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Sigma-algebra generated by a function

  1. May 12, 2010 #1
    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 sigma-algebra 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})

    so then is the sigma-algebra [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 sigma-algebra generated by a function.
    the i do it is find the pre-image 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. jcsd
  3. May 13, 2010 #2
    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(1-x) \, x \in [0,1][/tex] i then realised to find to generated sigma-field [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 sigma-algebra contains 'coarser' sets that those contained in [tex] \mathcal{B}([0,1]) [/tex]
    Last edited: May 13, 2010
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