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

# The 3 basic properties of Lebesgue outer measure

Reminder that the Lebesgue outer measure is denoted as $$m^{*}$$.

### The 3 basic propositions of Lebesgue outer measure

Empty set
$$m^{*}(\emptyset) = 0$$
Monotonicity
If $$E \subseteq F \subset \mathbb{R}^d$$, then $$m^{*}(E) \le m^{*}(F)$$.
If $$E_1, E_2, ... \subset \mathbb{R}^d$$ is a countable sequence of sets, then $m^{*}(\bigcup_{n=1}^{\infty}E_n) \le \sum_{n=1}^{\infty}m^{*}(E_n)$

These three ideas are very fundamental. The complex apprearance of the expressions obscures the simplicity of the ideas.

### Compact essence

Monotonicity
If you can cover a whole, you can cover a part.
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.

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 $$S$$.
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.]

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.
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.