Is there a "disjoint union" sigma algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:49:28Z http://mathoverflow.net/feeds/question/54923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54923/is-there-a-disjoint-union-sigma-algebra Is there a "disjoint union" sigma algebra? Neil Toronto 2011-02-09T20:31:59Z 2011-04-05T17:02:37Z <p>I'm looking for a measure-theoretic analogue to the disjoint union topology, or for work on the $\sigma$-algebra generated by canonical injections. More formally:</p> <blockquote>For an indexed family of sets $\{A_i\}_{i \in I}$, define $\psi_i : A_i \to A$, $a \mapsto (a,i)$ (the canonical injections), where $A = \bigcup_{i \in I} (A_i \times \{i\})$. Then the disjoint union $\sigma$-algebra $\mathcal{A}$ is the finest $\sigma$-algebra on $A$ such that for every $i \in I$, $\psi_i$ is $\mathcal{A}_i$-$\mathcal{A}$-measurable.</blockquote> <p>There are so many analogies between measure theory and topology that I've been surprised at how difficult it's been for me to find anything on this.</p> <p>I'd appreciate references to any related ideas as well.</p> http://mathoverflow.net/questions/54923/is-there-a-disjoint-union-sigma-algebra/54924#54924 Answer by Peter LeFanu Lumsdaine for Is there a "disjoint union" sigma algebra? Peter LeFanu Lumsdaine 2011-02-09T20:48:12Z 2011-04-05T17:02:37Z <p>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.</p> <p>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.</p> <p>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.)</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/54923/is-there-a-disjoint-union-sigma-algebra/54953#54953 Answer by Chris Heunen for Is there a "disjoint union" sigma algebra? Chris Heunen 2011-02-09T23:12:09Z 2011-02-09T23:12:09Z <p>David H. Fremlin's "Measure Theory", vol 2, 214K, gives this construction explicitly. He also proves some elementary properties, but unfortunately stops short of universal properties such as in Peter's insightful answer. </p>