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

Lebesgue Measure. Definition

The Jordan measure has limitations. Tweak the Jordan measure to arrive at the Lebesgue measure.

Jordan measure on \( \mathbb{R}^d \)

Recap. The development of Jordan measure proceeded as follows:

Boxes: First, one defines the notion of a box \( B \) and its volume \( |B| \).
Elementary sets: Define the notion of an elementary set \( E \) (a finite union of boxes) and define the elementary measure \( m(E) \) of such sets.
Jordan inner and outer measure: Define the inner and outer Jordan measures, \( m_{*,(J)}(F) \) and \( m^{*,(J)}(F) \), of an arbitrary bounded set \( F \subset \mathbb{R}^d \). These are limits of elementary measure of elementary sets that are either contained in (inner) \( F \) or contain (outer) \( F \).
Jordan measurability: If the inner and outer Jordan measures match for \( F \), we say that \( F \) is Jordan measurable and call \( m(F) := m_{*,(J)}(F) = m^{*,(J)}(F) \) the Jordan measure of \( F \).

Jordan measure limitations

This concept of measure is perfectly satisfactory for any sets that are Jordan measurable. However, not all sets are Jordan measurable: the classic example is the bullet riddled square, \( [0,1]^2 \setminus \mathbb{Q} \) and the bullets \( [0,1]^2 \cap \mathbb{Q} \)—both of these sets have Jordan outer measure 1 and Jordan inner measure 0.

More power to the Jordan outer measure

Trying to measure non-Jordan measurable sets leads us to develop the Lebesgue Measure.

Let's tinker with the Jordan outer measure to give it more power. The Jordan outer measure for a set \( F \subset \mathbb{R}^d \) is defined as:

\[ m^{*,(J)}(F) := \inf_{F \subseteq E; \, E \text{ is elementary}} m(E) \]

Jordan outer measure

As an elementary set is made up of boxes, we can rewrite the Jordan outer measure definition as:

\[ \begin{aligned} m^{*,(J)}(F) &:= \inf_{F \subseteq B_1 \cup ... \cup B_k;\, B_1, ..., B_k \text{ are boxes}} |B_1| + ... + |B_k| \\ &= \inf_{F \subseteq \cup_{n=1}^{k}B_n;\, B_1, ..., B_k \text{ are boxes}} \sum_{n=1}^{k} |B_n| \end{aligned} \]

Focus on the bit under then infimum. In words, the Jordan measure is the infimal cost (or volume) required to cover \( F \) by a finite union of boxes.

Lebesgue outer measure

The tweak: allow a countable union of boxes instead of just a finite union. This is the Lebesgue outer measure of \( F \):

\[ m^{*}(F) := \inf_{F \subseteq \cup_{n=1}^{\infty}B_n; \, B_1, ..., \text{ are boxes}} \sum_{n=1}^{\infty} |B_n| \]

Can you spot the tiny tweak?

In words, the Lebesgue outer measure is the infimal cost (or volume) required to cover \( F \) by a countable union of boxes.

Outer measures comparison

Some points of comparison between Jordan outer measure and Lebesgue outer measure.

Lebesgue outer measure is always less than Jordan outer measure

\( m^{*}(F) \le m^{*,(J)}(F) \), as we can pad out finite set of boxes with infinite empty boxes, which won't change the volume sum.

Countable sets have Lebesgue measure 0

Covering a countable set with zero-volume boxes, one for each element of the set, covers the set completely, and the volume sum is zero. For Jordan outer measure this is not the case. For example, if \( F = \mathbb{Q} \cap [-M, M] \), then \( m^{*,(J)}(F) = 2M \) (the closure of \( F \) is the set of reals \( [-M, M] \), which has outer measure \( 2M \), and for any set \( F \), \( m^{*,(J)}(F) = m^{*,(J)}(\bar{F}) \)).

No concept of Lebesgue inner measure

There is no increase in power gained by Jordan inner measure if the finite union is replaced by infinite union. Tao mentions that this boils down to the fact that elementary measure is subadditive rather than superadditive (which I'm yet to appreciate properly).

No more power to be added by switching to uncountable boxes

If the countable union of boxes of Lebesgue measure is replaced by an uncountable union of boxes, then any subset of \( \mathbb{R}^d \) would have outer measure 0, as we could place every element of the set in a box of zero volume.

13.06.2020