Need help on how to prove Prove that every n-dimensional rectangle on R^n is Lebesgue measurable

Prove that every n-dimensional rectangle on R^n is Lebesgue measurable

Linear Algebra
Aaron Chola Aaron Chola
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2 Answers

Every rectangle in $R^n$ is the product of n intervals, and intervals are measurable, so their product is measurable.

Martin Martin
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Let

$R = [a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_n,b_n] \subseteq \mathbb{R}^n,$


where $a_i≤b_ia_i \leq b_i $ for each $i=1,\dots,n$ .

1. Closedness.
Each interval $[a_i,b_i]$ is closed in $\mathbb{R}$. Since finite products of closed sets are closed in the product topology, $R$ is closed in $\mathbb{R}^n$.

2. Borel measurability.
Every closed set belongs to the Borel $\sigma-$ algebra of $\mathbb{R}^n$. Hence $R$ is a Borel set.

3. Lebesgue measurability.
By definition, the Lebesgue measure is the completion of the Borel measure with respect to outer measure. Therefore, every Borel set is Lebesgue measurable, in particular $R$.

4.Measure of $R$.
The Lebesgue measure of $R$ is given by
$m(R) = \prod_{i=1}^{n} (b_i - a_i)$.

Conclusion:
Thus every n-dimensional rectangle in $\mathbb{R}^n$ is Lebesgue measurable, with measure equal to the product of its side lengths.
Edward Kingman Edward Kingman
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