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In the following, consider the Lebegue 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?

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Every set of positive measure contains a non-measurable subset. Therefore such an $E$ doesn't exist. – Listing Dec 3 '11 at 23:45
Yes, sorry. I was talking about measurable subsets. I'll edit the post. Thanks @Listing – leo Dec 3 '11 at 23:48
The assertion that no such set exists is the assertion that Lebesgue measure is "non-atomic". – GEdgar Dec 3 '11 at 23:59
(I intended this comment to be here rather than after the answer, so here it is again.) "Non-atomic" is a lousy term for what is usually called that; "atomless" is much better. The problem is that "non-atomic" sounds so similar to "not atomic". And that's a different thing. – Michael Hardy Dec 4 '11 at 2:48
up vote 16 down vote accepted

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.

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In short, Lebesgue measure is non-atomic ( – Nate Eldredge Dec 3 '11 at 23:56
"Non-atomic" is a lousy term for what is usually called that; "atomless" is much better. The problem is that "non-atomic" sounds so similar to "not atomic". And that's a different thing. – Michael Hardy Dec 4 '11 at 0:03
@MichaelHardy: Excellent point. – Nate Eldredge Dec 4 '11 at 1:19
@leo That's accurate. In fact, I rewrote the answer to emphasise this point better. I hope the new proof is simpler. Thanks, – Srivatsan Dec 4 '11 at 3:25
For $x \in {\mathbb R}^d$ define $C_x = \{y: y_i \le x_i \text{ for } i=1 \ldots d\}$. Then $F(x) = m(C_x \cap E)$ is a continuous function of $x \in {\mathbb R}^d$ with $F([-r,\ldots,-r]) \to 0$ and $F([r,\ldots,r]) \to m(E)$ as $r \to +\infty$. In particular, for any $0 < t < m(E)$ there exists $B \subset E$ with $m(B) = t$. – Robert Israel Dec 4 '11 at 9:28

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