Math and science::Analysis::Tao, measure::02. Lebesgue measure

Outer Lebesgue measure of countable union of almost disjoint boxes

If a set is expressible as a countable union of almost disjoint boxes, then what is it's outer Lebesgue measure?

Outer Lebesgue measure of countable union of almost disjoint boxes

Let \( E = \bigcup_{n=1}^{\infty} B_n \) be a countable union of almost disjoint boxes. Then

\[ m^*(E) = \sum_{n=1}^{\infty}|B_n| \]

What else equals the RHS above? We can say the following:

For a countable union of disjoint boxes, the Lebesgue outer measure is equal to the Jordan inner measure.

Outer Lebesgue measure of countable union of almost disjoint boxes. Proof

Claim 1: \( m^*(E) \le \sum_{n=1}^{\infty}|B_n| \). Proof. This follows from countable sub-additivity:

\[ m^*(E) \le \sum_{n=1}^{\infty} m^*(B_n) = \sum_{n=1}^{\infty}|B_n| \]

The right equality follows from the Lebesgue outer measure of a box being the box's volume (Lemma 1.2.6 by Tao. Covered by another Anki card). It's interesting to note here that Jordan outer measure could not make this claim.

Claim 2: \( m^*(E) \ge \sum_{n=1}^{\infty}|B_n| \). Proof. For any natural \( N > 0 \), \( B_1 \cup ... \cup B_N = \bigcup_{n=1}^{N} B_n \) is a subset of \( E \). So, by monotonicity of Lebesgue outer measure, \( m^*(E) \ge m^*(\bigcup_{n=1}^{N} B_n) \); and by Lemma 1.2.6 \( m^*(\bigcup_{n=1}^{N} B_n) = m(\bigcup_{n=1}^{N} B_n) \); finally, this means that \( m^*(\bigcup_{n=1}^{N} B_n) \ge \sum_{n=1}^{N} |B_n| \), as the boxes are all disjoint, making the Jordan measure equal the sum of the box volumes. This is true for all \( N \), and taking the limit as \( N \to \infty \) we arrive at the real:

\[ \lim_{n \to \infty} \sum_{n=1}^{N}|B_n| = \sum_{n=1}^{\infty}|B_n| \]

Lebesgue outer measure equals Jordan inner measure (for countable almost disjoint boxes). Proof.

Let \( E = \bigcup_{n=1}^{\infty} B_n \) be a countable union of disjoint boxes. From above, we know that \( m^*(E) = \sum_{n=1}^{\infty} |B_n| \). Claim: \( m_{*,(J)}(E) = \sum_{n=1}^{\infty}|B_n| \). The limit of a monotone increasing sequence is its supremum (note that \( m_{*,(J)}(E) \) is a supremum), and as box volumes are always positive we must have

\[ m_{*,(J)}(E) = \lim_{N \to \infty} \sum_{n=1}^{N} |B_n| = \sum_{n=1}^{\infty} |B_n| \]

Visual intuition

Below is visual justification for why a countable union of disjoint boxes has a Jordan inner measure and Lebesgue outer measure which are equal and finite, yet it doesn't necessarily have a finite Jordan outer measure.

The Lebesgue outer measure can cover the countably infinite boxes with a 1-1 cover (for every box, cover the box with itself).
The Jordan inner measure is the limit as the number of finite boxes being filled approaches infinity. Effectively, the Jordan inner measure can be imagined to ignore filling in a countable number of boxes yet incur an arbitrarily small loss of volume compared to the Lebesgue outer measure.
The Jordan outer measure can't 'ignore' any boxes: it must cover the whole set.

A case where the Jordan outer approaches infinity yet the Jordan inner measure converges is where there is an infinite number of boxes spaced evenly apart in a line with decreasing volume, say \( |B_n| = \frac{1}{n^2} \)). The Jordan inner measure will represent a convergent limit of filling in these boxes, whereas the Jordan outer measure will be forced to cover an infinite volume.

Source

p23