Disjoint union of measures

Question

This is a sort of follow-up question to this old post 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?

Answer 1

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.

Answer 2

The first issue: In order to talk about a coproduct / disjoint union we need a category for our objects to live in.

A natural candidate seems to be the following category (result of a grothendieck construction):

• Objects are triples $X=(S_X, \Sigma_X, \mu_X)$ where $S_X$ is a set, $X_\Sigma$ is a $\sigma$-algebra and $\mu_X$ is a measure.
• A morphism $f:X\to Y$ are maps $f:S_X\to S_Y$ that are measurable and satisfy $f_*\mu_X\leq \mu_Y$ (edit) up to extensional equality on measures (i.e. $f_*\mu=g_*\mu$ for all $\Sigma_X$-measures).

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Now let's see if there is a coproduct for a family $X_i$ of such objects:

• Underlying set: We take the direct sum $\amalg_i S_{X_i}$.
• $\sigma$-algebra: We take the largest $\sigma$-algebra s.t. all the inclusions $S_{X_i}\to\amalg_i S_{X_i}$ are measurable (This should exist, if I am not mistaken, since the set of such $\sigma$-algebras is stable under ascending chains).
• The measure: We take the smallest measure s.t. all the inclusions are morphisms in our category. This measure exists; Explicitly: $$\mu_\amalg(E):=\inf_i \mu_{X_i}(E).$$ To show that this is a measure, I think the following should be true: We have $\sup_j\inf_i a_{i, j}=\inf_i(\sup_j a_{i, j})$ given the $a_{i, *}$ are countable, ascending sequences.

By construction, all the inclusions $e_i:S_{X_i}\to\amalg_i S_{X_i}$ are morphisms in our category.

Now let $f_i:X_i\to Y$ be morphisms in our category. Then the induced map $f:S_{\amalg_i X_i}\to S_Y$ is a morphism in our category:

• It is measurable since we have $e_i^*f^*\Sigma_Y=f_i^*\Sigma_Y\subseteq\Sigma_{X_i}$ and so $f^*\Sigma_Y\subseteq \Sigma_\amalg$ by construction.
• It is also compatible with the measures since the pushforward of measures respects the order.

The uniqueness of this morphism is inherited from the category $\mathcal{Set}$.

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Now, I am not sure if this is fully the answer you want since it is not specifically about Hausdorff spaces with finite borel measures. But the question could then be tackled further by comparing the convergences of the hausdorff topologies, borel $\sigma$-algebras and respective measures.

A starting point could be to investigate if the subcategory $\mathcal{Haus}$ we get when restricting to Hausdorff spaces is reflective or coreflective. This could be used to show a lax or oplax distributivity in the following sense: The identity map is most likely a morphism $\amalg_{\mathcal{Haus}}X_i\to \amalg X_i$ (or the other way around?) given the coproduct in $\mathcal{Haus}$ exists.