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I was reading a math textbook and the author gives the following without proof. I have no clue on how to proceed.

Let $(X, \mathcal{F}, \mu)$ be a measure space and $(Y,d)$ be a separable metric space ($d$ is the metric). If $f:(X,\mathcal{F}) \rightarrow (Y, d)$ is a $\mu$-measurable function prove that there exists an $\mathcal{F}$ measurable function which coincides with $f$ everywhere except on a $\mu$-negligible set.

Any help is greatly appreciated.

EDIT: The textbook is "Functions of Bounded Variation and Free Discontinuity Problems" by Luigi Ambrosio et. al.

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    See [this MO-question](http://mathoverflow.net/questions/54003/does-there-exist-a-measurable-function-which-is-not-a-e-strongly-measurable/54531) for references for the result you're looking for (given in the question there) and in his answer George Lowther gives a great explanation for a (very deep) extension with further restrictions on $(X,\mathcal{F},\mu)$ but for any metrizable space $Y$.2011-05-09
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    Which math textbook? Be precise in the reference.2011-05-09
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    Thanks for the super-fast response. I did not realize this was a difficult result as the authors gave this as an exercise in the first chapter.2011-05-09
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    I'm sorry, in my first comment I overlooked the separability assumption on $Y$. That's why I first said it is very deep. Indeed, it is not that complicated with assuming separability (start with choosing a countable base for the topology of $Y$ and use the fact that every $\mu$-measurable set is $\mathcal{F}$-measurable up to a null-set). Dudley's book should contain a proof in chapter III.2011-05-09
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    No Problem. Thanks for the Dudley reference.2011-05-09
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    It's the one referred to in the MO-question: Richard M. Dudley, *Real Analysis and Probability*, Wadsworth 1989, reprinted in Cambridge University Press.2011-05-09
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    Thanks. I have just found it.2011-05-09
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    @jpv: By the way: Add an `@username` tag when you want someone to see your comments. See [here](http://meta.math.stackexchange.com/questions/2063/ping-only-works-for-the-first) for some explanations.2011-05-09
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    @TheoBuehler: Thanks, hope this is working now.2011-05-09
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    Yes, it worked. The first three letters of the user's name suffice in fact.2011-05-09
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    What if two usernames begin with the same 3 letters?2011-05-09
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    @GEdgar: According to [these seemingly comprehensive rules](http://meta.stackexchange.com/questions/43019/how-do-comment-replies-work) only the last one who commented will be notified.2011-05-09
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    @The: I could not yet prove this; Dudley's book does not contain a proof. I had tried to prove it but could not understand how to define the new function $g$ which is a.e. equal to $f$ and is $\mathcal{F}$ measurable. Any directions for the proof are appreciated.2011-05-11
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    Okay, I'll add an answer in a while. I'm sorry, I was quite sure it was in Dudley.2011-05-11
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    Thanks a lot, I have been stuck here for two days and I would love to start reading the rest of the book. I found a proof that a countable base exists for a separable metric space in "Introductory Real Analysis" by Kolmogorov and Fomin but have been struggling after that.2011-05-11
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    @Byron: I think I can prove the statement with some further assumptions $(X,\mathcal{F},\mu)$ ($\sigma$-finite is enough). Would you be interested in seeing that argument, or do you want complete generality?2011-10-13
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    @t.b. As a probabilist, I am happy to assume that all my measure spaces are probability spaces!2011-10-13
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    @Byron: My argument would have been essentially the same as the one André gives below. For some reason I thought I needed more control of sets of infinite measure, but that's probably because it was late. So I won't write it up now, since I have little to add.2011-10-14

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