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In Folland's book Real Analysis: Modern Techniques and Their Applications, p. 39 has an explanation of how to construct a generalized Cantor set with positive measure. For reference, the construction of the generalized Cantor set involves starting with $K_0 = [0,1]$ removing the open interval of length $\alpha_1$ (for $\alpha_1 \in (0,1)$) centered at the midpoint, and at each step $j$, creating $K_j$ by removing the open middle $\alpha_j^{th}$ from each interval in $K_{j-1}$.

After we construct a generalized Cantor set $K$ with positive measure $\beta \in (0,1)$, I want to know how to calculate the measure of any open interval, call it $V$ intersected with $K$. I am guessing that it is just the length of the $V$ inside $[0,1]$ times $\beta$, and I would like to know how to rigorously show this, assuming that my guess is correct. Thanks.

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  • $\begingroup$ (1) The intersection can be empty. Since $K$ is a nowhere dense set, every open interval $I$ contains an open subinterval $V$ such that $V\cap K=\varnothing$. (2) The intersection can be nonempty with arbitrarily small measure. Start with an open interval $V$ such that $V\cap K=\varnothing$. Since $f(t)=m((V+t)\cap K)$ is a continuous function of $t$, given $\varepsilon\gt0$ we have $0\lt f(t)\lt\varepsilon$ for some value of $t$. (3). Since $m(K)\gt0$, there is an open interval $V$ such that $m(V\cap K)\gt(1-\varepsilon)m(V)$. $\endgroup$
    – bof
    Commented Apr 2, 2021 at 0:58

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The guess is not correct. Choose an odd $n=2m+1\in\Bbb Z^+$ large enough so that $\frac1n<\alpha_1$. Then

$$\beta=\sum_{k=0}^{n-1}m\left(K\cap\left(\frac{k}n,\frac{k+1}n\right)\right)\,,$$

so

$$m\left(K\cap\left(\frac{k}n,\frac{k+1}n\right)\right)>0$$

for some $k\in\{0,\ldots,n-1\}$, but

$$m\left(K\cap\left(\frac{m}n,\frac{m+1}n\right)\right)=m(\varnothing)=0,.$$

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  • $\begingroup$ Why was the choice of an odd $n$ important? I am having trouble seeing how the last statement follows. $\endgroup$ Commented Apr 2, 2021 at 0:27
  • $\begingroup$ @RichardKYu: When $n=2m+1$, the interval $\left(\frac{m}n,\frac{m+1}n\right)$ is centred at $\frac12$, and its length is less than $\alpha_1$, so it’s contained in the first open interval that is removed in the construction of $K$. $\endgroup$ Commented Apr 2, 2021 at 0:28

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