Does set with Lebesgue-Mass nonzero have almost surely an open subset Let $A$ be a arbitrary borel set in $\mathbb R^d$ with $\lambda(A)>0$.
Question: Does there exist a set $N$ with $\lambda(N)=0$, such that $A\cup N$ contains a nontrivial open subset?
 A: Consider any $A$ that is closed nowhere dense (e.g. a fat Cantor set in $ℝ$). If $A ∪ N$ contains a nonempty open set, then the same is true for $N \setminus A$, which is not possible.
A: No. For example, take $A$ to be the unit interval minus all balls of rational center $p/q \in (0,1)$ and radius $1/((q+1)2^{q+2})$. 
$$A=(0,1)-\cup_{p,q} B\left(\frac{p}q,\frac{1}{(q+1)2^{q+2}}\right).$$
Let's first check this has positive measure.
$$\lambda\left(\cup_{p,q} B\left(\frac{p}q,\frac{1}{(q+1)2^{q+2}}\right) \right)\leq \sum_{q\geq 1, 0\leq p \leq q} \frac{2}{(q+1)2^{q+2}},$$
so
$$\lambda(\cup_{p,q} B(p/q,1/((q+1)2^{q+2})\leq \sum_{q\geq 1} \frac1{2^{q+1}}=\frac12.$$
So $A$ has measure at least $1/2$. 
Now let $N$ be a Borel set such that $N\cup A$ contains a non-empty open subset. This non-empty open subset contains a rational number, say $p/q$, and a neighborhood of it, say $B(p/q,r)\cap(0,1)$ for some $r>0$. We can safely reduce $r$ and assume that $r<1/((q+1)2^{q+1})$. But then $B(p/q,r)\cap A=\emptyset$, so $B(p/q,r)\cap(0,1) \subset N$. This implies that $\lambda(N)\geq r>0$.
A: There is an open set $U$ containing $\mathbb Q$ such that $\lambda (U) <\infty.$ Let $A = \mathbb R\setminus U.$ Then $A$ is closed and $\lambda(A) >0.$ Suppose $V$ is nonempty and open. Then $V$ contains a rational, hence $V\setminus A$ is nonempty. But since $A$ is closed, $V\setminus A$ is also open. All nonempty open sets have positive measure. Thus if $\lambda (N)=0,$ $A\cup N$ does not contain $V.$ Since $V$ was arbitrary, we have a counterexample.
