Can we claim that $A \cap B$ or $A \cup B$ is not in $\mathcal{B}(\mathbb{R}^n))$?

Question

This question can be posed for more general sigma algebras, but let's stick to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ for now. Suppose $A \in \mathcal{B}(\mathbb{R}^n))$, and $B_n \notin \mathcal{B}(\mathbb{R}^n))$.

1. Can we claim that $A \cap B$ or $A \cup B$ is not in $\mathcal{B}(\mathbb{R}^n))$?

2. Then what about $\bigcup_{n\in \mathbb{N}} B_n$ or $\bigcap_{n\in \mathbb{N}} B_n$; can we claim that these sets are not measurable?

(I'm asking because I don't have a proof. This is a question that has crossed my mind several times, but I've never been able to satisfactorily prove/disprove.)

Answer 1

For 1: Suppose, to the contrary, that both $A\cup B$ and $A\cap B$ were measurable. Then $A \mathop{\triangle} B = (A\cup B) \setminus (A\cap B)$ is also measurable, where $A \mathop{\triangle} B$ denotes the symmetric difference. Therefore, $B = A \mathop{\triangle} (A \mathop{\triangle} B)$ would also be measurable, giving a contradiction. Thus, we conclude that if $A$ is measurable but $B$ is not measurable, then at least one of $A\cup B$ or $A\cap B$ must be non-measurable.

Answer 2

The answer to 1 is negative, $A\cap B$ can be measurable, for example if $A$ and $B$ are disjoint (or meet in a single point if you don't want them to be disjoint), while $A\cup B$ can be measurable as well, for example if $B\subseteq A$.

For 2 the answer is also negative, countable unions of nonmeasurable sets can be measurable, in fact even finite unions can, just pick a nonmeasurable set and its complement for example.

Edit: after reading Daniel's answer I realized I probably misread the "or" in your question. My example for question 2 however does show that both $\bigcup B_n$ and $\bigcap B_n$ can be measurable when none of the $B_n$ are, so I'll leave my answer here