$\mu$ measurable functions and separable metric spaces 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.

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.
