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Let $f : (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R)$ be a measurable map, then it is well-known that $f$ could be approximated by a sequence $(f_n)$ of simple measurable functions, such that (i) $0 \le f_n(\omega) \uparrow f(\omega)$ if $f(\omega) \ge 0$, and (ii) $0 \ge f_n(\omega) \ge f(\omega)$ if $f(\omega) \le 0$.

Now does there exists a similar result for arbitrary measurable maps $f : (X, \mathcal F) \to (Y,\mathcal G)$ (i.e. $f$ need not be a real or complex function)? For this the notion of simple function must also be generalised, but I think this is quite easy, as simple just means that $f(X)$ is finite and $\{ f = c \} \in \mathcal F$ for all $c \in f(X)$.

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  • $\begingroup$ What is $Y$? You need some sort of convergence in $Y$ for the question even to make sense. Do you want an "order" convergence as you wrote in the real-valued case? $\endgroup$
    – Gerald Edgar
    Commented Jun 24, 2015 at 13:29
  • $\begingroup$ Yes, thank you, you are right, I forgot that. No, I do not require the domain to be ordered, the most general situation I can think of is to equip $Y$ with a topology such that the sequence $(f_n)$ of simple functions should fulfill $f_n(x) \to f(x)$ for each $x$ (meaning that for each open set $U$ around $f(x)$ there exists an index $N$ such that for $n > N$ we have $f_n(x) \in U$). $\endgroup$
    – StefanH
    Commented Jun 24, 2015 at 13:51
  • $\begingroup$ For $Y$ a topological space, the answer is "no" in general. This is the wrong forum for the question, though. $\endgroup$
    – Gerald Edgar
    Commented Jun 24, 2015 at 14:00
  • $\begingroup$ Okay, thank you. I thought there would be some research around this question and some non-trivial generalisations, so I posted it here. Sorry if it is a trivial question, tell me if you want me to delete it... $\endgroup$
    – StefanH
    Commented Jun 24, 2015 at 14:05
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    $\begingroup$ If $Y$ is a Banach space, then a $f$ that is approximated by these kind of simple functions is sometimes called Bochner-measurable or strongly measurable. A Banach valued $f:X\to Y$ is Bochner measurable iff it measurable and $f(X)$ is separable. $\endgroup$
    – Johannes Hahn
    Commented Jun 24, 2015 at 16:47

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There are results of this sort for functions with values in a Banach space---see, for example, p. 42 of "Vector measures" by Diestel and Uhl.

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