Spring 2010 - Problem 3

Solution.

We will first prove the following covering lemma: let $\mathcal{C} = \set{B\p{x_\alpha, r_\alpha}}_{\alpha \in I}$ be a collection of open balls in $\R$ with $\set{r_\alpha}_\alpha$ bounded, where $I$ is some indexing set. Then there exists a pairwise disjoint subcollection $\mathcal{D} \subseteq \mathcal{C}$ such that for any $B\p{x_\alpha, r_\alpha} \in \mathcal{C}$, there exists $B\p{x_\beta, r_\beta} \in \mathcal{D}$ such that

To prove it, let $R = \sup_{\alpha \in I} r_\alpha$, which is finite by assumption. Partition $\mathcal{C}$ into $\set{\mathcal{C}_n}_n$, where $\mathcal{C}_n$ contains all balls $B\p{x_\alpha, r_\alpha} \in \mathcal{C}$ with radius $r_\alpha \in \poc{\frac{R}{2^{n+1}}, \frac{R}{2^n}}$.

We pick $\mathcal{D}_n$ as follows: let $\mathcal{U}_0 = \mathcal{C}_0$, and let $\mathcal{D}_0$ be a maximal disjoint subcollection of $\mathcal{U}_0$, which exists by, say, Zorn's lemma. Now suppose we have chosen $\mathcal{D}_0, \ldots, \mathcal{D}_n$ along with $\mathcal{U}_0, \ldots, \mathcal{U}_n$. Let

and let $\mathcal{D}_{n+1}$ be a maximal disjoint subcollection of $\mathcal{U}_{n+1}$. Set $\mathcal{D} = \bigcup_{n=0}^\infty \mathcal{D}_n$. We claim that this has the desired properties.

Let $B \neq C \in \mathcal{D}$. Then there exist $n_B$ and $n_C$ such that $B \in \mathcal{D}_{n_B}$ and $C \in \mathcal{D}_{n_C}$. If $n_0 \coloneqq n_B = n_C$, then by construction, $B \cap C = \emptyset$ since $\mathcal{D}_{n_0}$ was chosen to be a disjoint subcollection of $\mathcal{U}_{n_0}$. Otherwise, suppose without loss of generality that $n_B < n_C$. By construction, $C \in \mathcal{U}_{n_C}$ which was chosen to be disjoint from any ball in $\mathcal{D}_1 \cup \cdots \cup \mathcal{D}_{n_C-1} \supseteq \mathcal{D}_{n_B} \ni B$, so $B \cap C = \emptyset$, so $\mathcal{D}$ is a pairwise disjoint subcollection of $\mathcal{C}$.

As for the intersection property, let $B = B\p{x_\alpha, r_\alpha} \in \mathcal{C}$, so there exists $n$ such that $B \in \mathcal{C}_n$, since we had a partition. If $B \in \mathcal{U}_n$, then $B$ intersects a ball $C \in \mathcal{D}_n$, since $\mathcal{D}_n$ was a maximal disjoint subcollection of $\mathcal{U}_n$. Otherwise, by definition of $\mathcal{U}_n$, there exists $C \in \mathcal{D}_0 \cup \cdots \cup \mathcal{D}_{n-1}$ such that $B \cap C = \emptyset$. In either case, $B$ intersects some ball $B\p{x_\beta, r_\beta} \in \mathcal{D}_0 \cup \cdots \cup \mathcal{D}_n$, so there exists $y \in B\p{x_\alpha, r_\alpha} \cap B\p{x_\beta, r_\beta}$. By construction, $r_\beta > \frac{R}{2^{n+1}}$ and $r_\alpha \leq \frac{R}{2^n}$, so $2r_\beta \geq r_\alpha$. For any $x \in B\p{x_\alpha, r_\alpha}$, the triangle inequality gives

Thus, $B\p{x_\alpha, r_\alpha} \subseteq B\p{x_\beta, 5r_\beta}$, as desired.

Notice that $\mathcal{D}$ is at most countable: by density of $\Q$, each $B_\alpha \in \mathcal{D}$ contains some $q_\alpha \in \Q$. Since $\mathcal{D}$ is a pairwise disjoint collection, it follows that $B_\alpha \mapsto q_\alpha$ is an injective map, so $\mathcal{D}$ injects into $\Q$, hence countable.

We now prove the Hardy-Littlewood maximal inequality. By definition, for any $x \in E \coloneqq \set{\p{Mf}\p{x} > \lambda}$, there exists $h_x > 0$ such that

where $I_x = \p{x - h_x, x + h_x}$ and $m$ denotes the Lebesgue measure. By our covering lemma, there exists $\set{x_n}_n$ such that $\set{I_{x_n}}_n$ is a countable disjoint collection of intervals and such that for any $x \in E$, there exists $x_{n_0}$ with $I_x \subseteq 5I_{x_{n_0}}$, which implies $m\p{I_x} \leq 5m\p{I_{x_{n_0}}}$. Thus,

which completes the proof.