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How to prove that the Lebesgue measure has no atoms: $$\lambda:\mathbb{R}^n\to\mathbb{R}_+$$ (Recall the definition for atoms: Measure Atoms: Definition?)

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freishahiri
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Here is one way: Let $B_r$ be the box $\{x\in\mathbb{R}^n: |x_k|\le r\text{ for } k=1,\ldots,n\}$. You can easily prove that $\lambda(B_r\cap A)$ is a continuous function of $r$, for any Lebesgue measurable set $A$. From this, it follows immediately that $A$ is not an atom.

Edit: To explain the last step, $\lambda(B_r\cap A)$ is a continuous function of $r$ taking the value $0$ at $r=0$ and $\lambda(A)$ in the limit $r\to\infty$. But if $A$ is an atom, this function can only take two values.

Harald Hanche-Olsen
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  • Can you explain the last step? – freishahiri Oct 23 '14 at 20:19
  • I added an explanation. – Harald Hanche-Olsen Oct 23 '14 at 20:26
  • Hey that argument works for any metric space, indeed even more that proves that for Borel algebras induced by metric topology any atom of a Borel measure must arise from a singleton an therefore being a point mass, isn't it? – freishahiri Oct 23 '14 at 21:01
  • How do you check that it is continuous ad-hoc? Certainly it is continuous from 'below' and from 'above' 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 'below' and the 'above', too, 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. – freishahiri Oct 23 '14 at 23:22
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    @Freeze_S If $r<s$ then $B_r\cap A\le B_s\cap A$, so $|\lambda(B_s\cap A)-\lambda(B_r\cap A)|=\lambda((B_s\cap A)\setminus(B_r\cap A))=\cdots$ $\cdots=\lambda((B_s\setminus B_r)\cap A)\le\lambda(B_s\setminus B_r)=\lambda(B_s)-\lambda(B_r)=(2s)^n-(2r)^n$. – Harald Hanche-Olsen Oct 24 '14 at 06:48
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Assuming, the Lebesgue measure has an atom $A\in\mathcal{B}(\mathbb{R}^n)$.

First consider growing boxes: $$Q_k:=[-k,k]^n$$ By definition one of the following always applies: $$\lambda(A\cap Q_k)=0\lor\lambda(A\cap Q_k^c)=0$$ If always the former case applies then there's a contradiction as $0=\mu(A\cap Q_k)\to\lambda(A)>0$.

So the atom splits into one within a finite box $A':=A\cap Q_K$.

Next consider cutting the box in halves: $$Q_{K;i,\pm}:=[-K,K]^n\cap\left([-K,K]^{i-1}\times\mathbb{R}_\pm\times[-K,K]^{n-i}\right)$$ Then again one of the following applies: $$\lambda(A'\cap Q_{K;i,+})=0\lor\lambda(A'\cap Q_{K;i,-})=0$$ Proceeding this way one obtains at a certain point a box of size: $$\lambda(A)>\left(\frac{K}{2}\right)^L=\lambda(Q_{K;L})\leq\lambda(A'\cap Q_{K;L})=\lambda(A')=\lambda(A)\quad(\lambda(A)>0)$$ That is a contradiction!

Concluding, the Lebesgue measure has no atoms.

freishahiri
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  • This seems like a far more complicated argument than what this question ought to require. – Michael Hardy Oct 23 '14 at 20:07
  • What could be a different way? – freishahiri Oct 23 '14 at 20:12
  • My answer could be a different way. – Harald Hanche-Olsen Oct 23 '14 at 20:16
  • At least in one dimension, you can do the following. Let $A \subset \mathbb{R}$ with $m(A)>0$. Let $U \supset A$ be an open set. Then $U=\bigcup_{n=1}^\infty I_n$ where $I_n$ are disjoint (possibly empty) intervals. Then $m(A) = m(U \cap A) = m \left ( \bigcup_{n=1}^\infty I_n \cap A \right ) = \sum_{n=1}^\infty m(I_n \cap A) > 0$. Hence one of $I_n \cap A$ has positive measure. Now you only need to show that one of them has measure strictly less than that of $A$. But note that if they don't, then one of them has all the measure, which means you can cut it in half to get the result. – Ian Oct 23 '14 at 20:20
  • @Ian: Well that is precisely my argument ;) – freishahiri Oct 23 '14 at 23:26