Let X be a set and F a σ-algebra. Does there exist a topological space U and a map f : X → U such that f is F , B -measurable and σ ( f ) = F ? Here B is the Borel σ-algebra of U.Of course, this is trivial if every σ-algebra on a set is the Borel σ-algebra with respect to some topology on the set. But this needn't be true. This is a weaker problem.

Question

Let   X be a set and       F   a   σ-algebra. Does there exist a topological space   U and a map   f  :  X  →  U such that   f is       F    ,      B  -measurable and   σ  (  f  )  =      F  ? Here       B   is the Borel   σ-algebra of   U.Of course, this is trivial if every   σ-algebra on a set is the Borel   σ-algebra with respect to some topology on the set. But this needn&#039;t be true. This is a weaker problem.

Belen Bentley · Accepted Answer

MathJax(?): Can&#039;t find handler for document MathJax(?): Can&#039;t find handler for document If   &amp;#x03C4; is a topology on a set   U and   f  :  X  &amp;#x2192;  U is a function, then

Jackson Duncan · Accepted Answer

This works, thanks. It simply never occurred to me that we could generate a topology by taking preimages.

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