Let $x\in \mathbb{R}^n$ and $A\subset \mathbb{R}^n$ be a set of positive Lebesgue measure and for any $r>0$ the set $$B(x,r)\cap A$$ is non empty. Let $D^c$ be a measure zero set and hence $D$ is a dense set in $\mathbb{R}^n$ and $A\cap D$ is non empty. Is it true that for any $r>0$ the set $$B(x,r)\cap A\cap D$$ is non empty?
($B(x,r) $ denotes the open ball centered at $x$ with radius $r$ in $\mathbb{R}$)
Best Answer
This answer has been edited from the original version in order to match the current formulation of the problem.
First, I comment that the extra hypothesis on $A$ is rather weak as stated. Indeed, if $A$ is nonempty then there is some $x\in \mathbb{R}^n$ such that for all $r>0$, $B(x,r)\cap A$ is nonempty. Just let $x$ be any point in $A$.
We can also use this to see that the answer to the question is no as stated. Let $A=[0,1]^n\cup\{x\}$ where $x$ is some point not in $[0,1]^n$ (so the hypotheses are satisfied since $x\in A$). Let $D=\mathbb{R}^n\setminus \{x\}$. Then $B(r,x)\cap A\cap D=\emptyset$ for small enough $r$.
In light of the comments below (on the original answer), one could instead ask:
Now the previous example doesn't work since $x$ is not a limit point of $A$. But we can easily modify it to work by adding a convergent sequence outside of $[0,1]^n$, rather than an isolated point. For example, in $\mathbb{R}$, let $A=[0,1]\cup\{2-\frac{1}{n}:n>0\}$ (the interval $[0,1]$ is only there to ensure positive measure, it serves no other purpose). Let $D=\mathbb{R}\setminus \{2-\frac{1}{n}:n>1\}$. So $2$ is a limit point of $A$, but not a limit point of $A\cap D=[0,1]$.