The set $E= \{x\in [0,1]: \sum_{j=1}^\infty t^j|x−q_j|^{-r} <\infty\}$ does not contain all irrational numbers in $[0,1]$

Question

Let $q_1,q_2,q_3,...$ be an enumeration of $\mathbb{Q}\cap[0,1]$ and let $r,t \in (0,1).$ Consider the set $$E= \{x\in [0,1]: \sum_{j=1}^\infty t^j|x−q_j|^{-r} <\infty\} $$

(a) Show that $E\neq [0, 1]$ \ $\mathbb{Q}.$

The second part of the question is to show $m(E) = 1$. This is easy by showing the $L_1([0, 1])$ norm of $f(x) = \sum_{j=1}^\infty t^j|x−q_j|^{-r} < \infty.$ But I am at a loss of how to show $(a)$. Thanks a bunch.

Answer 1

I inductively construct $x \not \in E$.

Initialize the construction by defining $I_0 = [0,1]$ and $j_0 = 0$.

For $k=1,2,\dots$, find $j_k > j_{k-1}$ so that $q_{j_k} \in I_{k-1}$. This is always possible because of the density of the rationals. Choose $\delta_k$ small enough that $t^{j_k} \delta_k^{-r} > 1/k$, and define

$$I_k = I_{k-1} \cap [q_{j_k}-\delta_k,q_{j_k}+\delta_k].$$

Having defined all the $I_k$, define

$$I = \bigcap_{k=0}^\infty I_k.$$

You should be able to justify that $I$ is nonempty, and so any $x \in I$ is our candidate.

Because of how $\delta_k$ were chosen, if $x \in I$, then we have

$$\sum_{j=1}^\infty t^j |x-q_j|^{-r} \geq \sum_{k=1}^\infty t^{j_k} |x-q_{j_k}|^{-r} \geq \sum_{k=1}^\infty 1/k = \infty.$$

as desired.

I have not actually shown that $x \not \in \mathbb{Q}$. I suspect this is already the case with the construction as written, but it is easier to prove it by making the following modification. For $k \geq 2$, choose $\delta_k$ perhaps smaller, so that $q_{k-1} \not \in I_k$. Then $I \cap \mathbb{Q} = \emptyset$.

Answer 2

Let's, for the fun of it, throw some topology at the problem.

Let $X = [0,1] \setminus \mathbb{Q}$. For every $n\in\mathbb{N}$, the function

$$f_n \colon x \mapsto \sum_{j=1}^n t^j \lvert x-q_j\rvert^{-r}$$

is continuous on $X$. Hence, for every $k \in \mathbb{N}$, the set

$$A_k = \bigcap_{n\in\mathbb{N}} f_n^{-1}([0,k])$$

is closed (in $X$ of course). Further, $A_k$ has empty interior for all $k$, since for all $x\in X$ and $\varepsilon > 0$ the interval $(x-\varepsilon, x+\varepsilon)\cap (0,1)$ contains a rational point, say $q_m$, and the sequence $x_k = q_m + 2^{-k}\pi$ has its tail in the neighbourhood $(x-\varepsilon,x+\varepsilon)\cap X$ of $x$ in $X$, and $f_n(x_k) \to +\infty$ for all $n \geqslant m$.

Thus

$$E = \bigcup_{k\in\mathbb{N}} A_k$$

is a countable union of closed sets with empty interior, that is, a meagre subset of $X$ (a set of the first category).

But, $X$ is a $G_\delta$ set in the compact space $[0,1]$, hence a Baire space, and therefore

• $E$ has no interior point (in $X$),
• the complement of $E$ in $X$, $X\setminus E$ is a dense ($G_\delta$) subset of $X$ that is of the second category (it is itself a Baire space).

Since $[0,1]\setminus X$ is meagre in $[0,1]$, $E$ is also meagre in $[0,1]$, and thus $E$ is a topologically small (meagre) set of full measure.