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Let $I=[0,1]\times [0,1]$ and $E\subset \mathbb{R}^2,$ be a set of zero Lebesgue measure. Is it true that $$\overline{I\setminus E}=I?$$

I guess that the counterexample will be some form space filling curve.

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2 Answers 2

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If the complement of $E$ is not dense in $I$, then $E$ contains some open rectangle, so it cannot be of measure zero.

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  • $\begingroup$ Okay $E^c$ should be dense in $I.$ But since $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$ we get that $\overline{I\setminus E}\subset I$. How to show equality? $\endgroup$ Commented Jul 24, 2020 at 13:34
  • $\begingroup$ The complement of $E$ in $I$ is $I\backslash E$. By definition, a subset of $I$ is dense in $I$ if and only if its closure coincides with $I$. This is how we show the equality $\overline{I\backslash E}=I$. $\endgroup$ Commented Jul 24, 2020 at 13:46
  • $\begingroup$ So according to your answer we could replace $I$ by $A\times B$ where $A,B\subset \mathbb{R}$ are closed positive Lebesgue measure set. Isn't it? $\endgroup$ Commented Jul 24, 2020 at 16:12
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Yes, it is true. Proving that $\overline{I\setminus E}\subseteq I$ is trivial.

For proving $I\subseteq\overline{I\setminus E}$ let $(x,y)\in I$ and assume that $(x,y)\notin\overline{I\setminus E}$.

Then some open set $U$ must exist with $(x,y)\in U$ and $U\cap(I\setminus E)=\varnothing$ or equivalently $U\cap I\subseteq E$.

But $U\cap I$ has positive Lebesgue measure.

So this contradicts that $E$ is a set with Lebesgue measure zero and we conclude that our assumption must be wrong.

That means that $(x,y)\in I$ implies that $(x,y)\in\overline{I\setminus E}$ and we are ready.

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  • $\begingroup$ Since you are only using $U\cap I$ is a set of positive Lebesgue measure, this leads me to ask whether the above is true if $I=A\times B$, where $ A, B \subset \mathbb{R}$ are closed and positive Lebesgue measure? $\endgroup$ Commented Jul 24, 2020 at 13:46
  • $\begingroup$ In the original question, $I$ is the unit square, not the unit interval. $\endgroup$ Commented Jul 24, 2020 at 13:48
  • $\begingroup$ @uniquesolution Indeed, and I (meant to) treat it as unit-square. Did I overlook something? $\endgroup$
    – drhab
    Commented Jul 24, 2020 at 13:50
  • $\begingroup$ It isn't clear why your are looking at intervals $(x,y)$ inside a two-dimensional set, where they have measure zero. $\endgroup$ Commented Jul 24, 2020 at 13:52
  • $\begingroup$ @uniquesolution $(x,y)$ is not an interval in my answer but a point in $I$. $\endgroup$
    – drhab
    Commented Jul 24, 2020 at 13:52

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