# Property of Lebesgue measure in $\mathbb{R}^2$

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.

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

• 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? Commented Jul 24, 2020 at 13:34
• 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 $\overline{I\backslash E}=I$. Commented Jul 24, 2020 at 13:46
• 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? Commented Jul 24, 2020 at 16:12

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.

• 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? Commented Jul 24, 2020 at 13:46
• In the original question, $I$ is the unit square, not the unit interval. Commented Jul 24, 2020 at 13:48
• @uniquesolution Indeed, and I (meant to) treat it as unit-square. Did I overlook something? Commented Jul 24, 2020 at 13:50
• It isn't clear why your are looking at intervals $(x,y)$ inside a two-dimensional set, where they have measure zero. Commented Jul 24, 2020 at 13:52
• @uniquesolution $(x,y)$ is not an interval in my answer but a point in $I$. Commented Jul 24, 2020 at 13:52