Show $A=\bigcap_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})\subseteq\mathbb{R}$ is a $\lambda^1$ zero set

Question

I have the following task:

Let us write $\mathbb{Q}=\{q_k\}$ for $k\in\mathbb{N}$ and set:

$$A=\bigcap_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})\subseteq\mathbb{R}$$

Show:

(i) $A$ is a $\lambda^1$ zero set.

(ii) $\mathbb{R}\diagdown A$ is lean in $\mathbb{R}$

I really do not know how to prove the statements here. Can somebody help me with this task?

Hint: When $n=1,$ what is the sum of the lengths of the intervals $(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})$ for $k=1,2,3,\ldots$? When $n=2,$ what is the sum of the lengths of the intervals $(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})$ for $k=1,2,3,\ldots$? When $n=3,$ what is the sum of the lengths of the intervals $(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})$ for $k=1,2,3,\ldots$? (In each case you'll be finding the sum of an infinite geometric series.) — Dave L. Renfro, Dec 17 '23 at 13:56
Does this answer your question? Decomposing real line as a union of a nullset and a set of first category — Mittens, Dec 17 '23 at 14:18
@Mittens no not really, so the first part of the task is clear now. But how do I prove the second statement (ii) ? — Lukas Kretschmann, Dec 17 '23 at 14:35
This was very helpfull for the second part (ii) https://math.stackexchange.com/a/1705085/1258129 — Lukas Kretschmann, Dec 17 '23 at 18:56

Luigi Traino · Accepted Answer · 2023-12-17T15:01:12.930

Let's write $\lambda:=\lambda^1$ for simplicity (I suppose that $\lambda^1$ is the Lebesgue measure on $\mathbb{R}$) then for every $k,n\in \mathbb{N}$ we have that $$\lambda\left(\left(q_k-\frac{1}{2^{n+k}},q_k+\frac{1}{2^{n+k}}\right)\right)=\frac{1}{2^{n+k-1}}$$ So for $\sigma-$subadditivity of the measure we have that for fixed $n$ $$\lambda\left( \bigcup_{k\in \mathbb{N}}\left(q_k-\frac{1}{2^{n+k}},q_k+\frac{1}{2^{n+k}}\right)\right)\le \sum_{k=0}^\infty\lambda\left(\left(q_k-\frac{1}{2^{n+k}},q_k+\frac{1}{2^{n+k}}\right)\right)=\frac{1}{2^{n-1}}\sum_{k=0}^\infty\frac{1}{2^k}=\frac{1}{2^{n-1}}$$ Now let's define $A_n$ as $$A_n:=\bigcup_{k\in \mathbb{N}}\left(q_k-\frac{1}{2^{n+k}},q_k+\frac{1}{2^{n+k}}\right)$$ And so we have that $A_{n+1}\subseteq A_n$ for every $n$ and that $A=\bigcap_{n\in \mathbb{N}} A_n$. So by continuity of the measure we have that $\lambda(A_n)\longrightarrow \lambda(A)$ for $n \longrightarrow +\infty$. So if we take the limit in the inequality that we have proved we obtain $$\lambda(A_n)\le \frac{1}{2^{n-1}} \Longrightarrow \lim_{n\rightarrow \infty} \lambda(A_n)=\lambda(A)\le \lim_{n\rightarrow \infty}\frac{1}{2^{n-1}}=0$$ So $\lambda(A)\le 0$ which implies that $\lambda(A)=0$.

For the second point observe that: $$\mathbb{R}\setminus A=A^c=\bigcup_{n\in\mathbb{N}}\bigcap_{k \in \mathbb{N}}\left(q_k-\frac{1}{2^{n+k}},q_k+\frac{1}{2^{n+k}}\right)^c$$ So by definition of first category set (countable union of nowhere dense sets) you need to show that the $A_n^c$ are nowhere dense (the two ways are: since the $A_n^c$ are closed, you can show that the interior part of $A_n^c$ is empty, or you can show that $(a,b)\cap A_n^c$ is not dense for arbitrary $a,b \in \mathbb{R}$)

This looks very good! How do I show now that $\mathbb{R}\diagdown A$ is of first category in $\mathbb{R}$? — Lukas Kretschmann, Dec 17 '23 at 14:19
So I still need to show that in (ii) the set is of first category in $\mathbb{R}$ — Lukas Kretschmann, Dec 17 '23 at 14:27

Show $A=\bigcap_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}(q_k-\frac{1}{2^{k+n}},q_k+\frac{1}{2^{k+n}})\subseteq\mathbb{R}$ is a $\lambda^1$ zero set

1 Answers1