Prove that the Cantor ternary set has Jordan content 0. Additionally, prove that the Cantor ternary set has uncountably many points.
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you mean measure zero ? – May 03 '14 at 13:41
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yea, but jordan measure. – user147566 May 03 '14 at 13:41
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what is your definition of jordan measure ? – May 03 '14 at 13:42
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change countable covering in lebesgue measure into finite covering – user147566 May 03 '14 at 13:47
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Alternatively, the Jordan measure is the Riemann integral of the characteristic function. So this is more than saying the Lebesgue measure is zero. For example the rationals in $[0,1]$ have Lebesgue measure zero, but not Jordan measure zero. – GEdgar May 03 '14 at 14:07
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Denote by $I_k$ the amount of the interval $[0,1]$ that remains after removing the middle thirds. Notice $|I_0| = 1 = (\frac{2}{3})^0, |I_1| = (\frac{2}{3})^1, |I_2| = ( \frac{2}{3} )^2$. By an easy induction argument, we observe that $|I_n| = (\frac{2}{3})^n$. This is so because $I_n$ is the union of $2^n$ closed intervals of length $\frac{1}{3^n}$. Next, we know $\mathcal{C} = \bigcap I_n \subseteq I_n $. This shows that $\mathcal{C}$ is covered by a union of $2^n$ closed intervals. Next, from calculus we know $|I_n| = ( \frac{2}{3})^n \to 0 $, hence for $\epsilon > 0$, we can find $N \in \mathbb{N}$ such that if $n > N$, then $|I_n|< \epsilon $. Therefore, $\mathcal{C}$ has content zero.
$\mathcal{C}$ denotes the cantor ternary set