Proof about outer measure. For an interval $I$, $|I|_e=v(I)$? 
My question is when proving $|I|_e \ge v(I)$, why cannot I conclude from $S={I_k}_{k=1}^\infty$ is a cover of $I$, then $v(I)\le \sigma(S)$, so $v(I)\le inf \sigma(S)=|I|_e$?
Why do we need the $I_k^*$?
Can anyone help? Thanks so much!
The following is the definition of $\sigma (S)$.

 A: For example, $ [a,b] \subset \mathbb{R}$ and $\{I_k\}$ is a countable collection of open intervals that covers of $[a,b]$. Let us show 
$$\sum_{k=1}^\infty v(I_k) \geq b-a.$$
Remark 1: I assumed $I_k$ are open so we do not have to redefine $I_k^*$, thus we skip the $\epsilon$ argument. 
Remark 2: In order to prove above inequality (which as you said, looks trivial), we have to do the following argument.
Proof: By Heine-Borel theorem, there exists a finite subcover $\{I_{k_n}\}_{n=1}^N$ (lets call them $I_n$ instead of $I_{k_n}$ and we understand that they are a subcollection) and we show that 
$$\sum_{k=1}^\infty v(I_k) \geq \sum_{n=1}^N v(I_n)\geq  b-a.$$
Remark 3: In your book's proof, it says that $\sum_{n=1}^N v(I_n)\geq  b-a$ is "clearly" true, but for completeness, I will prove it. (and maybe you will see that $\sum_{k=1}^\infty v(I_k) \geq b-a$ is not so "trivial"...)
The end point $a$ has to be in an open interval in this finite subcollection $\{I_n\}$ let us call it $(a_1, b_1)$ and we have $a\in (a_1, b_1)$, now if $b\in (a_1, b_1)$ then we are done, since 
$$\sum_{n=1}^N v(I_n)\geq v((a_1,b_1)) = b_1 - a_1 \geq b-a.$$
Else if $b\notin (a_1, b_1)$, we must have $b_1\in [a,b]$. And there is an open interval $(a_2,b_2)$ in $\{I_n\}$ such that  $b_1\in (a_2,b_2)$. If $b\in (a_2,b_2)$ then we are done, 
$$\sum_{n=1}^N v(I_n)\geq  v((a_1,b_1)) + v((a_2,b_2)) \geq b-a.$$
Else, we continue until it terminates, and there must exists $N'\leq N$ such that the subcollection $\{(a_i,b_i)\}_{i=1}^{N'}$ of $\{I_n\}_{n=1}^N$ for which 
$$a_1 < a \quad\quad b_{N'}  > b$$
and
$$a_{i+1} < b_i\quad \text{ for } 1\leq i\leq N'-1.$$
Then 
$$\sum_{n=1}^N v(I_n)\geq \sum_{i=1}^{N'} v((a_i,b_i)) = (b_{N'} - a_{N'}) + (b_{N'-1} - a_{N'-1}) + ...+ (b_2 - a_2) + (b_1 - a_1) \\ = b_{N'} - (a_{N'} - b_{N'-1}) - ... - (a_2 -  b_1) - a_1 \geq b_{N'} - a_1  \geq b-a.$$
