Proof of $\mu^*$ is an outer measure

Question

Why is $\sum_j l(C_{ij}) \leq \mu^*(A_i) + \epsilon/2^i$?

Isn't $A_i$ covered by $C_{ij}$?

Answer 1

Given $\varepsilon>0$, the number $\mu^*(A_i)+\varepsilon/2^i$ is not the infimum of the set $$\{\sum_{j=1}^{\infty}l(B_j): B_j\in{\cal C} \quad A_i\subset\cup_jB_j\}$$ so there exists an element of the set which is not larger than it, namely, with $B_j=C_{ij}$, the number $\sum_{j=1}^{\infty}l(C_{ij})$ with $A_i\subset\cup_j C_{ij}$ is a member of the set which is not larger than $\mu^*(A_i)+\varepsilon/2^i$.

Answer 2

All they're saying is that there exists such $C_{ij}$. If no such $C_{ij}$ exists then

$$\forall C_{ij}: A_i \subset \cup_j C_{ij}$$

We have this:$$ \sum_jl(C_{ij}) \gt \mu^{*}(A_i) + \frac{\epsilon}{2^i}$$

Label $$ \mathbb{A} = \{\sum_{j=1}^{\infty}l(B_j): B_j\in{\cal C} \quad A_i\subset\cup_jB_j\}$$ Since $\mu^{*}(A_i) + \frac{\epsilon}{2^i}$ is not equal to $\inf(\mathbb{A})= \mu^{*}(A_i)$ then that means there must exist $a\in \mathbb{A}$ such that

$$\mu^{*}(A_i) \le a < \mu^*(A_i) + \frac{\epsilon}{2^i}$$,
Otherwise $\forall x \in \mathbb{A}, x\ge \mu^{*}(A_i) + \frac{\epsilon}{2^{i}} \implies x = \inf(\mathbb{A}) = \mu^{*}(A_i) \implies \mu^{*}(A_i) \ge \mu^{*}(A_i) + \frac{\epsilon}{2^i} \implies $
$$ 0 \ge \frac{\epsilon}{2^i}$$
a contradiction.

So let $a$ be as above. Then this means that there are $B_i$ that witness $$ \sum_i l(B_i) = a < \mu^{*}(A_i) + \frac{\epsilon}{2^i} $$

But that violates this

Answer 3

Let $$ m_i := \mu^*(A_i) = \inf \underbrace{\left \{\sum_{k=1}^\infty \ell(B_k) : B_k \in \mathcal C \textit{ for each }k \textit{ and } A_i \subset \cup_{k=1}^\infty B_k \right \}}_{M_i} $$

Since $m_i$ is the infimum of $M_i$, there cannot be a greater lower bound on $M_i$. Therefore, if we increase $m_i$ slightly by $\epsilon/2^i$, there will be an element of $M_i$ between $m_i$ and $m_i+\epsilon/2^i$. Therefore $\forall i\exists (C_{ij})_j\subset \mathcal C$ such that the $C_{ij}$ cover $A_i$ and $$ \sum_{j=1}^\infty \ell(C_{ij}) \leq m_i + \epsilon/2^i = \mu^*(A_i) + \epsilon/2^i. $$