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The Smith-Volterra-Cantor set SVC or the so-called "Fat Cantor set" is a nowhere dense set with a positive Lebesgue measure.

The following is a question from A radical approach to Lebesgue’s theory of integration by Bressoud $Q\, 4.4.6$ pg $118$.

Define $K_1$ to be the fat Cantor set and $K_2$ to be the subset of $[0,1]$ formed by putting a copy of $K_1$ inside every open interval in $[0,1] \setminus K_1$. Similarly define $K_3$ to be the subset of $[0,1]$ formed by putting a copy of $K_1$ inside every open interval in $[0,1] \setminus (K_1 \cup K_2)$. In general define $K_n$ to be the subset of $[0,1]$ formed by putting a copy of $K_1$ inside every open interval in $[0,1] \setminus (K_1\cup K_2\cup \ldots \cup K_{n-1})$. Define $$K=\bigcup_{n=1}^{\infty} K_n \, .$$ Show that $K$ is of full measure.

Q1. What does it mean to put a copy of $K_1$ inside every open interval in $[0,1] \setminus K_1 $?

My understanding:- The open interval $(\frac{3}{8}, \frac{5}{8}) \in [0,1] \setminus K_1$. I guess to put a copy of $K_1$ inside it is to repeat the process by which we got the fat Cantor set from $[0,1]$ on $(\frac{3}{8}, \frac{5}{8})$ with shorter intervals being removed now. Am I correct?

Q2. How to show that $K$ is of full measure?

Thanks!

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  • $\begingroup$ @KaviRamaMurthy I'm not sure about that. Though this question is from a very standard book (Cambridge University Press) and it's highly unlikely it's wrong. $\endgroup$
    – miyagi_do
    Commented Aug 7, 2018 at 6:53
  • $\begingroup$ Just look at what happens to the measure of the complement every time you iterate the procedure. $\endgroup$
    – shalop
    Commented Aug 7, 2018 at 20:11

1 Answer 1

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You get the fat Cantor set by deleting countably many open intervals of total measure $\frac12$ so that the fat Cantor set has measure $\frac12.$ In each of the deleted open open intervals you do the fat Cantor set construction, as you described for $(\frac38,\frac58),$ so that you delete countably many open interval (countable union of countable sets is countable) of total length $1/4$ and therefore $\mu(K_1)=1/2-1/4=1/4.$ Lather, rinse, repeat.

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