"Nice" sigma-algebra on set of measurable functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:22:59Z http://mathoverflow.net/feeds/question/104305 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104305/nice-sigma-algebra-on-set-of-measurable-functions "Nice" sigma-algebra on set of measurable functions Avi Steiner 2012-08-08T21:12:31Z 2012-08-13T23:55:23Z <p>In topology, given topological spaces $X$ and $Y$, the compact-open topology is considered, under the relatively mild requirement that $X$ be locally compact Hausdorff, to be the most "natural" topology on the set $\mathcal{C}(X,Y)$ of continuous functions $X\to Y$. (I'm not going to even attempt to define "natural" here---take it to mean whatever seems most appropriate).</p> <p>Is there something analogous, in the sense of being somehow "natural", for the set of measurable functions between two measure spaces?</p> http://mathoverflow.net/questions/104305/nice-sigma-algebra-on-set-of-measurable-functions/104315#104315 Answer by Gerald Edgar for "Nice" sigma-algebra on set of measurable functions Gerald Edgar 2012-08-09T00:07:05Z 2012-08-09T00:07:05Z <p>There is the <em>Effros Borel structure</em>. But that is a $\sigma$-algebra for the collection $F(S)$ of closed sets in a Polish space $S$. But it is again a standard Borel structure, so maybe it meets your criterion of "nice"... </p> <p>Google "effros borel structure" for more information.</p> http://mathoverflow.net/questions/104305/nice-sigma-algebra-on-set-of-measurable-functions/104656#104656 Answer by Michael Greinecker for "Nice" sigma-algebra on set of measurable functions Michael Greinecker 2012-08-13T23:47:36Z 2012-08-13T23:55:23Z <p>There is an impossibility theorem: If you let $\mathcal{L}$ be the the space of Borel-measurable functions $f:[0,1]\to[0,1]$, and $e:\mathcal{L}\times [0,1]\to[0,1]$ the evaluation given by $e(f,x)\mapsto f(x)$, then there is no $\sigma$-algebra on $\mathcal{L}$ such that the evaluation is jointly measurable. The result is a consequence of the rather complicated classification result in R. Aumann, <a href="http://11.%20%22Borel%20Structures%20for%20Function%20Spaces,%22%20Illinois%20Journal%20of%20Mathematics%205%20%281961%29,%20pp.%20614-630.%20%20/" rel="nofollow">Borel Structures for Function Spaces</a>, Illinois Journal of Mathematics 5 (1961), pp. 614-630. Easier proofs of the main results can be found in the paper "Borel Structures for Function Spaces" (yes, same title) by B.V. Rao, Colloquium Mathematicum, 1971.</p> <p>A $\sigma$-algebra on measurable functions I have actually seen used is the following: If $(S,\mathcal{S})$ and $(T,\mathcal{T})$ are measurable spaces, we endow the family of measurable functions between them with the $\sigma$-algebra generated by sets of the form $\{f:f(s)\in B\}$ with $s\in S$ and $B\in\mathcal{T}$. The author used this $\sigma$-algebra to show that to each Markov kernel from $S$ to $T$, there corresponds a certain probability measure on this $\sigma$-algebra. The paper is H. v. Weizsäcker <a href="http://www.digizeitschriften.de/dms/img/?PPN=PPN365956996_0011&amp;DMDID=dmdlog9" rel="nofollow">Zur Gleichwertigkeit zweier Arten der Randomisierung</a>, Manuscripta Mathematica 11 (1974).</p>