I am reading "Analysis on Manifolds" by James R. Munkres.
We begin by defining the volume of a rectangle. Let $$Q=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]$$ be a rectangle in $\mathbb{R}^n$. Each of the intervals $[a_i,b_i]$ is called a component interval of $Q$. The maximum of the numbers $b_1-a_1,\dots, b_n-a_n$ is called the width of $Q$. Theire product $$v(Q)=(b_1-a_1)(b_2-a_2)\cdots (b_n-a_n)$$ is called the volume of $Q$.
Theorem 10.4. Every constant function $f(x)=c$ is integrable. Indeed, if $Q$ is a rectangle and if $P$ is a partition of $Q$, then $$\int_Q c=c\cdot v(Q)=c\sum_R v(R),$$ where the summation extends over all subrectangles determined by $P$.
Proof. If $R$ is a subrectangle determined by $P$, then $m_R(f)=c=M_R(f)$. It follows that $$L(f,P)=c\sum_R v(R)=U(f,P),$$ so the Riemann condition holds trivially. Thus $\int_Q c$ exists; since it lies between $L(f,P)$ and $U(f,P)$, it must equal $c\sum_R v(R)$.
This result holds for any partition $P$. In particular, if $P$ is the trivial partition whose only subrectangle is $Q$ itself, $$\int_Q c=c\cdot v(Q).$$
Munkres proved $$\int_Q c=c\cdot v(Q)=c\sum_R v(R)$$ elegantly.
But I thought this theorem is obvious by the distributive law of multiplication.
Is my elementary proof of this fact ok?
My Proof.
Let $P$ be a partition of $Q$.
If $R$ is a subrectangle determined by $P$, then $m_R(f)=c=M_R(f)$.
It follows that $$L(f,P)=c\sum_R v(R)=U(f,P).$$ By the distributive law of multiplication, $\sum_R v(R)=(b_1-a_1)(b_2-a_2)\cdots (b_n-a_n)=v(Q)$ holds.
So, $L(f,P)=U(f,P)=c\cdot v(Q)$ for any partition $P$ of $Q$.
So, $\sup_P \{L(f,P)\}=c\cdot v(Q)=\inf_P\{U(f,P)\}$.
So, $\int_Q c=c\cdot v(Q)=c\sum_R v(R)$ holds.