TITLE: If $X\subset(\delta,\infty)$ and $ Y = \{ \sqrt x : x \in X \} $ then $ m^*Y \leq m^*X/(2\sqrt \delta) $
QUESTION [0 upvotes]: Problem:
  Let $\delta \in \mathbb{R}$ such that $\delta > 0$ and  $X \subset (\delta, \infty )$. Define the set $ Y = \{ \sqrt x : x \in X \} $. Show that $$ m^*Y \leq \frac{m^*X}{2\sqrt \delta} $$

Here's my approach but I'm stuck at one part.
Let $\{I_n\}_{n=1}^\infty \in \mathcal{C}(X)$ where $ \mathcal{C}(X)$ is the set of all open coverings of $X$ and $I_n$'s are open intervals. If I have proven that $\left\lbrace \frac{1}{2\sqrt \delta}I_n\right\rbrace_{n=1}^\infty \in \mathcal{C}(Y)$, then by taking the infimum over all open coverings of $Y$ yields (where $\mathcal{l}$ denotes the length of the open interval)
\begin{align*}
m^*Y &\leq \sum_{n=1}^\infty \mathcal{l}\left( \frac{1}{2\sqrt\delta} I_n \right) \\
&= \frac{1}{2\sqrt\delta} \sum_{n=1}^\infty \mathcal{l}\left( I_n \right) \\
&= \frac{m^*X}{2\sqrt\delta}
\end{align*} where the last inequality follows by taking the infimum over all open coverings of $X$.Thus, conclusion follows.
Can anyone provide hints to show that $\left\lbrace \frac{1}{2\sqrt \delta}I_n\right\rbrace_{n=1}^\infty \in \mathcal{C}(Y)$? Or maybe from the very start, my approach is wrong?

REPLY [1 votes]: Start with a covering $\{I_n\}$ of $X$ by open intervals $I_n = (a_n,b_n)$. You can assume without loss of generality that $a_n > \delta$ for each $n$. Instead of scaling the intervals by a factor of $\frac{1}{2 \sqrt \delta}$, define new intervals $J_n$ as follows.
Let $J_n = (\sqrt{a_n},\sqrt{b_n})$.  If $y \in Y$, then $y^2 \in X$ so that $a_n < y^2 < b_n$ for some $n$. Thus $\sqrt{a_n} < y < \sqrt{b_n}$ for the same $n$ implying $y \in J_n$ for some $n$.  Thus $\{J_n\}$ is a covering of $Y$. Note 
$$ \ell(J_n) = \sqrt{b_n} - \sqrt{a_n} = \frac{b_n - a_n}{\sqrt{b_n} + \sqrt{a_n}} \le \frac{b_n - a_n}{2 \sqrt{a_n}} \le \frac{\ell(I_n)}{2 \sqrt{\delta}}$$
since $\sqrt{b_n} > \sqrt{a_n} \ge \sqrt{\delta}$. Consequently $$m^*Y \le \sum_n \ell(J_n) \le \frac{1}{2\sqrt{\delta}} \sum_n \ell(I_n).$$
{"set_name": "stack_exchange", "score": 0, "question_id": 1456701}