📂Measure Theory

The Maximal Lemma

Theorem¹

Let $\mathcal{B}$ be a collection of open balls in $\mathbb{R}^n$. Let $U=\bigcup \limits_ { B\in \mathcal{B}} B$. Then, for any constant $c \lt m (U)$, there exist finitely many mutually disjoint $B_{j} \in \mathcal{B}$ satisfying the following condition.

$$ \dfrac{c}{3^{n}} \lt \sum \limits_{j=1}^{k} m(B_{j}) $$

Here, $m$ is the $n$-dimensional Lebesgue measure.

Explanation

This theorem is not actually named the maximal lemma; the name was assigned loosely because it is used as a lemma in the maximal theorem.

It guarantees that a finite set ${B_{j}}$ whose measure lies between $m(U)$ and $c/3^{n}$ must exist.

Proof

First, there exists a compact set $K \subset U$ satisfying $c< m (K) \le m (U)$². Then, by the definition of compactness, there exists a sub cover $\left\{ A_{i} \right\}_{1}^l$ of $K$. Now, among these, let the largest³ one be $B_{1}$. Among the $A_{i}$ that are disjoint from $B_{1}$, let the largest be $B_2$. And among the $A_{i}$ disjoint from $B_{1}$ and $B_2$, let the largest be $B_{3}$. In this way, we can construct a finite collection $\left\{ B_{j} \right\}$.

For an $A_{i}$ that could not belong to $\left\{ B_{j} \right\}$, there exists a $j$ satisfying $A_{i} \cap B_{j} \ne \varnothing$. Moreover, for the smallest such $j$⁴, the radius of $A_{i}$ is at most that of $B_{j}$. That is, it cannot be larger than the radius of $B_{j}$. If it were, then $A_{i}$ would have taken the name $B_{j}$ when constructing $\left\{ B_{j}\right\}$⁵.

Now let $B^{\ast}_{j}$ be the open ball with the same center as $B_{j}$ and three times the radius. Then $A_{i}$ has a radius no larger than that of $B_{j}$ and overlaps with $B_{j}$, so it is necessarily contained in $B^{\ast}_{j}$. Therefore $K \subset \bigcup A_{j} \subset \bigcup B^{\ast}_{j}$.

$$ \begin{align*} c \lt m (K) & \lt m \left( \bigcup \nolimits_{1}^k B^{\ast}_{j}\right) \\ &= \sum \limits_{1}^{k} m (B^{\ast}_{j}) \\ &= \sum \limits_{1}^{k} 3^{n} m (B_{j}) \end{align*} $$

$$ \implies \dfrac{c}{3^{n}} \lt \sum \limits_{j=1}^{k} m(B_{j}) $$

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Definition

For every bounded measurable set $K \subset \mathbb{R}^n$, a function $f : \mathbb{R}^n \rightarrow \mathbb{C}$ satisfying

$$ \int_{K} |f(x)|dx<\infty $$

is said to be locally integrable (with respect to the Lebesgue measure), and the set of locally integrable functions is denoted by $L^{1}_{\mathrm{loc}}$.

Let $f \in L^1_{\mathrm{loc}}$, $ x\in \mathbb{R}^n$, $r>0$. Let us denote the open ball with center $x$ and radius $r$ by $B(r,x)=B_{r}(x)$. Then we define the average of the function values $A_{r}f(x)$ of $f$ over $B_{r}(x)$ as follows.

$$ A_{r} f(x) := \frac{1}{m \big( B_{r}(x) \big)} \int _{B_{r}(x)}f(y)dy $$

$A_{r}$ is called the averaging operator.