Functions measurable with respect to a particular measure In a text I was reading, I saw the author refer to a function $f$ being $\mu$-measurable.  Does this mean something different than saying that $f$ is measurable with respect to the $\sigma$-algebra that $\mu$ acts on, or does it imply something more about $f$?
 A: A general notion of $\mu$-measurability is as follow: suppose that $f:E\to F$ where $(E,\mathcal{E},\mu)$ is a measure space and $F$ is a Banach space equipped with it Borel $\sigma $-algebra $\mathcal{F}$. Then we says that $f$ is $\mu$-measurable if and only if there is a sequence of simple functions $\{f_n\}_{n\in\mathbb{N}}$ such that $f_n\to f$ $\mu$-almost everywhere, where a function $g$ is simple if and only if

*

*$g(E)$ is a finite set

*$g^{-1}(y)\in \mathcal{E}$ for each $y\in F$

*and $\mu(g^{-1}(F\setminus \{0\})<\infty $
It can be shown that the above is equivalent to say that $f$ is Borel measurable (that is, $f^{-1}(A)\in \mathcal{E}$ for every $A\in \mathcal{F}$) and that there is some $\mu$-null set $N$ such that $f(N^\complement )$ is separable.
Then, in particular, $f$ is $\mu$-measurable when it is complex valued, or take values in any Euclidean space.
The above definition of $\mu$-measurability can be extended, in an obvious way, to the case where $f$ take values in the extended real line or in the Riemann sphere (or more generally to the one point compactification of any Euclidean space).
