Let $\mathfrak{A}$ be an uncountable collection of Borel sets in $\mathbb{R}^d$ such that for any $A,B\in\mathfrak{A}$, either $A\subset B$ or $A\supset B$. Then is it necessarily true that the union $\bigcup\_{A\in\mathfrak{A}}A$ is Borel measurable?



Ordered union of Borel sets
### Construction


Consider some non-Borel set $Y \subset [0,1]$ (e.g. Vitali set).


Enumerate $Y$ using ordinals as $Y=\{x\_\alpha\}\_{\alpha < \beta}$.
Let $m$ denote the smallest ordinal such that $X=\{x\_\alpha\}\_{\alpha < m}$ is non-Borel. Note that $m\ge \omega\_1$, since otherwise $X$ would be at most countable. 


Then the family $\mathfrak A = \{A\_\gamma\}\_{\gamma<m}$, where $A\_\gamma = \{x\_\alpha\}\_{\alpha<\gamma}$, has the properties desired in the OP, but $\bigcup\_{A \in \mathfrak A} A$ is non-Borel.


For more details about ordinals see e.g. *Set theory* by T. Jech (2006).


### Discussion of cardinality


Let us prove that in fact $|X| = \aleph\_1$.


Indeed, by construction $\aleph\_1 \le |X| \le 2^{\aleph\_0}$.
Any uncountable Borel set has cardinality $2^{\aleph\_0}$ (e.g. by Theorem 13.6 in *Classical Descriptive Set Theory* by A.S. Kechris or Corollary 2C.3 in *Descriptive Set Theory* by Y.M. Moschovakis.). Therefore, if CH fails then $|X|\le \aleph\_1$, since otherwise there would exist a Borel subset with cardinality $\aleph\_1$ (by minimality of $m$). On the other hand if CH holds then immediately $|X| = \aleph\_1$.


*Most of this answer comes from the very useful comments in this thread.*


