
#1
May1210, 04:37 AM

P: 27

suppose we have a [tex] X = [0,1] [/tex] and a function [tex] f\colon X \to \Re [/tex] where
[tex] f(x) = 1  2x 1 [/tex]. i'm bit confused on finding the sigmaalgebra generated by this function. This is what i did [tex] f(x)= \begin{cases} 2 2x & x \in [\frac{1}{2},1] , \\ 2x& x \in [0, \frac{1}{2}) \end{cases} [/tex] so then is the sigmaalgebra [tex] \sigma(f(x)) = \mathcal{B}([\frac{1}{2},1] \bigcup \mathcal{B}([0, \frac{1}{2}) = \mathcal{B}([0,1]) [/tex] ? some thing about this doesnt feel quite right to me, could someone show me where i have made a mistake. Also what is a systematic way or method of finding the sigmaalgebra generated by a function. the i do it is find the preimage of the function of any open set in [tex] \Re [/tex] it far to easy for me to make mistakes when doing it this way. are alternative methods ? any comments, help much appreciated 



#2
May1310, 11:30 AM

P: 27

i think i may have figured it out. i graphed the function [tex] f(x) [/tex] and realised it was symmetrical, [tex] f(x) = f(1x) \, x \in [0,1][/tex] i then realised to find to generated sigmafield [tex] \sigma(f(x)) = \{ f^{1}(B) \colon B \in \mathcal{B} \} [/tex] the inverse image for any borel set is the union of two intervals in [0,1] since the function symetrical.
[tex] \sigma(f(x) = \{[\frac{1}{2},1] \bigcap \{1\frac{B}{2} \colon B \in \mathcal{B} \} \bigcup [0, \frac{1}{2}] \bigcap \{\frac{B}{2} \colon B \in \mathcal{B} \} [/tex] where [tex] 1\frac{B}{2} = \{ 1\frac{x}{2} \colon x \in B\} [/tex] This seems right to me, since the sigmaalgebra contains 'coarser' sets that those contained in [tex] \mathcal{B}([0,1]) [/tex] 


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