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 $f : (X, \mathcal F) \to (Y,\mathcal G)$? 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 $c \in f(X)$.

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 $c \in f(X)$.