Generalized Cantor set removing the open middle $\alpha_n$-th

Question

I'm working on some Measure Theory problems and came across this relatively interesting problem (to me, at least.)

Fix a sequence $\{\alpha_j\}_{j=1}^\infty \subset (0,1)$, and consider the generalized Cantor set $C_\alpha$ constructed by starting with $K_0=[0,1]$, and removing open middle $\alpha_1$-th of $K_0$ to produce $K_1$ and so on. Define $C_\alpha=\bigcap_{j=1}^\infty K_j$.

There are 3 problems associated with this:

1. Compute $m(C_\alpha)$.
2. Show that for any $\beta \in [0,1]$, some choice of $\{\alpha_j\}$ results in $m(C_a)=\beta$.
3. Show that $C_\alpha$ is nowhere dense.

For 1., I think the measure would be in the form of $(1-\alpha_1)(1-\alpha_2)(1-\alpha_3)...$ but I'm not sure if that intuition is right, or if that sums up to a 'nicer looking' term.

For 2., I have no idea how to go about this one, and would like some help on this part.

For 3., Since $C_\alpha$ is a countable intersection of closed sets, it should be closed as well. Now since it is closed, $C_\alpha$ is equal to its closure, and remains to show that it has an empty interior. But clearly, the set should have an empty interior, as $C_\alpha$ cannot contain an interval.
Now, I was hoping to use monotonicity to reach a contradiction if $C_\alpha$ is not nowhere dense, but I came to a point where I would need to know $m(C_\alpha)$ to do that, so came to a dead end since I wasn't able to solve 1.

I would greatly appreciated some help on these questions, thank you!

Answer 1

The only issue here is $(2).$ The rest is OK. I think we can do $(2)$ directly, without iteration. Say you want $m(C)=\beta$ and you need to choose $(\alpha_j)_j$ so that $\prod^\infty _{j=1}(1-\alpha_j)=\beta.$ Or equivalently, $\sum^{\infty}_{j=1}\ln (1-\alpha_j)=\ln \beta.$ (Note, if the product makes sense, then so does the sum). We can accomplish this by solving the equations $\ln (1-\alpha_j)=\frac{\ln \beta}{2^{j}},\ $ and if we do this, we get $\alpha_j=1-\exp\left ( \frac{\ln \beta}{2^{j}} \right ).$

Answer 2

Your answer to 1) is correct, because at step n the value $(1-\alpha_1)\ldots(1-\alpha_n)$ is clearly the total length of $\cap_{j=1}^n K_j$, and the measure of n intersection of a decreasing sequence of sets is equal to the limit of the measures, i.e. it is the limit of the partial products, i.e. it is the infinite product.

For 2), it's time to explore. Can you produce an infinite product equal to $1/2$? to $1/3$? Can you generalize?