Let $A_1, A_2$ be non-Borel sets in $\mathbb{R}^n$, with the property that $A_1 \cap A_2 = \emptyset$. It seems to me that the union $A = A_1 \cup A_2$ should also be non-Borel. Is this the case?
Clearly if $A_1, A_2$ were Borel, then $A$ would be as well. Additionally, I can see that if $A$ is Borel and one of $A_1, A_2$ is Borel, then so is the other, since (without loss of generality) $A \cap A^c_1 = A_2$. It seems to me like the finite union of disjoint non-Borel sets should be non-Borel, but I cannot seem to prove it, nor come up with a counterexample.
My thinking is generally that $A_1$ cannot be generated by open sets, and neither can $A_2$, so their union cannot be generated by open sets, but this is fuzzy, and ignores the idea that maybe the right pairing of $A_1$ and $A_2$ combine to make a Borel set. I suppose that this is somewhat in the same way that $(a,b]$ and $[b,c)$ are neither open nor closed, but their union $(a,c)$ is certainly open.
If anyone could provide a proof or counterexample, I would greatly appreciate it! Thank you!
Let $X$ be a topological space, and suppose $N \subseteq X$ is not a Borel set. Since the Borel sets are closed with respect to complements, the set $N^\complement = X \setminus N$ is also not a Borel set. But then $$ N \cup N^\complement = X $$ is the union of non-Borel sets. Therefore the union of non-Borel sets could be a Borel set (since $X$ is necessarily a Borel set).