Hardy-Littlewood maximal operator is defined by
$$ Mf(x):= \sup_{B\ni x} \frac{1}{|B|} \int_{B} |f(y)| dy. $$
Here, the supremum is taken over balls $B$ in $\mathbb{R}^n$ which contain the point $x$ and $|B|$ denotes the measure of $B$.
If the balls in the definition is open, then we denote the operator $M_{o}$.
If the balls in the definition is closed, then we denote the operator $M_{c}$.
Can we say that $M_{o}=M_{c}$?