# if $A$ has Lebesgue outer measure $0$ then so does $B=\left\{x^2: x\in A \right\}$

Let $A$ be a subset of $\mathbb{R}$ such that $\mu^*(A)=0$, where $\mu^*$ is the Lebesgue outer measure. Prove that if $B=\left\{x^2: x\in A \right\}$, then $\mu^*(B)=0$.

Recall that the Lebesgue outer measure of a subset $A$ of $\mathbb{R}$ is defined as

$$\mu^*(A)=\inf \left\{\sum_{i \geq 1} (b_i-a_i): (a_i, b_i) \subseteq \mathbb{R}, A\subseteq \cup_{i\geq 1} (a_i, b_i) \right\}.$$

Since $B=\left\{x^2: x\in A \right\}$, then if $A\subseteq \cup_{i\geq 1} (a_i, b_i)$, then $B\subseteq \cup_{i\geq 1} (a_i^2, b_i^2)$. Then

$$\mu^*(B)=\inf \left\{\sum_{i \geq 1} (b_i^2-a_i^2): (a_i^2, b_i^2) \subseteq \mathbb{R}, A\subseteq \cup_{i\geq 1} (a_i, b_i) \right\}=\inf \left\{\sum_{i \geq 1} (b_i+a_i)(b_i-a_i): (a_i^2, b_i^2) \subseteq \mathbb{R}, A\subseteq \cup_{i\geq 1} (a_i, b_i) \right\}.$$

From here I am not sure of what to do, although it seems like I am on the verge of getting the solution.

• a less than or equal x less than or equal b implies x^2 lands in between a^2 and b^2 only if a, b are nonnegative. So you need to first consider the case where A a subset [0,n] for some fixed n. Dec 10 '15 at 6:21

You have to get in control of those terms $(b_i+a_i)$ and here is a way to do so: It suffices to show that $B_n := B \cap [-n,n]$ has measure zero for all $n$. For a fixed $n$, the set $B_n$ is bounded, hence you can assume your $b_i$ to be bounded.
• What is the set $B$? Dec 10 '15 at 18:31