Due March 29 Problem 1 $latex A\subset\mathbb R^d$ is measurable if and only if, for all $latex B\subset\mathbb R^d$, $latex |B|_* = |B\cap A|_* + |B\setminus A|_*$. Problem 2 Let $latex A\subset\mathbb R^d$. The following are equivalent. $latex A$ is measurable. $latex A = P\setminus M$, where $latex P$ is a $latex G_\delta$ set and $latex |M|=0$. $latex A = Q\cup N$, where $latex Q$ is an $latex F_\sigma$ set and $latex |N|=0$. Problem 3 Let $latex A\subset\mathbb R^d$ be a measurable set. For $latex \delta > 0$, let $latex \delta A = \{ \delta x: x\in A\}$. Then $latex \delta A$ ls measurable and $latex |\delta A| = \delta^d |A|$. For a $latex d$-tuple $latex \bar{\delta} = (\delta_1, \ldots, \delta_d)$ with each $latex \delta_j>0$, $latex j=1,\ldots,d$, define $latex \bar\delta A = \{(\delta_1 x_1, \ldots, \delta_d x_d): (x_1, \ldots, x_d)\in A\}$. Then $latex \bar\delta A$ is measurable and $latex |\bar\delta A| = \delta_1\cdots\delta_d |A|.$