$v(Q)\le\sum_{i=1}^k v(Q_i)$ where $Q_1,...,Q_k$ are rectangles that cover the rectangle $Q$ What shown below is a reference from "Analysis on manifolds" by James R. Munkres.

Well I don't formally understand why each $R\subseteq Q$ is contained in at least one of the rectangles $Q_1,...,Q_k$ so I ask to prove this formally. Could someone help me, please?
 A: Let $a = (a_x, a_y)$ be any interior point of $R$ (say the center of $R$ for sake of concreteness), and let $Q_j$ be any rectangle of $\{Q_1, \ldots, Q_k\}$ that contains $a$. We will show that $Q_j$ contains $R$. Let $Q_j$ be defined by the equations $x=l$, $x=r$, $y=d$, and $y=u$. Suppose for contradiction that $b = (b_x, b_y)\in R$ is not in $Q_j$. Without loss of generality suppose that $b_x > a_x$, that is $b$ is to the right of $a$ (the cases $b_x < a_x$, $b_y > a_y$, and $b_y < a_y$ are similar). Then the right vertical edge of $Q_j$ must pass between $c$ and $d$, i.e. $a_x \leq r < b_x$. But this means that $R$ would not be a subrectangle of $P$ (since the line $x=r$ would further partition $R$). 
A: As stated in the beginning, "... each of the rectangles $Q,Q_1,\ldots, Q_k$ is a union of subrectangles determined by $P$".
Thus,  $Q_j = \bigcup_{l=1}^{m_j} R_{jl}$ for each $j=1,\ldots,k$ and since $Q_1,\ldots, Q_k$ cover $Q$, we have
$$Q \subset \bigcup_{j=1}^k Q_j = \bigcup_{j=1}^k\bigcup_{l=1}^{m_j}R_{jl}$$
If $R \subset Q$, then as a member of the partition $P$ it must belong to the set $\{R_{jl}\}$ and so is contained in at least one of the rectangles $Q_1, \ldots , Q_k$.
