Sigma-algebra generated by a function

[cappadonza]

suppose we have a X = [0,1] and a function f\colon X \to \Re where
f(x) = 1 - |2x -1|.
i'm bit confused on finding the sigma-algebra generated by this function. This is what i did

f(x)= \begin{cases} <br /> 2 -2x &amp; x \in [\frac{1}{2},1] , \\<br /> 2x&amp; x \in [0, \frac{1}{2})<br /> \end{cases}<br />

so then is the sigma-algebra \sigma(f(x)) = \mathcal{B}([\frac{1}{2},1] \bigcup \mathcal{B}([0, \frac{1}{2}) = \mathcal{B}([0,1]) ?

some thing about this doesn't 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 \Re it far to easy for me to make mistakes when doing it this way. are alternative methods ?

any comments, help much appreciated

 

[cappadonza]

i think i may have figured it out. i graphed the function f(x) and realized it was symmetrical, f(x) = f(1-x) \, x \in [0,1] i then realized to find to generated sigma-field \sigma(f(x)) = \{ f^{-1}(B) \colon B \in \mathcal{B} \} the inverse image for any borel set is the union of two intervals in [0,1] since the function symetrical.
\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} \}
where 1-\frac{B}{2} = \{ 1-\frac{x}{2} \colon x \in B\}

This seems right to me, since the sigma-algebra contains 'coarser' sets that those contained in \mathcal{B}([0,1])