real analysis - The set $E= {xin [0,1]: sum_{j=1}^infty t^j|x−q_j|^{-r}

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

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 $I \cap \mathbb{Q} = \emptyset$.