I recently started studying Lebesgue measure, and I have a doubt :
Let $A\subset \Bbb{R}^n $ . The Lebesgue outer measure is defined as $ m^*(A) = \inf \{\sum_{k=1}^{\infty}l (I_k) : A \subseteq \cup_{k=1}^{\infty}I_k\}. $
Now, it is also true that if we have $A\subset \Bbb{R}^2 $, the Lebesgue outer measure of $A$ is the product of the lenght of the $x_1$ interval multiplied by the lenght of $x_2$ interval.
My problem is that I dont see exactly what is the relationship between the definition of of Lebesgue outer measure and the example of $A\subset \Bbb{R}^2 $. I would like some clarification about this. Thanks.
For example, let { $ (x,0) : 1 \leq x \leq 2$ } = A. I understand that the Lebesgue outer measure of this is zero. But I dont see how to apply the definition.
It's important to grasp what the general definition of the outer measure is. $$ m^*(A) = \inf\left\{\sum_1^\infty l(I_k):A \subseteq \cup_1^\infty I_k\right\} $$ This says that the outer measure is the infimum of many overestimations for volumes of coverings of $A$, which indicates why we call it an outer measure, it determines the volumes by measuring from the outside.
In your example $A = \{(x,0)\mid 1 \leq x \leq 2\} = [1,2]$, i.e it is a line segment in the plane. You can certainly cover it with a rectangle defined by say $[0,5]\times[-3,3]$, but is this a covering with the smallest possible volume? As Andrew Zhang writes, if you take a rectangle with width $\varepsilon>0$ and cover the line segment with it, you can shrink the width down to $0$ completely covering the line segment, but what is the area of such a rectangle?