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

Title: 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.
BackOnlyText: 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.

Title: The 3 basic properties of Lebesgue outer measure
Q: Reminder that the Lebesgue outer measure is denoted as \( m^{*} \).
The 3 basic propositions of Lebesgue outer measure
Empty set
[...]
Monotonicity
If \( E  \subseteq F \subset \mathbb{R}^d \), then [...].
Countable subadditivity
If \( E_1, E_2, ... \subset \mathbb{R}^d \) is a countable sequence of sets, then [...]
These three ideas are very fundamental. The complex apprearance of the expressions obscures the simplicity of the ideas.
A: \( m^{*}(\emptyset) = 0 \); \( m^{*}(E) \le m^{*}(F) \); \[ m^{*}(\bigcup_{n=1}^{\infty}E_n) \le \sum_{n=1}^{\infty}m^{*}(E_n) \]
BackOnlyText: Compact essence
Monotonicity
If you can cover a whole, you can cover a part.
Subadditivity
If you can cover two parts of something, you can cover the whole.
Proof outline & essense
Empty set
The empty set is equal to any degenerative box, such as \( (1,1) \), and all such boxes have zero volume. Box volume can't be negative, so the covering with the least volume sum must have sum = 0.
This applies to finite sets of boxes (Jordan measure) and countable sets (Lebesgue measure).
Monotonicity
If \( E \subseteq F \), then any set of boxes covering \( F \) will cover \( E \), so \( m^{*}(E) \le m^{*}(F) \).
This line of reasoning would also work for proving monotonicity of Jordan measure in the outer or inner cases separately.
Subadditivity—Essence
First, a few words on the intuitive and abstract meaning of subadditivity. Imagine for a collection of sets, for each set we have a lower bound on the volume needed to cover it with boxes. Taking the union over the collection, we get another set, say \( S \). We know that if we combine all the boxes used to cover the individual sets we will cover the set \( S \). So we have found an upper bound to the smallest volume needed to cover \( S \).
Subadditivity—Finite subcase
Consider first, a finite union of arbitrary sets. Let \( J = J_1 \cup J_2 \cup ... \cup J_k \) for finite \( k \). Each set \( J_i \) has a Jordan outer measure, \( m^{*,(J)}(J_i) \). For any number \( M_i \) larger than this, we can find an elementary set containing \( J_i \) whose elementary measure is less than \( M_i \).
So, lets be strategic: let \( \varepsilon > 0 \). For every \( 0 < i \le k \), choose an elementary set \( E_i \) that has measure \( m^{*,(J)}(J_i) + \frac{\varepsilon}{k} \). Then, the union of all of these elementary sets will contain \( J \) and will have elementary measure \[ \sum_{i=1}^{k}(m^{*,(J)}(J_i) + \frac{\varepsilon}{k} ) = (\sum_{i=1}^{k} m^{*, (J)}(J_i)) + \varepsilon \] \( \varepsilon \) is arbitrary, so the lower limit for the elementary measure of elementary sets containing \( J \) must be less than this construction. Thus, we have subadditivity: \[ m^{*, (J)}(J) \le \sum_{i=1}^{k}m^{*,(J)}(J_i) \] We have divided our \( \varepsilon \) over each \( J_i \) and managed to thus transfer a lower limit from the individual sets to the union of these sets.
[Note, we could have unwrapped the definition of elementary measure. If we did, the Jordan formulation would look more the setup below—dealing with individual boxes.]
Subadditivity
Now consider a countably infinite union of sets. Let \( L = \bigcup_{i \in \mathbb{N}}L_i \) where \( (L_i)_{i \in \mathbb{N}} \) is a countable collection of sets. Each set \( L_i \) has a Lebesgue outer measure, \( m^*(L_i) \). For any number \( M_i \) larger than this, we can find a countable collection of boxes \( (B_{ij})_{j \in J \subseteq \mathbb{N}} \) such that their union contains \( L_i \) and the sum of their volumes is less than \( M_i \).
So, lets be strategic: let \( \varepsilon > 0 \). For every \( i \in \mathbb{N} \) choose a family of boxes \( (B_{ij})_{j\in J} \) such that the sum of box volumes is \( \frac{\varepsilon}{2^i} \). Then, the union of all of these sets of boxes will be another set of boxes that will cover \( L \). The sum of box volumes for this set will be: \[ \sum_{i=1}^{\infty}(m^{*}(L_i) + \frac{\varepsilon}{2^i} ) = (\sum_{i=1}^{\infty} m^{*}(L_i)) + \varepsilon \] Thus, we have subadditivity: \[ m^{*}(L) \le \sum_{i=1}^{\infty}m^{*}(L_i) \]
More on the essense of subadditivity
It's important to note that the subadditivity notion doesn't use any visually appealing notions of overlapping of sets—these ideas are not needed. They are needed in additivity, but not subadditivity. Instead, the essense of subadditivity is the sense of transferring of the infimums through the sum.
Another thought on the essense is that the less than here should be seen more like a "not greater than"; this is clear in the proof were we deal with equality which then becomes a less-than due to the nature of supremums.
Compared to Jordan and elementary measure formulation route
Elementary measure and Jordan measure possesses finite additivity: \( m(E \cup F) = m(E) + m(F) \) for disjoint elementary sets, \( E, F \). So when treating the above properties, finite additivity was used as it was available and convenient. However, I feel the use of the more powerful finite additivity obscures the essence of monotonicity and subadditivity. Using finite additivity, showing monotonicity and subadditivity for Jordan measure proceeds like so:
Monotonicity
Considering \( E \subset F \), we can break this into \( m(F) = m(E) + m(F\setminus E) \), and \( m(F \setminus E) \ge 0\) so monotonicity follows.
Subadditivity
For elementary measure, we can say \[ \begin{aligned} m(E \cup F) &= m(E \cup (F \setminus E)) \qquad \text{just set manipulation} \\ &= m(E) + m(F \setminus E)  \quad \text{from finite additivity} \\ &\le m(E) + m(F) \qquad \text{from monotonicity} \end{aligned} \] Induction extends this idea to a finite union of sets. The result for Jordan measure proceeds just like this (as finite additivity and monotonicity are both available). The Lebesgue measure case is a bit more involved and is shown on the back.

Title: Lebesgue outer measure. Finite additivity (for separated sets)
Q: For Lebesgue outer measure, we left Pallet Town, so to speak, with just three properties: the measure of an empty set is zero, monotonicity and [something]. Our next step is additivity, though a restricted version of additivity: restricted to the union of just two sets, and the sets must be [something].
Separated sets in \( \mathbb{R}^d \)
Let \( E, F \subset \mathbb{R}^d \) be such that \( \operatorname{dist}(E, F) > 0 \), where \( \operatorname{dist}(E, F) \) is the distance between \( E \) and \( F \), defined as:
\[ \operatorname{dist}(E, F) := \inf(\{ |x - y| : x \in E, y \in F \} ) \]
When \( \operatorname{dist}(E,F) > 0 \), \( E \) and \( F \) are said to be separated sets.
Let \( E, F \subset \mathbb{R}^d \) be separated sets (\( \operatorname{dist}(E, F) > 0 \)). Then [what can we say about the Lebesgue outer measure?].
A: countable subadditivity; separated; \( m^*(E \cup F) = m^*(E) + m^*(F) \)
BackOnlyText: Essense contained within proof
Please appreciate in the following proof that no set of boxes is converted to a disjoint set of boxes. Also appreciate that \( m^* \) is a greatest lower bound, not just any ol' real. Read something like \( m^*(E \cup F) \le m^*(E) + m^*(F) \) as: there must be box covers of \( E \cup F \) having a volume less-equal to the greatest lower bound required to cover \( E \) plus the greatest lower bound required to cover \( F \). Why? Well, we can box-cover \( E \) with any volume less that \( m^*(E) + \varepsilon \), and box-cover \( F \) with any volume less than \( m^*(F) + \varepsilon \), so we can combine these covers to cover \( E \cup F \). In the 'worst case' with no overlapping boxes, the cover will have a volume of \( m^*E) + m^*(F) + 2 \varepsilon \).
Proof
Goal: for two separated sets \( E, F \subset \mathbb{R}^d \), show that \( m^*(E \cup F) = m^*(E) + m^*(F) \).
From countable subadditivity we have \( m^*(E \cup F) \le m^*(E) + m^*(F) \). So we are left to show that \( m^*(E) + m^*(F) \le m^*(E \cup F) \).
Utilize \( m^*(E \cup F) \) to select a countable family of boxes, \( (B_n)_{n=1}^{\infty} \) that covers \( E \cup F \):
\[ \sum_{n=1}^{\infty}|B_n| \le m^*(E \cup F) + \varepsilon \]
Now, make an assumption that we will later remove: suppose that each box intersects at most one of \( E \) and \( F \).
We can then divide the family of boxes \( (B_n)_{n=1}^{\infty} \) into two at most countable sub-families: \( (B'_i)_{i \in I} \) that covers \( E \) and \( (B''_j)_{j \in J} \) that covers \( F \). But for \( m^*(E) \) and \( m^*(F) \) we have:
\[ m^*(E) \le \sum_{i \in I}^{}|B'_i| \]
and
\[ m^*(F) \le \sum_{j \in J}|B''_j| \]
Thus,
\[ m^*(E) + m^*(F) \le \sum_{i \in I}|B'_i| + \sum_{j \in J}|B''_j| = \sum_{n=1}^{\infty} |B_n| \]
And in particular,
\[ m^*(E) + m^*(F) \le m^*(E \cup F) + \varepsilon \]
Removing the disjoint cover assumption
The assumption that each box of the cover of \( E \cup F \) intersects at most one of \( E \) and \( F \) can be removed by noting that as \( \operatorname{dist}(E, F) = r > 0 \), all boxes in a cover of \( E \cup F \) can be partitioned such that the diameter of each box is less than \( r \), and in this case, none of the smaller boxes will intersect both \( E \cup F \).
Jordan outer measure finite additivity
The argument above applies regardless of whether the box cover is finite or countable, so it could be used for the finite additivity for Jordan outer measure too. Jordan outer measure already has finite additivity established, but only for when the sets involved are Jordan measurable. The Jordan measurability allowed us to infer the finite additivity of the Jordan outer measure simply by equating it to the inner measure, for which the finite additivity is easily proved due to any inner cover necessarily being disjoint.
Example: Disjoint closed sets, with one being compact ⟹ dist > 0
Let \( E, F \subset \mathbb{R}^d \). If \( E \) and \( F \) are disjoint closed sets, and at least one of them is compact, then \( \operatorname{dist}(E, F) > 0 \). In this case, we have finite additivity for Lebesgue outer measure.
Rough proof. If \( \operatorname{dist}(E, F) = 0 \), then we could create a sequence in the compact (and thus bounded) set which converges to an element of the other set. But sequences in a closed bounded set must converge within the set. So we would have an element that belongs to both sets, which is a contradiction.
Source: p19

Title: Lebesgue outer measure is between the Jordan inner and outer measures
Q: This relation is useful to ground one's mental model of Lebesgue measure.
\[ m_{*,(J)}(F) \le m^*(F) \le m^{*,(J)}(F) \]
The proof outline for the rightmost inequality is simple. The leftmost inequality is not simple, it is somewhat profound.
1. Rightmost inequality, proof outline
\[ m^*(F) \le m^{*,(J)}(F) \]
In words: the Lebesgue outer measure is less-equal to the the Jordan outer measure.
Consider the set of all at-most-countable box covers for the set \( F \subset \mathbb{R}^d \); denote it by \( \mathcal{S} \). The real \( m^*(F) \) is the [something bound] for the volume of these covers. Contained as a subset off these covers is the set of all finite box covers; denote it as \( \mathcal{C} \). \( m^*(F) \) is a lower bound (the greatest) for \( \mathcal{S} \), so it must also be a lower bound for the subset \( \mathcal{C} \). The real \( m^{*,(J)}(F) \) is the greatest lower bound for \( \mathcal{C} \), so we have \( m^*(F) \le m^{*,(J)(F) } \).
2. Leftmost inequality. It's not trivial.
\[ m_{*,(J)}(F) \le m^*(F) \]
In words: the Lebesgue outer measure is not less than the Jordan inner measure.
The same logic as above seems to intuitively fit, but it doesn't.
Not a Lebesgue inner measure
Any finite set of boxes contained in \( F \) is not in the set of countable covers of \( F \). We are not comparing the Jordan inner measure \( m_{*,(J)}(F) \) to some countable extension. Such a real would be a supremum (not an infimum like the outer measure) and might look like:
\[ m_{*}(F) := \sup \left( \{ m \in \mathbb{R} : m = \sum_{b \in I} |b| \text{, for some countable collection of boxes, } I \text{, contained in } F \} \right) \]
If we were talking about such a measure, then the inequality \( m_{*}(F) \ge m_{*,(J)}(F) \) would follow easily, as the collection of all countable box sets contained within some set \( F \subset \mathbb{R} \) includes all finite box sets. This progression of ideas is a symmetric match to how we arrived at \( m^*(F) \le m^{*,(J)}(F) \) above; however, it is not the inequality we are trying to prove.
So, remember then that we are comparing an outer measure (Lebesgue) with an inner measure (Jordan).
The back side of the card continues to explain the rightmost inequality.
A: greatest lower bound
BackOnlyText: Failed attempt (and motivation for Heine-Borel)
Trying to prove this inequality in the same way we thought about \( m^*(F) \le m^{*,(J)}(F) \) doesn't work:
Consider all finite sets of boxes that are contained within \( F \subset \mathbb{R}^d \). The volume of any of set union of these sets (i.e. the volume of any elementary set), must be less than the real \( m_{*,(J)}(F) \). Now consider any arbitrary set of boxes (including countably or uncountably infinite) contained within \( F \). Each of these sets has a finite subset whose union must have a volume less than \( m_{*,(J)}(F) \). Is this useful? No. \( m_{*, (J)}(F) \) is an upper bound and there is nothing (yet) preventing an infinite collection of boxes both covering \( F \) and having a volume less than \( m_{*, (J)}(F) \). Case in point: an uncountable set of boxes can have volume zero and cover \( F \). The fact that such a collection that has a finite sub-collection contained within \( F \) has no significance.
The question is: can we achieve the same zero-volume covering with just a countable set of boxes? If we can, then we can't make the assertion of the inequality: \( m_{*,(J)} \le m^*(F) \). However, it turns out we can't. And to show this, we need to draw on some fundamental qualities of compact spaces.
Why compactness comes into play
Consider the following chain of thought.
- Try to cover an uncountable set in \( \mathbb{R}^d \) with boxes.
- If we are allowed to use an arbitrary set of boxes, use uncountably infinite degenerate boxes with 0 volume.
- But, now only countable boxes are allowed. If we can do it then...
- The boxes can't all be zero volume anymore.
- The boxes are (or are close enough to being) open sets. And the union of open sets is open.
- \( F \) may not be closed, but consider any closed set \( D \) contained in \( F \). If our set of boxes covers \( F \), then it covers \( D \).
- \( D \) is a closed set in a Hausdorff space, so it is compact and any open cover of \( D \) has a finite sub-cover.
- So, we can cover \( D \) with a finite subset of our boxes.
- Our box-cover of \( F \) must necessarily be larger than the finite box-cover of any closed \( D \subset F \).
- And here's some magic. For any real \( V_E \) less than \( m_{*,(J)}(F) \) there is an elementary set \( E \) contained within \( F \) with volume \( V_E \). Any countable box-cover must cover \( E \) and must have a volume at least that of \( V_E \). So we can't cover \( F \) with a cover less than \( m_{*,(J)}(F) \).
In Tao's book, he first shows that \( m^{*}(E) = m(E) \) for an elementary set \( E \) with elementary measure \( m(E) \). There you can find a proper proof. From this result along with monoton
Countable boxes can't cover an elementary set with any less volume than a finite set of boxes
Can we cover an elementary set \( E \) with say uncountable boxes and get a smaller volume? Yes-cover each point with a zero volume box, for a total cover of zero volume.  So it's not a trivial fact that a countable set cannot improve upon a finite set.
Can we cover an elementary set with countable boxes and get a smaller volume? No. Why? The reason stems from a box being compact (uncountably many yet bounded elements), and every open cover of a compact space has a finite sub-cover. The back side goes into more detail, but the gist is that no matter how we cover \( E \) with open boxes, there will be a finite set that also covers \( E \). A few details clean up cases where boxes are not open. So, we can't do better than a finite cover, and no set of boxes then can fit within \( F \) \( m_{*,(J)(F)} \) is the
An infinite box cover for an elementary set would allow for a cover consisting of closed, single element sets; the idea of a finite open cover from any open cover doesn't apply to a non-open cover, so it doesn't apply here.
Insight into uncountable box covers
It's interesting to take a detour to consider the exact same chain of thought, but starting from the set of arbitrary box covers for the set \( F \). We know that the union of boxes around each element of \( F \) is a cover for \( F \), and the total volume will be a sum of zero-volume boxes, making the total volume zero. This completes the zoo of possible outer measures when categorized as finite, countable and infinite.
Insert diagram for compactness.

Title: Outer Lebesgue measure of countable union of almost disjoint boxes
Q: 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
[\[ m^*(E) = \quad ? \]]
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.
A: \[ m^*(E) = \sum_{n=1}^{\infty}|B_n| \]
BackOnlyText: 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

Title: Lebesgue outer measure (and Jordan inner measure) for any open set
Q: Recall the result:
Let \( E \subset \mathbb{R}^d \) be expressible as the countable union of almost disjoint boxes, \( E = \bigcup_{n=1}^{\infty} B_n \). Then the Lebesgue outer measure of \( E \) is equal to [...]. In other words:
\[ m^*(E) = m_{*,(J)}(E) = \lim_{N \to \infty} \sum_{n=1}^{N} |B_n| = \sum_{n=1}^{\infty} |B_n| \]
The leftmost equality is covered in Lemma 1.2.9. The next equality follows from \((\sum_{n=1}^{N}|B_n|)_{N} \) being a monotone increasing sequence, so the limit is also the supremum. There is no way to rearrange the sequence so that it doesn't converge, like having tiny boxes first (see Definition 8.2.1 in Tao Analysis I for a mention of this). The last equality is simply the definition of an infinite sum.
Theomem. Every open set can be expressed as a countable union of disjoint boxes.
So, the outer Lebesgue measure and Jordan inner measure of any open set \( E \) is the supremum of the volume of finite boxes filling the set. If we have a collection of boxes \( (B_n)_{n=1}^{\infty} \) that union to \( E \), then we can consider instead the limit of a sequence \( (\sum_{n=1}^{N})_N |B_n| \) where \( \bigcup_{n=1}^{\infty} B_n = E \).
We can't just consider any sequence boxes within \( E \), as you could union together a countable set of boxes that are too small to fill \( E \).
Dyadic mesh
A dyadic mesh is used for the proof of the theorem above. It is recalled here.
A dyadic mesh is a way of specifying cubes in \( \mathbb{R}^d \) such that \( \mathbb{R}^d \) is divided into disjoint cubes at multiple resolutions, and all boxes in a finer resolution combine to fit tightly into a a cube at the next level.
Countable and almost disjoint
The important magic of these dyadic cubes is that there is only countably many of them (across all resolutions!). So anything that can be represented as a set of dyadic cubes can be represented as a countable set of cubes. The formulaic organisation of the dyadic cubes allows for any set of them to be decomposed such that all dyadic cubes representing a set are all almost disjoint.
Countable disjoint dyadic cubes can represent any open set. Proof outline
Let \( E \subseteq \mathbb{R}^d \) be an open set. Consider a dyadic mesh where each cube \( Q \) is of the form:
\[ Q = [\frac{i_1}{2^n}, \frac{i_1 + 1}{2^n}] \times ... \times [\frac{i_d}{2^n}, \frac{i_d + 1}{2^n}] \]
Where the integer \( i \) specifies the coordinates of the cube and \( n \) the resolution level the cube exists in.
For every \( x \in E \) there must be an open ball centered at \( x \) (by the definition of open sets). Within an open ball centered at \( x \) we can find a closed dyadic cube containing \( x \) (how?). If we consider all \( x \in E \), we can construct a set \( \mathcal{S} \) of dyadic cubes that union to \( E \). As there are only countable dyadic cubes, \( \mathcal{S} \) is at most countable. Finally, the properties of dyadic cubes allows us to construct another set \( \mathcal{S}' \) of dyadic cubes such that all cubes are almost disjoint and \( \bigcup \mathcal{S} = \bigcup \mathcal{S}' = E \).
A: the Jordan inner measure of \( E \)
Source: p24

Title: Outer regularity. Theorem.
This property connects the outer Lebesgue measure of an arbitrary set \( E \subset R^d \) to the outer Lebesgue measure of open sets which contain \( E \).
Outer regularity. Theorem.
Let \( E \subset \mathbb{R}^d \) be an arbitrary set. Then one has:
\[ m^*(E) = \inf_{E \subset U, U \text{is open} } m^*(U). \]
The proof can be intuitively grasped by noticing that \( m^*(E) \) can be used to specify an open cover of \( E \) with measure arbitrarily close to \( m^*(E) \).
The prove is illustrative of the \( \frac{\varepsilon}{2^n} \) trick.
Proof
From monotonicity of outer Lebesgue measure, we have:
\[ m^*(E) \le \inf_{E \subset U, U \text{is open} } m^*(U). \]
So we are left to show the reverse:
\[ \inf_{E \subset U, U \text{is open} } m^*(U) \le m^*(E). \]
Reading this second less-equal as "not greater than" can motivate intuition.
Tao points out specifically that the inequality is trivial if \( m^*(E) \) is infinite, and we can assume that \( m^*(E) \) is finite.
What follows is a exemplary use of the \( \frac{\varepsilon}{2^n} \) trick.
Let \( \varepsilon > 0 \) be a real. The definition of outer Lebesgue measure affords us the ability to assert the existance of a countable family of boxes covering \( E \) such that it's outer measure is slightly greater than the infimum:
\[ \sum_{i=1}^{\infty} |B_i| \le m^*(E) + \varepsilon \]
We can enlarge each box \( B_i \) to become an open box \( B'_i \) such that \( |B_i'| \le |B_i| + \frac{\varepsilon}{2^i} \). The union of these open boxes is also open and also contains \( E \). In particular, we have:
\[ \sum_{i=1}^{\infty}|B'_i| \le m^*(E) + \varepsilon + \sum_{i=1}^{\infty} \frac{\varepsilon}{2^i} = m^*(E) + 2\varepsilon. \]
Applying subadditivity (the outer measure of the union is less than the sum of the outer measures) we have:
\[ m^*\left( \bigcup_{i=1}^{\infty} B'_i \right) \le m^*(E) + 2\varepsilon. \]
Denote this union as \( B' = \cup_{i=1}^{\infty}B'_i \). We have found an open set \( B' \) that contains \( E \) and has outer measure \( m^*(B') \le m^*(E) + 2\varepsilon \). The infimum of measure over open covers of \( E \) must be less than this one instance of an open cover:
\[ \inf_{E \subset U, U \text{is open}} m^*(U) \le m^*(B') \le m^*(E) + 2\varepsilon. \]
As \( \varepsilon \) was arbitrary, we have:
\[ \inf_{E \subset U, U \text{is open} } m^*(U) \le m^*(E), \]
and combined with our initial subadditivity claim, we have:
\[ \inf_{E \subset U, U \text{is open}} m^*(U) =  m^*(E). \]
Source: p25

Title: Lebesgue measurability. Definition.
Q: We want to describe a class of sets such that the Lebesgue outer measure obeys nice properties. Every subset of \( \mathrm{R}^d \) has a Lebesgue outer measure (possibly infinite). Unfortunately, it turns out that if we include all sets, we lose properties like the union of some disjoint sets has measure which is the sum of the individual sets.
With this in mind, we wish to choose a criteria which will create a restricted class of sets. The below definition is one way to choose the criteria. Note how it both acts as a criteria and as a useful property describing the limits/capabilities of Lebesgue measure for such sets.
Lebesgue measurability. Definition.
A set \( E \subset \mathbb{R}^d \) is said to be Lebesgue measurable iff [for every ? there exists ? such that ?]. If \( E \) is Lebesgue measurable, we refer to \( m(E) := m^*(E) \) as the Lebesgue measure of \( E \).
A: for every real \( \varepsilon > 0 \) there exists exists an open set \( U \subset \mathbb{R}^d \) containing \( E \) such that \( m^*(U \setminus E) \le \varepsilon \)
BackOnlyText: Appeal to intuition
This definition hints at Littlewood's first principle—that measurable sets are almost open.
It is worth comparing Lebesgue and Jordan measure:
Jordan measure
Jordan measurable sets can be efficiently contained in elementary sets, with an error that has small Jordan outer measure.
Lebesgue measure
Lebesgue measurable sets can be efficiently contained in open, with an error that has small Lebesgue outer measure.
A later card covers other criteria for measurability which are equivalent to the above.
Source: p17

Title: Lebesgue measurable sets.
The following sets meet the criteria to be Lebesgue measurable:
1. Every open set is Lebesgue measurable.
2. A countable union of Lebesgue measurable sets is Lebesgue measurable.
3. Every closed set is Lebesgue measurable.
4. The complement \( \mathbb{R}^d \setminus E \) of a Lebesgue measurable set \( E \) is Lebesgue measurable.
5. A countable intersection of Lebesgue measurable sets is Lebesgue measurable.
Two others are:
- Every set of Lebesgue outer measure zero is measurable. These sets are called null sets.
- The empty set \( \emptyset \) is Lebesgue measurable.
The proofs are on the reverse side, along with a repeat of the definition of Lebesgue measurability.
BackOnlyText: Definition recap.
Lebesgue measurability. Definition.
A set \( E \subset \mathbb{R}^d \) is said to be Lebesgue measurable iff for every real \( \varepsilon > 0 \) there exists exists an open set \( U \subset \mathbb{R}^d \) containing \( E \) such that \( m^*(U \setminus E) \le \varepsilon \). If \( E \) is Lebesgue measurable, we refer to \( m(E) := m^*(E) \) as the Lebesgue measure of \( E \).
The proofs are below. We start with the easier ones.
Every set with Lebesgue outer measure zero is measurable. Proof.
Let \( E \subset \mathrm{R}^d \) be a set with \( m^*(E) = 0 \). Recall that outer regularity allows allows us to choose, for any \( \varepsilon > 0 \), an open set \( \mathcal{O} \) that contains \( E \) and has measure \( m^* (\mathcal{O}) < m^*(E) + \varepsilon \). As \( m^*(E) = 0 \), this means \( m^*(\mathcal{O}) < \varepsilon \). We are free to choose \( \varepsilon \) and so, the measurability criteria above is easily met.
The empty set \( \emptyset \) is Lebesgue measurable. Proof.
This follows from the above proposition. Alternatively, repeat the idea of the previous proof for the specific case of \( \emptyset \).
1. Every open set is Lebesgue measurable. Proof.
This proposition it follows directly from the definition. Any open set has itself as a set which can be used to satisfy the criteria.
2. A countable union of Lebesgue measurable sets is Lebesgue measurable. Proof.
This proof uses a classic application of the \( \frac{\varepsilon}{2^n} \) trick. Let \( E = \cup_{i=1}^{\infty} E_i \) be the countable union of measurable sets, \( E_1, E_2, ... \). Let \( \varepsilon > 0 \) be a real. For every \( E_i \), choose an open set \( \mathcal{O}_i \) with \( E_i \subset \mathcal{O}_i \) and \( m^*(O_i - E_i) < \frac{\varepsilon}{2^i} \). Then consider the sum of these open sets: \( \mathcal{O} = \cup_{i=1}^{\infty} \mathcal{O}_i \). \( \mathcal{O} \) is open, \( E \subset \mathcal{O} \) and \( \mathcal{O} - E \subset \cup_{i=1}^{\infty} \mathcal{O}_i - E_i \). By monotonicity and sub-additivity we have:
\[ m^*(\mathcal{O} - E) \le \sum_{i=1}^{\infty} m^*(\mathcal{O}_i - E_i) \le \sum_{i=1}^{\infty} \frac{\varepsilon}{2^i} = \varepsilon \]
The reasoning behind the next 2 propositions is a bit more involved.
3. Every closed set is Lebesgue measurable. Proof.
Proof essence: consider a closed compact set. We can cover it with an open set with just \( \varepsilon \) extra measure. An open set minus a closed set is open. The remaining space within the open set but not in the closed set ends up having at most \( \varepsilon \) measure, as we can approximate it by finite disjoint boxes which are also disjoint from the closed set (we are able to apply finite additivity).
Refer to Tao or Stein for the detailed proof. The following diagram tries to evoke some intuition behind the reasoning.
4. The complement \( \mathbb{R}^d \setminus E \) of a Lebesgue measurable set \( E \) is Lebesgue measurable. Proof.
The proof of this proposition involves approximating the complement from within by closed sets, then noting that the left over space has measure zero, and then the complement is measurable as closed sets are measurable and so too is any set with measure zero.
Proof. Let \( E \) be measurable. For every \( n \subset \mathbb{N} \) we choose an open set \( \mathcal{O}_n \) containing \( E \) such that \( m^*(O_n - E) \le \frac{1}{n} \). The complement \( \mathcal{O}_n^c \) of each set is closed, and the union of all of them, \( S = \cup_{i}^{n} \mathcal{O}_n^c \) is measurable (it's a union of measurable sets). Now consider the set \( E^c - S \). Elements of this set are in \( E^c \) and not in \( S \), so they are not in \( E \) and are in every \( \mathcal{O}_n \). In other words:
\[ E^c - S \subseteq \mathcal{O}_n - E, \text{ for any } n > 0 \]
By monotonicity, we have:
\[ m^*(E^c - S) \le m^*(\mathcal{O}_n - E) \le \frac{1}{n}, \text{ for all } n \} \]
Therefore \( m^*(E^c - S ) = 0 \), and \( E^c - S \) is measurable (all sets with outer Lebesgue measure zero are measurable–see above). Now note that \( E^c = S \cup (E^c - S) \), both of which sets are measurable, so \( E^c \) must also be measurable, as it is the union of two measurable sets.
The following diagram tries to stimulate some visual intuition for these steps.
5. A countable intersection of Lebesgue measurable sets is Lebesgue measurable.
This follows from the measurability of the complement of a countable union of measurable sets.
Source: Stein, p16
Tao, p26