If $A$ is measurable then $A/2$, $-A$ are measurable sets?

Question

Is it true that if $A \subseteq \mathbb{R}$ is Lebesgue measurable set, then

i) the set $A/2$ is Lebesgue measurable,

ii) the set $-A$ is Lebesgue measurable.

The definition I have for Lebesgue measurability of a set $A$ is: Let $\lambda^*()$ be the Lebesgue outer measure. A set $A$ is called Lebesgue measurable if $\lambda^*(X) = \lambda^*(X \cap A) + \lambda^*(X \cap A^c)$, for any arbitrary set $X$.

I think both are true but I can't show them. A proof sketch or a counter example would be very helpful. Thanks!

Answer 1

(Literally copy pasted)

• Let us say you are working with $-A$. Fix any $X$, and note that $X \cap -A$ = $-(-X \cap A)$. You can do something similar with $A^c$.

• To use the fact that $A$ is Lebesgue measurable, one correlates covers by rectangles of $-X \cap A$ with covers of $X \cap -A$. This is because they are the same set up to translation.

Now, for $\frac A2$, maybe you will find some other relation like $X \cap \frac{A}{2} = ? \cap A$, and something similar for $A^c$, and the same sort of tricks will apply.