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My problem is the following: Let's take a line segment. Question is, does it have an area (2D Jordan measure)?

I think it has $0$ area, because we know that $\emptyset$ has $0$ area, and if we construct a sqaure with sidelength $\delta>0$, which contains our line segment then the following must be true: $area\{\emptyset\} = 0 \leq area\{line$ $segment\} \leq area\{rectangle\} = \delta^2$. Let's say there is a number $b \geq 0$, that is between the area of the line segment and our square for any $\delta \geq 0$. Since the area approaches $0$ as $\delta$ shrinks smaller and smaller, this $b$ cannot be anything else than $0$, thus the area of the line segment is $0$ too.

I think here I made the assumption that the line segment is measurable. Can I make this? Why can I make this? If I cannot than what is a correct way of proving it?

Thanks in advance!

math_inquiry
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  • A line segment is Borel/Lebesgue measurable, as you can write it as a countable intersection of a sequence of open rectangles, as you have essentially already done in your construction. – Christopher A. Wong Jun 16 '20 at 06:57
  • Oh ok. Thank you! – math_inquiry Jun 16 '20 at 06:59
  • So, I'm not entirely clear on Lebesgue measure, as I haven't encountered it "officially" yet. But from what I've read having 0 Lebesgue measure doesn't mean that it has 0 Jordan content as well. It means it's either 0 or non-measurable. This is that I have a problem with. Sorry for not being clear in the first place! – math_inquiry Jun 16 '20 at 07:13
  • Sorry, I did miss that you were actually asking about Jordan measure. However, the argument is still the same here. Jordan measure is taken by approximating a set from below and above by unions of rectangles. A set is measurable if these two approximations coincide in the limit. Your argument is therefore correct. You're approximating from below with the empty set, and from above with rectangles of size $\delta$. At the limit both are 0, hence the set is measurable. – Christopher A. Wong Jun 16 '20 at 07:20
  • Thank you for the quick answers! – math_inquiry Jun 16 '20 at 07:23

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