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Jan 11, 2016 at 19:57 vote accept StefanH
Jun 24, 2015 at 16:47 comment added Johannes Hahn 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.
Jun 24, 2015 at 16:43 history closed Gerald Edgar
Stefan Kohl
András Bátkai
Joonas Ilmavirta
Johannes Hahn
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Jun 24, 2015 at 14:59 answer added corserine timeline score: 1
Jun 24, 2015 at 14:05 comment added StefanH 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...
Jun 24, 2015 at 14:00 comment added Gerald Edgar For $Y$ a topological space, the answer is "no" in general. This is the wrong forum for the question, though.
Jun 24, 2015 at 13:51 comment added StefanH 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 $f_n(x) \in U$).
Jun 24, 2015 at 13:47 review Close votes
Jun 24, 2015 at 16:43
Jun 24, 2015 at 13:29 comment added Gerald Edgar 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?
Jun 24, 2015 at 12:40 history edited StefanH CC BY-SA 3.0
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Jun 24, 2015 at 11:17 history asked StefanH CC BY-SA 3.0