal function bounded below by original function

Question

Is is true that for a locally integrable function, we always have $Mf(x)\geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.

Answer 1

If $x$ is a Lebesgue point of $f$, then $Mf(x) \geq |f(x)| $. Indeed, we have $$ \tag{1} Mf(x) = \sup_{x\in B} \frac{1}{|B|} \int_B |f(y)| dy \geq \lim\limits_{r\to 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| dy = |f(x)|, $$ where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).