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I'm stucked in this question:

If $E$ is a Lebesgue measurable set contained in the interval $[0,1]$ and there is a $\alpha>0$ such that $m(E \cap I) \geq \alpha m(I)$ for all open intervals $I \in [0, 1]$ then $m(E) = 1$.

How should I proceed to prove this?

heaven-of-intensity
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João Costa
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  • $m(x)$ is a measure of $x$? – LRDPRDX Nov 16 '17 at 18:16
  • Doesn't it have any further assumption? If you multiply the measure by a positive constant it is still consistent with the assumption. – heaven-of-intensity Nov 16 '17 at 18:16
  • What have you tried to do.? When answering questions, we like to know that the questioner has put some thought into it. Also please try to learn MathJax and use more and more of it in the future. – Stephen Meskin Nov 16 '17 at 18:19
  • Oh, maybe you mean the set $E$ is a Lebesgue measurable set. Look at here: https://math.stackexchange.com/questions/33819/question-on-measure-theory – heaven-of-intensity Nov 16 '17 at 18:24
  • I've proceeded like in this post. However using the Lebesgue Density Theorem that α should be equal zero. From there I can't conclude. – João Costa Nov 16 '17 at 18:33

3 Answers3

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Outline: (1) Show that $m(E\cap U)\ge\alpha m(U)$ for every open subset of $[0,1]$; (2) Use the approximation theorem for general Lebesgue measurable sets to show that $m(E\cap B)\ge \alpha m(B)$ for every Lebesgue measurable subset of $[0,1]$; (3) Consider what happens in the special cae $B=E^c$.

John Dawkins
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  • For which measurable set should (2) be applied? – Twnk Jun 20 '20 at 07:17
  • Take $B=E^c$ in (2). – John Dawkins Jun 22 '20 at 15:46
  • I mean, to prove (2), do I have to use the fact that for every Lebesgue measurable set $B$ and every $\epsilon>0$ there is an open set $U \supset B$ such that $m(U)<m(B)+\epsilon$? – Twnk Jun 23 '20 at 16:54
  • Or the fact that for every measurable set $B$ and every $\epsilon>0$ there is an open set $U \supset E \cap B$ such that $m(U)<m(E \cap B)+\epsilon$? I tried with both facts but I'm not getting the conclusion. – Twnk Jun 23 '20 at 16:55
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    Choose open $U$ containing $B$ so that $m(U\setminus B)<\epsilon$. Then $(U\cap E)\setminus (B\setminus E)\subset U\setminus B$, so that $$ \epsilon+m(B\cap E)>m(U\setminus B)+m(B\cap E) \ge m(U\cap E)\ge\alpha m(U)\ge \alpha m(B). $$ – John Dawkins Jun 25 '20 at 16:40
  • Did you mean $(U \cap E) \setminus (B \cap E) \subset U \setminus B$? – Twnk Jun 25 '20 at 19:34
  • I did! Thanks for the correction. – John Dawkins Jun 26 '20 at 15:07
  • Thanks for your help. – Twnk Jun 27 '20 at 04:30
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We prove it by contradiction. Assume the following holds: $m(E)=1-\epsilon$ for some $\epsilon>0$. Clearly $m(E^c)=\epsilon$. Now you would actually like to choose some open interval $I$ such that $I$ is around $E^c$. If you could do that then $m(E\cap I) $ would be small and $\alpha m(I)$ would be large and that gives you the contradiction. But how you choose such$I$? It can be the case that $E^c$ is scattered around the whole interval $[0,1]$ so finding an open interval $I$ "near" $E^c$ cannot be achieved. Luckily we have a supertool to fix that, namely Littlewood's First Principle.

Littlewood's First Principle (LFP): Let $A\subset\mathbb{R}$ be Lebesgue measurable with $m(A)<\infty$ and let $\delta>0$. Then there exists a finite collection of disjoint open intervals $\{B_1,...,B_k\}$ such that $m(A\triangle \bigcup_{i=1}^kB_i)<\delta$.

The triangle means symmetric difference, i.e. $A\triangle B=[A\cup B]\setminus [A\cap B]$. The theorem as stated above works also if we would take $[0,1]$ instead of $\mathbb{R}$ and the open intervals that we will get will be in $[0,1]$.

We use LFP for $E^c$. Let $\delta=\alpha\epsilon/2>0$, then we get $\{B_1,...,B_k\}$ with the mentioned property. Define $O:=\bigcup_{i=1}^NB_i$. So we have $m(E^c \triangle O)<\delta$.

Note that we have $\alpha m(B_j)\leq m(E\cap B_j )$ for all $j\in \{1,...,k\}$. Since the sets $B_j$ are mutually disjoint we get: \begin{align} \alpha m(O) \leq m(E\cap O) \end{align} Now do some little algebra: \begin{align} m(O) = m( E\cap O) + m(E^c \cap O) \end{align} Furthermore: \begin{align} m(E^c\triangle O) = m(E^c \cup O) - m(E^c\cap O) \end{align} So: \begin{align} m(O) = m( E\cap O)+m(E^c \cup O)-m(E^c\triangle O) \end{align} So now we get: \begin{align} \alpha ( m(E\cap O) +m(E^c \cup O)-m(E^c\triangle O)) &\leq m(E\cap O)\\ \alpha ( m(E^c \cup O)-m(E^c\triangle O)) &\leq (1-\alpha) m(E\cap O) \end{align}

Note also that $m(E\cap O)\leq m(E^c\triangle O)$ so: \begin{align} \alpha (m(E^c \cup O)-m(E^c\triangle O)) &\leq (1-\alpha) m(E^c\triangle O)\\ \alpha m(E^c \cup O) &\leq m(E^c\triangle O) \end{align} We finally get: \begin{align} m(E^c \cup O) < \frac \epsilon 2 \end{align} Remember that $\epsilon=m(E^c) \leq m(E^c\cup O)<\epsilon/2$, a contradiction. So $m(E)=1$.

Shashi
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Following your comment about the Lebesgue Density Theorem, if you want to use it, you can argue as follows:

It is clear that $0<\alpha <1.$ Consider $F:=[0,1]\setminus E.$ Since $I\cap (E\cup F)=(I\cap E)\cup (I\cap F)=I,$ we have $m(I\cap F)=m(I)-m(I\cap E)\le (1-\alpha )m(I).$ Then, $\frac{m(I\cap F)}{m(I)}\le 1-\alpha.$ Since this is true for all intervals $I\in [0,1],$ we conclude from the Lebesgue Density Theorem that $m(F)=0$ which now implies that $m(E)=1.$

Matematleta
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