# What is the Lebesgue measure of a open interval intersected with a generalized Cantor set with positive Lebesgue measure?

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.

• (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 $m(V\cap K)\gt(1-\varepsilon)m(V)$.
– bof
Commented Apr 2, 2021 at 0:58

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

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

• Why was the choice of an odd $n$ important? I am having trouble seeing how the last statement follows. Commented Apr 2, 2021 at 0:27
• @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 $K$. Commented Apr 2, 2021 at 0:28