Is the intersection of two non-open/non-measurable sets non-open/non-measurable? Very simple question. If I have two sets A and B that are both non-open in some topological space, is their intersection necessarily non-open?
And similarly, is the intersection of two non-measurable sets non-measurable?
I think the answer is that it could go either way. But I can't find any good discussion of this on the internet, and I am having trouble thinking of counterexamples...
 A: In both cases, make the intersection empty, which is both open and measurable.
For other examples, $$(1,2]\cap [0,2)=(1,2). $$ And, take disjoint  non-measurable sets $E_1,E_2$ and a disjoint (to both) measurable set $D $. Then $$A=E_1\cup D,\ \ \ \ B=E_2\cup D $$ are non-measurable and $$A\cap B=D $$ is measurable. 
A: Onsider $A=(0,1)\cup[2,3]$ and $B=(0,1)\cup[4,5]$. They are both not open but their intersection surely is. 
A: No. Take $[-1,1)$ and $(0,2]$ They are both non-open in $\mathbb R$ with respect to the usual topology, but their intersection is $(0,1)$, which is open.
And of course, for any two disjoint non-measurable sets, their intersection is empty and therefore measurable.
A: In the same style as the counterexample AJ gave for non-openness, you can construct a counterexample for measurability. Let $C$ be a non-measurable subset of $[2,3]$ and $D$ be a non-measurable subset of $[4,5]$, then $A=]0,1[\cup C$ and $B=]0,1[\cup D$ are non-measurable with a clearly measurable intersection.
A: For a large swathe of counterexamples, let $U \subset \mathbb{R}^n$ be an open subset, and let $x \ne y \in \mathbb{R}^n-U$ be two points. The subsets $U \cup \{x\}$ and $U \cup \{y\}$ are not open, and their intersection $U$ is open. A similar construction will work in many topological spaces.
A: Answers of the both question is NO.

*

*Take two disjoint non-open sets. Then their intersection is empty set which is open.


*Similarly for non measurable sets, take two disjoint non measurable sets (infact we can choose a non measurable set and it's complement).Then their intersection is empty set which is measurable.
Non trivial examples:

*

*$A=[0, 2) $ and $B=(1, 2]$ . Then both are non-open in the euclidean space but $A\cap B=(1, 2) $ is open.


*Let $\mathcal{B}\subset \Bbb{R}$ is a Bernstein set.Then $\mathcal{B}$ is non measurable( see here).
Let $A=\mathcal{B}\cup [0, 1]$ and $B=\mathcal{B}^c \cap [0, 1]$
Then $A\cap B=[0, 1]$ is measurable.
