Weird measurable set In the following, consider the Lebesgue measure in $\mathbb{R}^d$.
Consider $E\subseteq \mathbb{R}^d$ measurable, with $0\lt m(E)\lt\infty$, such that any measurable subset $F$ of $E$ satisfies $m(F)=m(E)$ or $m(F)=0$. What can we say about $E$? Does there exist such a set?
 A: You are asking if the Lebesgue measure has any atoms $E$, i.e., a minimal measurable set of positive measure. The answer is negative: the Lebesgue measure is atomless. [My answer assumes we're working with $\mathbb R$, but the main idea carries over for all dimensions.]
Proof 1. Fix any measurable $E$ of positive measure, and a real $r$ such that $0 < r < m(E)$. For $n \in \mathbb Z$, define $J_n := [n r, (n+1) r)$ and $E_n := E \cap J_n$. Thus the $J_n$'s (resp. $E_n$'s) partition $\mathbb R$ (resp. $E$) into sets of measure at most $r$.


*

*For any $n$, $E_n$ is a measurable subset of $E$.

*From $E_n \subseteq J_n$, we have $m(E_n) \leqslant m(J_n) = r < m(E)$. [In particular, $E_n$ has finite measure even if $m(E) = \infty$.]

*From subadditivity, we have $m(E) \leqslant \sum_n \ m(E_n)$. Thus, assuming $m(E) > 0$, there exists some $n$ such that $m(E_n) > 0$. Such an $E_n \subseteq E$ satisfies $0 < m(E_n) < m(E)$.  



Proof 2. (Slightly modified from Robert Israel's comment.) Let $B_r$ denote the (open or closed) ball of radius of $r$ about the origin. Then the function $h : [0, \infty) \to [0, \infty)$ defined by $h(r) := m(B_r \cap E)$ is both monotonically increasing and continuous.* Further, $h(0) = 0$, and $h(r) \to m(E)$ as $r \to \infty$. Therefore, by the intermediate value theorem, we can conclude that for every $0 \leqslant t \lt m(E)$, there exists some $0 \leqslant r \lt \infty$ such that $h(r) = m(E \cap B_r) = t$. 
*Proof left as exercise.

Edit. I rewrote my answer so as not to appeal to contradiction. I hope this is simpler. Thanks to Nate, GEdgar and Michael for their comments regarding the terminology. Thanks to Robert for his comment.
