If $f$ is a continuous function on $X$ and E a Lebesgue measurable set, can we conclude that $f^{-1}(E)$ is measurable?
Does a continuous function preserve measurability?
3
$\begingroup$
measure-theory
functions
-
1I presume $f : X \to Y$ is a continuous function between topological spaces equipped with their Borel $\sigma$-algebras? – 2012-09-28
-
0I know that "The inverse image of any Borel set is measurable". But I want to see if the general case (any measurable) holds. – 2012-09-28
-
4No. See here for the standard counterexample: http://math.stackexchange.com/questions/104519/a-question-concerning-the-cantor-ternary-function – 2012-09-28
-
4What do you mean by "the general case" here? What is the codomain of $f$ and what $\sigma$-algebra are you using on it? – 2012-09-28
-
0Actually the example in the link provided by Chris is used to proof that "measurable" is really more general than "Borel": there is a measurable set $B$ such that its inverse image by a continuous map is not measurable so is not Borel. The same function in that post serves to show that measurability is not a topological property. – 2012-09-28