MHB Why Is a Rectangle Considered Closed and Bounded in Volume Proofs?

[mathmari]

Hey!

I am looking at the proof of the theorem that for any rectangle the outer measure is equal to the volume.

At the beginning of the proof there is the following sentence:

It is enough to look at the case where the rectangle R is closed and bounded.

Why does it stand?? (Wondering)

Is it maybe as followed??A closed rectangle is
$$[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_d,b_d]$$

An open rectangle is
$$(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_d,b_d)$$
which can be written as a union of closed intervals.

 

[mathmari]

Is it maybe as followed??A closed rectangle is
$$[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_d,b_d]$$

An open rectangle is
$$(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_d,b_d)$$
which can be written as a union of closed intervals.

Or is there an other reason, why it is sufficient to suppose that $R$ is closed and bounded to prove that $m^*(R)=v(R)$?? (Wondering)