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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 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 $Y$. – t.b. May 9 '11 at 10:11
Which math textbook? Be precise in the reference. – Glen Wheeler May 9 '11 at 10:11
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. – jpv May 9 '11 at 10:21
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. – t.b. May 9 '11 at 10:26
No Problem. Thanks for the Dudley reference. – jpv May 9 '11 at 10:32

Edit: I have just figured a much easier way. So, I edited the answer.

Let $\mathcal{V} = \{V_n : n = 1, 2, \dotsc\}$ be a countable base for the topology of $Y$.

For each $V_n$, choose a negligible $E_n \subset X$ such that $f^{-1}(V_n) \setminus E_n \in \mathcal{F}$. It may happen that $\bigcup E_n \not \in \mathcal{F}$. But since it is a negligible set, there is a negligible $Z \in \mathcal{F}$ such that $\bigcup E_n \subset Z$.

Fix some $y \in Y$, and then define $$ g(x) = \left\{ \begin{array}{} f(x), & x \not \in Z \\ y, & x \in Z \end{array} \right. $$

Notice that for any $V_n \in \mathcal{V}$, if $y \not \in V_n$, $$ \begin{align*} g^{-1}(V_n) &= f^{-1}(V_n) \setminus Z \\&= (f^{-1}(V_n) \setminus E_n) \setminus Z \in \mathcal{F}. \end{align*} $$ And if $y \in V_n$, $$ \begin{align*} g^{-1}(V_n) &= f^{-1}(V_n) \cup Z \\&= (f^{-1}(V_n) \setminus E_n) \cup Z \in \mathcal{F}. \end{align*} $$ That is, $g^{-1}(\mathcal{V}) \subset \mathcal{F}$. All open sets of $Y$ are (countable) union of elements in $\mathcal{V}$. Therefore, $\mathcal{V}$ generates the $\sigma$-algebra of Borel sets $\mathcal{B}$. And so, $g$ is $\mathcal{F}$-measurable. In fact, $$ g^{-1}(\mathcal{B}) = g^{-1}(\sigma(\mathcal{V})) = \sigma \left(g^{-1}(\mathcal{V})\right) \subset \mathcal{F}. $$

Since it is evident that $g$ and $f$ are equal almost everywhere, the proof is complete.

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Thanks! $\hphantom{zzzzzz}$ – Byron Schmuland Oct 14 '11 at 19:12
@Byron: two trivial remarks: 1. metrizability wasn't really used, just in guise of second countability. 2. If you need extra characters, ${}{}{}{}{}{}{}{}$ (as many {} as you need) gives somewhat nicer results. – t.b. Oct 14 '11 at 22:03
@t.b. Thanks for the tips about spaces. I did notice that the topology of $Y$ is not used at all. I suppose this proof goes through for any measurable space $(Y,{\cal Y})$, provided $\cal Y$ is countably generated. – Byron Schmuland Oct 14 '11 at 22:23
@Byron: Indeed. – t.b. Oct 15 '11 at 5:23

Let $(V_n) \in Y$ be a countable base for the separable metric space $Y$. Then define $V_n^\prime = V_n \cap V_{n-1}^c \cap V_{n-2}^c... \cap V_1^c$ and $V_1^\prime = V_1$. Clearly, $V_n^\prime$ are all measurable. Now $f^{-1}(V_n^\prime) = E_n \cup N_n = E_n + (E_n^c \cap N_n) = E_n + N_n^\prime$, where $E_n$ is $\mathcal{F}$-measurable and $N_n, N_n^\prime$ are $\mu$-negligible. As all $V_n^\prime$ are all pairwise disjoint, $f^{-1}(V_n^\prime) = E_n + N_n^\prime$ are all disjoint. Define a new function $g: X \rightarrow Y$ such that $g(x) = f(x)$ on all $x \in E_n$ and on $x \in (\cup_{i=0}^\infty E_i)^c$ to be any value. Then $g$ is $\mathcal{F}$-measurable and is a.e. equal to $f$.

I hope this proof is correct.

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This proof cannot be correct: $V_1$ might even be all of $Y$. But even if that's not the case, you only know that the preimage of $E_1$ is in $\mathcal{F}$, but this doesn't say anything about preimages of subsets of $E_1$. – Florian Jul 14 '11 at 9:27

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