Definition of measurable function: If $X$ is measurable space, $Y$ is topological space, then $f:X\to Y$ is measurable provided that $f^{-1}(V)$ is measurable set in $X$ for every open set $V$ in $Y$.Definition of Borel measurable function: If $f:X\to Y$ is continuous mapping of $X$, where $Y$ is any topological space, $ (X,\mathfrak B)$ is measurable space and $f^{-1}(V)\in\mathfrak B$ for every open set $V$ in $Y$, then $f$ is Borel measurable function.
Both functions are mapping from measurable space to topological space what's the difference between the two definition?
$\endgroup$2 Answers
$\begingroup$A Borel measurable function is a measurable function but with the specification that the measurable space $X$ is a Borel measurable space (where $\mathfrak B$ is generated as the smallest sigma algebra that contains all open sets). The condition "$f$ is continuous" is equivalent to "$f^{-1}(V)$ is open (and thus Borel measurable) for every open set $V\subseteq Y$".
But not every measurable function is Borel measurable, for example no function that takes arguments from $(\mathbb R,\{\emptyset,\mathbb R\})$ is Borel measurable, because $\{\emptyset,\mathbb R\}$ is not a Borel sigma algebra.
$\endgroup$ $\begingroup$The difference is in the $\sigma$-algebra that is part of the definition of measurable space. In your first definition you say that $X$ is measurable but you don't specify the $\sigma$-algebra; in the second you specify that the $\sigma$-algebra is the collection of all Borel sets.
The $\sigma$-algebra is a collection of sets closed under countably-infinite unions (that's rather rough and ready, but if you need to know more you're better off looking up a definition and some examples).
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