I have an elementary question on the definition of Hardy-Littlewood maximal function (which is a question on the definition of $\sup$ really).
Do we have that
$$Mf(x) = \sup_{r>0}\frac{1}{m(B_r(x))}\int_{B_r(x)} f(y) dy $$
is equal to $$Mf(x) = \sup_{R>0}\sup_{r \in (0,R)}\frac{1}{m(B_r(x))}\int_{B_r(x)} f(y) dy ? $$
Let $f:(0,\infty) \rightarrow \mathbb{R}$. We want to show that the following two quantities are equal:
\begin{align} A &= \sup \{ f(r) : 0 < r < \infty \} \\ B &= \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} \end{align}
First remark that by definition, $A$ is an upper bound of $\{ f(r) : 0 < r < \infty \}$. Since this set includes $ \{ f(r) : 0 < r < R \}$ as a subset, $A$ is also an upper bound for this set for any $R$. But since the supremum of a set is less or equal than any upper bound, we have
$$ \sup \{ f(r) : 0 < r < R \} \leq A $$
for every $0 < R < \infty$. Hence $A$ is an upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$.
To show that $A$ is the supremum of this set, we must show that it is the least upper bound. Suppose that $D$ is another upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$. Let $0 < r < \infty$, and pick $R=r+1$. Then we know that we must have $D \geq \sup \{ f(s) : 0 < s < R \}$. We know $f(r)$ is in this set, since $0 < r < r+1$, and since the supremum is an upper bound, it is greater than $f(r)$. Therefore, $D \geq f(r)$, and this holds for every $0 < r < \infty$.
Thus we conclude that $D$ is an upper bound on $\{ f(r) : 0 < r < \infty \}$. But since $A$ is the supremum of that set, we have that $A \leq D$. Hence we have shown that $A$ is less or equal to any upper bound of $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$, and since we have also shown that $A$ itself is an upper bound of this set, we get, by definition:
$$ A = \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} = B $$
And this is what we wanted to show.