This is a sort of follow-up question to [this old post](https://mathoverflow.net/questions/54923/is-there-a-disjoint-union-sigma-algebra) I came across.


**Setup:**




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Let $\{X\_n\}\_{n \in \mathbb{N}}$ be a collection of Hausdorff topological spaces and let $\{\Sigma\_n\}\_{n \in \mathbb{N}}$ be their respective Borel $\sigma$-algebras. Let $\left\{\mu\_n\right\}\_{n \in \mathbb{N}}$ be a family of *finite* Borel measures on $X\_n$, respectively. 


**Question:**




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Can one define the disjoint union of $\left\{(X\_n,\Sigma\_n,\mu\_n)\right\}\_{n \in \mathbb{N}}$? 


**What I have So far:**




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1. Then one can define the disjoint union $\sigma$-algebra on $\bigsqcup\_{n \in \mathbb{N}} X\_n$ as the subset $\Sigma$ of $\bigsqcup\_{n \in \mathbb{N}} \Sigma\_n$ consisting of all sets $S$ satisfying
$$
S \cap X\_n \in \Sigma\_n \,(\forall n \in \mathbb{N}).
$$
Therefore, one may define the *coproduct* measurable space $(X,\Sigma)$ to be $\left(\bigsqcup\_{n\in \mathbb{N}} X\_n, \Sigma\right)$. 


**Note:** *(I'm curious, does this correspond to the Borel $\sigma$-algebra on the disjoint union somehow.)*
2. Does the following:
$$
\mu\triangleq \sum\_{n =1}^{\infty} \mu\_n,
$$
work?


I can't find a problem as $\Sigma$ restricted to (the copy of) $\Sigma\_n$ (within $\Sigma$) is contained in $\Sigma\_n$. So (informally written) $\mu\rvert\_{\Sigma\_n}=\mu\_n$ but is defined on a (potentially) smaller $\sigma$-algebra than $\Sigma\_n$ ….
 Is this correct?



Disjoint union of measures
1. As written your definition doesn't quite make sense because $\bigsqcup\_{n \in \mathbb{N}} \Sigma\_n$ doesn't contain nearly enough sets (e.g., $X \mathrel{:=} \bigsqcup\_n X\_n$ itself is not in it). It seems better to me just to set
$\Sigma = \left\{ \bigsqcup\_n A\_n : A\_n \in \Sigma\_n \right\}.$ Equivalently, this is the set of all $A \subset X$ for which $A \cap X\_n \in \Sigma\_n$ for every $n$. You can easily check that this is a $\sigma$-algebra and that it equals the Borel $\sigma$-algebra of the disjoint union topology on $X$.
2. This definition of $\mu$ works fine, and in fact your restriction statement is true on the nose: for every $A \subset X$ with $A \subset X\_n$, we have $A \in \Sigma$ iff $A \in \Sigma\_n$, and in this case we have $\mu(A) = \mu\_n(A)$. The $\sigma$-algebra is not smaller.


Note that everything here would also work just fine for uncountable disjoint unions, though of course the resulting measure $\mu$ need not be $\sigma$-finite.


