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 $\mathbb R^2$.
Your approach of picking a small $\delta$ so that when you enlarge every side of the box by $2\delta$, you increase the area by a factor of less than $(1 + \epsilon)$ is certainly the right idea! (Although I think you have a slight mistake in the case $\Bbb R$ - it should be $v(I_2) = v(I_1) + 2\delta$. Similarly you should have $v(I_3) = b_1 - a_1 + 2\delta$, and so on. It might have helped you here to double-check your algebra with the picture you drew :).) It can be fiddly if you don't know the right way to lay out a proof like this.
For clarity, I will re-state what you're trying to prove. I will refer to "$n$-dimensional intervals" as "boxes", because I think it helps to have some visualisation in mind of the geometry of what's going on.
Lemma. Let $n \in \Bbb N$, let $\varepsilon > 0$, let $a_j \le b_j$ be real numbers for $j = 1, \dotsc, n$, and let $I \subseteq \Bbb R^n$ be the box $\{\mathbf x: a_j \le x_j \le b_j, j = 1, \dotsc, n\} = [a_1, b_1] \times [a_2, b_2] \times \dotsb \times [a_n, b_n]$. Furthermore, assume that $v(I) > 0$, ie $a_j < b_j$ for all $j$.
Then there is some box $I^* = [a_1^*, b_1^*] \times \dotsb \times [a_n^*, b_n^*]$ with $I \subseteq \mathrm{int}(I^*)$ and $v(I^*) \le (1 + \varepsilon)v(I)$.
Proof 1. We will argue lazily, by constructing a sequence of boxes containing $I$ whose volumes converge from above to $v(I)$. Consider the sequences $a_j^{(k)} = a_j - \tfrac 1k$ and $b_j^{(k)} = b_j + \tfrac 1k$ for $j = 1, \dotsc n$. Let $I^{(k)}$ be the box $[a_1^{(k)}, b_1^{(k)}] \times \dotsb \times [a_n^{(k)}, b_n^{(k)}]$. Clearly $I \subseteq \mathrm{int}(I^{(k)})$ for all $k$. Also, $v(I^{(k)}) = (b_1^{(k)} - a_1^{(k)})\dotsm(b_n^{(k)} - a_n^{(k)})$. Clearly we have $b_j^{(k)} - a_j^{(k)} \to b_j - a_j$ as $k \to \infty$, for each $j = 1, \dotsc, n$ (the limit of a sum is the sum of the limits).
Hence, $v(I^{(k)}) \to (b_1 - a_1) \dotsm (b_n - a_n) = v(I)$ as $k \to \infty$ (the limit of a product is the product of the limits). Therefore, we can find some $K$ for which $\lvert v(I^{(K)}) - v(I)\rvert \le \varepsilon v(I)$ (this part requires $v(I) > 0$!) In particular, then we have $v(I^{(K)}) \le (1 + \varepsilon) v(I)$. So take $I^* = I^{(K)}$, and we are done.
Remarks. In the proof of the theorem from your book, they don't state the assumption that $v(I) > 0$ when they make use of this lemma. This assumption is actually necessary here! Can you see why? Can you also see why the theorem is still true when $v(I) = 0$ anyway?
In general, I think the moral of this argument is "don't bother too much with finding explicit bounds when you don't need to". At this point in your analysis career, you have proved a lot of nice analysis results in the past - and you should use them liberally! It's important to start realising the "bigger-picture", intuitive reason why something is true, and also be able to translate that into a "bigger-picture", intuitive proof that doesn't involve too much messing around.
I suspect that if you asked the authors of your book to justify this fact, they would just say "it's obvious, because you can make the box a tiny bit bigger". If you pressed them on it, they might say "making the box bigger changes the volume continuously". If you put a gun to their heads I reckon they might write down the proof I give, but really this level of detail is no longer necessary if you're at the point of learning measure theory. (Of course, you should always make sure that you believe the facts that are supposed to be obvious, and that you could write down a proof like this if you wanted. So very well done for being critical enough of the proofs you read to end up asking this question!)
Nevertheless, I will remark that it is possible to find an explicit $\delta$ if you want to. In fact, all the facts about sequences that I use in my proof above give you perfectly constructive bounds, so you can work out what $K$ will be in terms of $\varepsilon$. I don't think this is a terribly productive exercise. I will sketch another proof that just directly picks a $\delta$:
Proof 2. Let $\delta > 0$ and consider the intervals $[a_j - \delta, b_j + \delta]$ for $j = 1, \dotsc, n$. Observe that \begin{equation*} \frac {v([a_1 - \delta, b_1 + \delta] \times \dotsb \times [a_n - \delta, b_n + \delta])} {v(I)} = \frac {(b_1 - a_1 + 2\delta)\dotsm(b_n - a_n + 2\delta)} {(b_1 - a_1)\dotsm(b_n - a_n)} \end{equation*} Now we want to make this ratio less than or equal to $1 + \varepsilon$. To achieve this, we'll divide the $(1 + \varepsilon)$ across the $n$ factors $(b_j - a_j + 2\delta)/(b_j - a_j)$ - in particular it will suffice if each of these factors is less than or equal to $(1 + \varepsilon)^{\frac 1n}$. In other words, we need to pick $\delta$ such that $b_j - a_j + 2\delta \le (b_j - a_j)(1 + \varepsilon)^{\frac 1n}$, or in other words, $\delta \le \tfrac 12 (b_j - a_j)[(1 + \varepsilon)^{\frac 1n} - 1]$.
So take $\delta = \min\{\tfrac 12 (b_j - a_j)[(1 + \varepsilon)^{\frac 1n} - 1] : j = 1, \dotsc, n\}$, and we are done.
I will emphasise again that this proof really isn't any better than the first one I gave. The exact algebra really isn't the point of this lemma, in my opinion, and getting bogged down in $\varepsilon-\delta$ manipulations kind of obscures what's really going on.
Lastly, let me remark - in your algebra for the case $\Bbb R^2$, you can finish what you started there by considering what happens in the limit $\delta \to 0$, very much in the spirit of my first proof.