Can a Ball in ℝn Be Proven as a Jordan Region with Zero Volume?

[sammycaps]

I was trying to show that any ball in ℝⁿ is a Jordan region, and this amounts to showing that its boundary has volume zero (Jordan content 0).

My TA proposed that you break the circle up into pieces of lines, and then adapt the following proof.

[
Pf: A Horizontal line of length 1 has content 0

A horizontal line of length 1 in ℝ² has volume 0 because if you cover it by n intervals of length ε, you see that nε=1 (or something like 1+ε*γ for some γ<1, if ε doesn't divide 1).

So, then define a grid of squares with sides ε. Then the volume of the grid, which bounds the volume of the circle, is ε²*n but since nε=1 (or about 1), you get ε²*(1/ε)=ε, and so we can bound the volume by an arbitrary ε, so the boundary of the ball has content 0.
]

The TA actually did this with balls instead of squares (i.e. balls of radius ε), but since we're trying to prove that a ball is a Jordan region, I wasn't sure about this.Anyway, I was not sure how to adapt this proof to the curved line, since there is no guarantee that the curved line will only pass through n of the squares of volume ε². I've seen at least 1 other way to do this with the compactness of the circle, but I was hoping that I could understand the TA's way.

Thanks.

 

[fresh_42]

I'm sure you can adapt the proof. In such a case, where we have a compact set (the boundary) which is a Lebesgue null set, it is also a Jordan null set and vice versa. The two measures on ball, open ball, and boundary are the same. The Jordan measurable sets don't form a $\sigma-$algebra, so in order to find a set which is of a different measure or not Jordan measurable at all, we have to find a countable union of Jordan measurable sets which is not Jordan measurable.

In the end it is the same difference than between Riemann integrability (Jordan measure) and Lebesgue integrability (Lebesgue measure) - and the ball isn't an example where the two are different.