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correction as pointed out in comments
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This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$$S \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $S \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

added accidentally omitted word
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This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have no references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?

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This does exist, and has a nice explicit description. Treating the sets $A_i$, for convenience, as disjoint subsets of $A$, take a subset $S \subseteq A$ to be measurable exactly if $S \cap A_i$ is a measurable subset of $A_i$, for each $i$. The proof that this is a sigma-algebra making each $\psi_i$ measurable, and is the finest such, is reasonably straightforward.

From a categorical point of view, one can find this description by saying: if such a σ-algebra exists, one would hope that it should make $A$ a coproduct of the $A_i$'s in the category of measurable spaces.

But measurable subsets $S \subseteq A$ must correspond to measurable functions $f \colon A \to 2$ (this holds for any measurable space); hence, by the universal property of $A$, to families of functions $f_i \colon A_i \to 2$; hence to families of measurable sets $S_i \subseteq A_i$; thinking about naturality shows that this correspondence has to be via $A \mapsto (A \cap S_i)_{i \in I}$, and so leads to the description above. (And one can check then that this does indeed give a coproduct.)

One can see this as talking about a duality between the category of measurable spaces and a suitable category of lattices: the coproduct as spaces corresponds to the product of the lattices of measurable subsets.

I have references, I’m afraid, since I don’t know of any categorically-minded treatments of measure theory. But any such book would surely include this construction; I’m hopeful that there’s one out there that I don’t know of?