Hardy–Littlewood maximal inequality on measurable sets

Question

The Hardy–Littlewood maximal inequality states that for $d ≥ 1$, $1 < p ≤ ∞$, and $f ∈ L^p(\mathbb{R}^d)$, there is a constant $C_{p,d} > 0$ such that

$$||Mf||_{L^p(\mathbb{R}^d)}\leq C_{p,d}||f||_{L^p(\mathbb{R}^d)},$$ where

$$Mf(x)=\sup_{r>0}{\dfrac{1}{|B(x,r)|}}\int_{B(x,r)}|f(y)|dy.$$ Is it true that $$\Big(\int_A |Mf(x)|^pdx\Big)^{1/p}\leq C_{p,d}\Big(\int_A |f(x)|^pdx\Big)^{1/p} $$

where A is a measurable set?

Answer 1

No if e.g. $f$ has support outside of $A$ the RHS is zero but the LHS is non-zero if $f$ has non-vanishing integral in ${\Bbb R}^d$ and $A$ has positive measure.