how to prove volume is less or equal the outer measure of any closed interval, $v(I)\leq m^*(I)$ 
Exercise 26: Let
$$
I=\left\{(x, y) \in \mathbb{R}^2 \mid a \leq x \leq b, c \leq y \leq d\right\}
$$
be a closed interval in $\mathbb{R}^2$. Let
$$
\begin{aligned}
&a=a_0<a_1<\ldots<a_m=b \quad \text { and } \\
&c=c_0<c_1<\ldots<c_n=d .
\end{aligned}
$$
For $i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, define the rectangle $I_{i j}$ by
$$
I_{i j}=\left\{(x, y) \in \mathbb{R}^2 \mid a_{i-1} \leq x \leq a_i, c_{j-1} \leq y \leq c_j\right\} .
$$
(This can be thought of as subdividing $I$ into subrectangles along the vertical lines $x=a_1, x=a_2, \ldots, x=a_{m-1}$ and the horizontal lines $y=c_1, y=c_2, \ldots, y=c_{n-1}$.) Using the definition of volume, prove
$$
\sum_{i=1}^m \sum_{j=1}^n v\left(I_{i j}\right)=v(I) .
$$ 
Exercise 27: Let $$I=\left\{(x, y) \in \mathbb{R}^2 \mid a \leq x \leq b, c \leq y \leq d\right\}$$ be a closed interval in $\mathbb R^2$. Let $J_1, J_2, \ldots, J_n$ be a finite collection of closed intervals that cover $I$. That is,
$$
I \subseteq \bigcup_{k=1}^n J_k .
$$
By carefully subdividing $I$ and the $J_k$ 's into subrectangles, and use the previous exercise to show that
$$
v(I) \leq \sum_{k=1}^n v\left(J_k\right) .
$$

I start with Exercise 26 (thanks to @Balajisb pointing out the telescoping sum),
$$
\begin{align}
\sum_{i=1}^m \sum_{j=1}^n v\left(I_{i j}\right)&=\sum_{i=1}^m (v(I_{i1})+v(I_{i2})+\cdots+v(I_{in}))\\
&= \sum_{i=1}^m (a_i-a_{i-1}) ((c_1-c_0)+\cdots+(c_n-c_{n-1}))\\
&= ((c_1-c_0)+\cdots+(c_n-c_{n-1})) \sum_{i=1}^m (a_i-a_{i-1})\\
&= (c_n-c_0) ((a_1-a_0)+\cdots+(a_m-a_{m-1})) \\
&= (d-c)(b-a) \\
&= v(I)
\end{align}
$$
Now, from here I couldn't see how exercise 26 can be used in exercise 27. What they mean by "By carefully subdividing $I$ and the $J_k$ 's into subrectangles..."?
Any help will be appreciated. TIA.
 A: Let's note $I = [a,b]\times [c,d]$. Each $J_k$ can be written as $J_k = [A_k,B_k]\times [C_k,D_k]$ with $A_k \leq B_k$ and $C_k \leq D_k$.
Let $Abs$ be the set containing all the $A_k$ and $B_k$ as well as $a$ et $b$, and $Ord$ be the set containing all the $C_k$ and $D_k$ as well as $c$ and $d$. We can order the elements of $Abs$ and write $Abs = \{a_0,a_1,\cdots, a_p\}$ with $a_0,\leq a_1 \leq \cdots \leq a_p$. Similarly, we can write $Ord = \{c_0,c_1,\cdots, c_q\}$ with $c_0 \leq c_1 \leq \cdots \leq c_q$. Now, we can define as in exercise 26 : $I_{ij} = [a_{i-1},a_{i}] \times [c_{j-i},c_{j}]$ for all $i \in \{1,...,p\}$ and $j \in \{1,...,q\}$. Here is a picture to visualize :

Adding up the volume of the $J_k$ is the same as adding up the volume of the small rectangles inside each $J_k$. In doing so, we add up every small rectangle inside $I$ (in black) at least once. Therefore the added volume of all the $J_k$ is greater than the volume of $I$.
Let's prove it correctly. Given $k \in \{1,..,n\}$, $J_k$ can therefore be written as
$$J_k = [A_k,B_k]\times [C_k,D_k] = [a_{i_k},a_{i'_k}] \times  [c_{j_k},c_{j'_k}]$$
for some indexes $i_k \leq i'_k$ and $j_k \leq j'_k$. Thus, $J_k$ is subdivided into disjoint rectangles by the family of rectangles $(I_{ij})_{\substack{i_{k}+1 \leq i \leq i'_k \\j_{k}+1 \leq j \leq j'_k}}$ and therefore by exercise 26 :
$$v(J_k)= \sum\limits_{\substack{i_{k}+1 \leq i \leq i'_k \\j_{k}+1 \leq j \leq j'_k}} v(I_{ij}).$$
Similarly, we get
$$v(I) = \sum\limits_{\substack{i_a+1 \leq i \leq i_b \\ j_c+1 \leq j \leq j_d}} v(I_{ij})$$
where $i_a,i_b$ (resp. $j_d,j_d$) are the indices of $a$ and $b$ (resp. $b$ and $c$) in $Abs$ (resp $Ord$).
Note that, as the rectangles $J_k$ cover $I$, every interval $I_{ij}$ included in our last sum (for $I$) must be contained in one of the $J_k$ and therefore appear in the sum we found for $v(J_k)$. (see Appendix for more precision)
Thus, as we have
$$\sum\limits_{k=1}^n v(J_k)= \sum\limits_{k=1}^n \sum\limits_{\substack{i_{k+1} \leq i \leq i'_k \\j_{k+1} \leq j \leq j'_k}} v(I_{ij}),$$
every $v(I_{ij})$ present in the sum for $v(I)$ is present at least once in this gigantic sum (one can see that some rectangles appear several times, corresponding to the overlaps between the $J_k$). As all the terms are positive, we can deduce that
$$\sum\limits_{k=1}^n v(J_k) \geq v(I).$$
Appendix
Let $I_{ij} = [a_{i-1},a_i]\times[c_{j-1},c_j]$ be a rectangle obtained in our subdivision of $I$. Let $x \in I_{ij}$ be the middle point of the rectangle $I_{ij}$. Then $x$ is in $I$ and therefore in some $J_k$. We already know that $J_k$ can be written as $J_k=[a_{i_k},a_{i'_k}] \times  [c_{j_k},c_{j'_k}]$. We must have $i_k \leq i-1$ : else we would have $i_k \geq i$ and then $x$ can't be the middlepoint of $I_{ij}$. Similarly, $'_k \geq i$, $j_k \leq j-1$ and $j'_k \geq j$. Thus $I_{ij}$ is one of the intervals of the subdivision we found for $J_k$.
