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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.2011-12-03
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    Yes, sorry. I was talking about measurable subsets. I'll edit the post. Thanks @Listing2011-12-03
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    The assertion that no such set exists is the assertion that Lebesgue measure is "non-atomic".2011-12-03
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    (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.2011-12-04

1 Answers 1

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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 (http://en.wikipedia.org/wiki/Atomic_measure).2011-12-03
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    "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.2011-12-04
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    @MichaelHardy: Excellent point.2011-12-04
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    Thanks by your answer. So, we can conclude that for any measurable $E\subseteq \mathbb{R}^d$ with $0\lt m(E)\lt \infty$ there exist $B\subset E$ (strictly contained) such that $0\lt m(B)\lt m(E)$.2011-12-04
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    @leo That's accurate. In fact, I rewrote the answer to emphasise this point better. I hope the new proof is simpler. Thanks,2011-12-04
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    Thanks to you, I'm completely satisfied :)2011-12-04
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    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 $m(B) = t$.2011-12-04
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    @Robert, I like that approach a lot. (Actually, I was thinking of a similar idea when I woke up in the morning.) But I don't like that this is buried inside the comments. If you write it as an answer, I will upvote it. If not, then I can include it in my answer, acking you. =)2011-12-04
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    @RobertIsrael, I also will upvote you if you elaborate this as an answer. What do you mean by $C_x=\{y:y_i\leq x_i \text{ for } i=1,\ldots,d\}$?2011-12-04
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    @leo I have expanded my answer incorporating a variant of Robert's argument.2011-12-06
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    You are very kind. Thank you @Srivatsan.2011-12-06
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    @Srivatsan Thanks for your great answer. May I know whether it's possible to prove there is a Borel set inside a measurable set with positive measure using your method?2014-10-19
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    @RobertIsrael: How do you check that it is continuous ad-hoc? Certainly it is continuous from the 'left' and from the 'right' but both together? I doubt it can be seen without going in a circular argument already implicitly using that Lebesgue was atomless. Think for example about the Dirac measure it is continuous from the 'left' and the 'right' but not continuous in 'total'. But to know that it was not continuous in 'total' was sort of equivalent of knowing it has an atom.2014-10-23
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    Given $\epsilon > 0$, there is some $R$ such that $m(E \backslash [-R,R]^d) < \epsilon/2$. $|m(C_x \cap [-R,R]^d) - m(C_y \cap [-R,R]^d)| \le d (2R)^{d-1} |x - y|$. So $|F(x) - F(y)| < \epsilon/2 + d (2R)^{d-1} |x - y| < \epsilon$ if $|x - y| < \ldots$.2014-10-24
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    @RobertIsrael: Thanks. :)2014-10-24