question on sigma-fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:20:02Z http://mathoverflow.net/feeds/question/3912 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3912/question-on-sigma-fields question on sigma-fields alex 2009-11-03T07:13:14Z 2010-07-24T07:37:22Z <p>Let X,Y to be mappings from the sample space &Omega; to R and suppose Y is measurable with respect to &sigma;(X), the smallest &sigma;-field that makes X measurable. </p> <p>Does it follow that there exists some Borel-measurable function f: R &rarr; R such that Y=f(X)? </p> http://mathoverflow.net/questions/3912/question-on-sigma-fields/3925#3925 Answer by Piotr Miłoś for question on sigma-fields Piotr Miłoś 2009-11-03T12:07:14Z 2009-11-03T12:07:14Z <p>The answer is yes. The proof is quite standard. <strong>1.</strong> If Y=1_A, where A is in \simga(X), then by the definition of \sigma(X) there exists a Borel set B such that A = X^{-1}(B) and therefore</p> <p>Y(\omega) = 1_A(\omega) = 1_B(X(\omega)) = f(X(\omega)),</p> <p>where we put f:= 1_B (of course f is now a Borel function).</p> <p><strong>2.</strong> If now Y is a simple r.v. i.e. it can be written in form Y = \sum_{i=1}^n c_i 1_{A_i}, where A_i are sets in \sigma(X), then using the previous point we can find Borel functions f_i such that 1_{A_i} = f_i(X) and obviously in this case f = \sum_{i=1}^n c_i f_i.</p> <p><strong>3.</strong> Finally, any r.v. Y measurable w.r. to \sigma(X) can be approximated by a sequence of simple r.v. Y_n measurable w.r. to \sigma(X) i.e. Y_n -> Y almost surely. By the previous point there exist f_n such that Y_n = f_n(X). Now we can define f(x) = \lim_n f_n(x) if the limit exists and put f(x)=0 otherwise. It is easy to check that f is a Borel function (basically it is a limit of Borel functions), and that Y = f(X).</p> http://mathoverflow.net/questions/3912/question-on-sigma-fields/3931#3931 Answer by Gerald Edgar for question on sigma-fields Gerald Edgar 2009-11-03T12:38:01Z 2009-11-03T12:38:01Z <p>If R is the reals, yes (as already explained). But if R is just some other Borel space, then perhaps not.</p> http://mathoverflow.net/questions/3912/question-on-sigma-fields/6239#6239 Answer by loloa for question on sigma-fields loloa 2009-11-20T09:31:20Z 2009-11-20T09:31:20Z <p>write the following in the form of sigm 5+11+17+23+...= and the second is 3+9+27+81+...= please help me</p> http://mathoverflow.net/questions/3912/question-on-sigma-fields/6330#6330 Answer by Jeff Hussmann for question on sigma-fields Jeff Hussmann 2009-11-20T22:56:39Z 2010-07-24T07:37:22Z <p>It is trivially true but maybe worth noting that the converse is also true - if there exists such an f, then Y is &sigma;(X)-measurable.</p> <p>This and the question asked are theorem 20.1(ii) in Billingsley's <em>Probability and Measure</em>, 3rd edition.</p>