Definition and notation: Consider closed $n$-dimensional intervals $I=\{x:a_j \leq x_j \leq b_j, j=1,2,3,...,n\}$ and their volumes $v(I) =\prod_{j-1}^{n}(b_j - a_j)$.We define the outer measure of an arbitrary subset of $E$ of $\mathbb R^n$,covered by a countable collection $S$ of intervals $I_k$ is $S=\{I_1, I_2, I_3, \cdots\}$ ,that is, $E\subset \cup_{I_j\in S}$ ,and let $\sigma(S)=\sum_{I_k\in S}v(I_k)$.The lebesgue outer measure of $E$ , denoted $|E|_e$,is defined by $|E|_e=\inf\sigma(S)$.
Theorem: For an interval $I$, $|I|_e=v(I)$.
Proof: Since $I$ covers itself, $|I|_e \le v(I)$. Let $I_k^*$ be an interval containing $I_k$ in its interior such that $v(I_k^*)\le(1+\varepsilon)v(I_k)$. Then, $I \subset \cup_k(I_k^*)$. Since $I$ is compact, there exists $N$ such that $I\subset \cup_{k=1}^N I_k^*$ by the Heine-Borel theorem. Clearly, $v(I)\le \sum_{k=1}^N v(I_k^*)$. Hence, $v(I)\le (1+\varepsilon)\sum_{k=1}^Nv(I_k)\le(1+\varepsilon)\sigma(S)$. Since $\varepsilon$ can be chosen arbitrarily small, it follows $v(I)\le\sigma(S)$ and $v(I)\le|I|_e$.
My question: The part "Let $I_k^*$ be an interval containing $I_k$ in its interior such that $v(I_k^*)\le(1+\varepsilon)v(I_k)$." .Why $I_k^*$ exist? In particular, how can i prove that $v(I_k^*)\le(1+\varepsilon)v(I_k)$ for $I_k$ containing in $I_k^*$ in $\mathbb R^n$?
Reference: Wheeden and Zygmund : Measure and Integration: An Introduction to Real Analysis
Edited (18/10 4pm,taiwan) :
My Attempt for my question :
*Correction made, thank you @Izaak van Dongen for pointing that out *
Let $\epsilon >0$ be given.
In the case of $\mathbb R$, set $I_1 = [a,b]$ and $I_2 = [a-\delta,b+\delta]$ ,so ,$I_1 \subset I_2$, In $\mathbb R$, the "volume of an interval" is actually the length of the interval. By the diagram i have drawn:
We have $v(I_2) = (a-(a-\delta))+(b-a)+((b+\delta)-b) = v(I_1)+2\delta $. If we choose $\delta \leq \frac{(b-a)\epsilon}{2}$ ,then we have the desired result $v(I_2) \leq (1+\epsilon)v(I_1)$.
In the case of $\mathbb R^2$,the interval would be a rectangle with $I_1$ and $I_2$.If one of the intervals shrinks down into a point, for example $(a_2,a_2)$, turning the rectangle into a line ,then the case is similar to the case of $\mathbb R$. Suppose that the length of each interval $I_1$ and $I_2$ are greater than 0. In this case, the "volume of the interval" in $\mathbb R^2$ is actually the area of the rectangle. Set $v(I_1) = |I_1| |I_2|$. We want to find a larger rectangle containing $I_1$ and $I_2$ such that this larger rectangle has at most the area of $(1+ \epsilon) v(I_1)$. I have drawn a diagram: (Of course i know that there are many rectangles with different sides can cover the rectangle,but this is the easiest case that what i can think of) I want to find a $\delta$ that involves $\epsilon$ like in the case of $\mathbb R$ so i can also get the similiar result .I dont know how to do it.
In the case of $\mathbb R^3 $, I know that it is a box and I need to create a bigger box to cover it and the volume of the bigger box has to be smaller or equal to $(1+\epsilon)$(the volume of the smaller box). I also dont know how to do it
Update 1 (18/10 6:00pm,taiwan): In the case of $\mathbb R^2$ , from the diagram i have drawn, set $I_3=[a_1-\delta,b_1+\delta]$ and $I_4=[a_2-\delta,b2+\delta]$. The volume of the bigger square Is $v(I_2)=|I_3||I_4|$.We see that $|I_3|=(b_1+\delta)-(a_1-\delta)=b_1-a_1+2\delta$ and $|I_4|=(b_2+\delta)-(a_2-\delta)=b_2-a_2+2\delta$. So We obtain $v(I_2)=|I_3||I_4|=((b_1-a_1)+2\delta)((b_2-a_2)+2\delta)=(|I_1|+2\delta)(|I_2|+2\delta)=|I_1||I_2|+2\delta |I_1|+2\delta |I_2|+4\delta^2=v(I_1)+2\delta(|I_1|+|I_2|)+4\delta^2$.
I dont know how to continue it .
Update 2 (18/10 10:00pm,taiwan): Choose $\delta \leq \sqrt{\frac{\epsilon v(I_2)}{2} + \frac{(|I_1|+|I_2|)^2}{16}}-\frac{(|I_1|+|I_2|)}{4}$, so we have:
$2\delta(|I_1|+|I_2|)+4\delta^2$ $\leq 2(\sqrt{\frac{\epsilon v(I_2)}{2} + \frac{(|I_1|+|I_2|)^2}{16}}-\frac{(|I_1|+|I_2|)}{4})(|I_1|+|I_2|)+4(\sqrt{\frac{\epsilon v(I_2)}{2} + \frac{(|I_1|+|I_2|)^2}{16}}-\frac{(|I_1|+|I_2|)}{4})^2$
$\leq 2(|I_1|+|I_2|)\sqrt{\frac{\epsilon v(I_2)}{2} + \frac{(|I_1|+|I_2|)^2}{16}}-\frac{(|I_1|+|I_2|)^2}{2}+ \frac{\epsilon v(I_1)}{2}+\frac{(|I_1|+|I_2|)^2}{4}-2\sqrt{\frac{\epsilon v(I_2)}{2} + \frac{(|I_1|+|I_2|)^2}{16}}+\frac{(|I_1|+|I_2|)^2}{4}$
$=\epsilon v(I_1)$
Hence we have $v(I_2) \leq (1+\epsilon) v(I_1)$ for the case $\mathbb R^2$.