Let $A_n\subset\mathbb{R}^N$ a bounded and Peano-Jordan measurable set for all $n\in\mathbb{N}$. Furthermore $A_n\cap A_k=\emptyset$ $n\ne k$. On $\mathbb{R}^N$ we consider the usual topology.
We suppose that \begin{equation} \bigcup_{n\in\mathbb{N}}A_n \end{equation} is bounded and Peano-Jordan measurable.
Can we say that \begin{equation} cl\big(\bigcup_n A_n\big)=\bigcup_ncl(A_n)\quad ? \end{equation} Thanks!
Let $\{x_1,\,x_2,\,x_3,\,\ldots\}$ be a countable subset of the Cantor middle thirds set (i.e. the Cantor ternary set). For example, such a set could be the set of endpoints of the maximal-size open intervals whose union is the complement of the Cantor set. Assume the enumeration is injective. That is, assume $i \neq j$ implies $x_i \neq x_j.$
For each positive integer $n,$ let $A_n = \{x_n\}.$ Then (a) each $A_n$ is Peano-Jordan measurable (with measure zero); (b) the $A_n$'s are bounded and pairwise disjoint; (c) the union of the $A_n$'s is a subset of the Cantor set, and hence the union is bounded; (d) the union of the $A_n$'s is Peano-Jordan measurable (with measure zero); (e) the equality you asked about is FALSE, because the closure of the union of the $A_n$'s is the entire Cantor set, which is uncountable and hence not equal to the union of the closure of the $A_n$'s.
The only nontrivial statement above is (d). Since the union is a subset of the Cantor middle thirds set, the outer Peano-Jordan measure of the union is less than or equal to the outer Peano-Jordan measure of the Cantor middle thirds set, which equals zero. Any set whose outer Peano-Jordan measure is zero will be Peano-Jordan measurable. This is because for any such set, its inner Peano-Jordan measure has to be less than or equal to zero, and hence equal to zero, since the inner Peano-Jordan measure cannot be negative, and therefore the outer Peano-Jordan measure equals the inner Peano-Jordan measure.
For readers whose native language is not English, the following pertains to (e):
$$ cl \left(\bigcup_{n=1}^{\infty} A_n \right) \; = \; cl \left(\bigcup_{n=1}^{\infty} \, \{x_n\} \right) \; = \; cl \left(\;\{x_1,\,x_2,\,x_3,\,\ldots\} \; \right) \; = \; \text{Cantor set} $$
$$ \bigcup_{n=1}^{\infty} \, cl(A_n) \; = \; \bigcup_{n=1}^{\infty} \, cl(\{x_n\}) \; = \; \bigcup_{n=1}^{\infty} \,\{x_n\} \; = \; \{x_1,\,x_2,\,x_3,\,\ldots\} \; \neq \; \text{Cantor set} $$
Incidentally, for EACH positive integer $N,$ a similar example exists in ${\mathbb R}^N$ by choosing an uncountable perfect nowhere dense set having $N$-Lebesgue measure zero and letting $\{x_1,\,x_2,\,x_3,\,\ldots\}$ be a countable dense subset of this perfect nowhere dense set. Or, when $N>1,$ just take the Cartesian product of all the sets in the previous example with ${\mathbb R}^{N-1}.$