$(X,A,\mu)$ complete measure space. $f$ $\mu$-measurable on $X$. Then every function $\bar{f} = f$ $\mu$-almost everywhere $\mu$-measurable.
This is written in my script. In my understanding: $D, \bar{D} \in A$ and $f: D\to \mathbb{\bar{R}}$ and $\bar{f}: \bar{D} \to \mathbb{\bar{R}}$. $f$ is $\mu$-measurable.
If $f=\bar{f}$ then $D=\bar{D}$ in my understanding. This is trivial. So does that mean: $f=\bar{f}$ on $D\cap \bar{D}$ $\mu$-measurable? And therefore it is $\bar{f}$ almost everywhere $\mu$-measurable?
If needed I can write the definition for $\mu$-measurable down. Definition of $\mu$-measurable:
$(X,A,\mu)$ measure space, $D\in A$ and $f$ defined on D as $f:D \to$ $\mathbb{\bar{R}}$ is $\mu$-measurable (on X) if $\mu (X\setminus D) =$ $0$ and $f$ $A|_{D}$-measurable. (With $A|_{D} := \{B \cap D | B \in$ $A\}$)
Consider $f: X \rightarrow \overline{\mathbb{R}}$ a $\mu-$measurable function on the measure space $(X, \mathcal{A}, \mu).$ Also consider $g: X \rightarrow \overline{\mathbb{R}}$ such that $g = f$ on $E \in \mathcal{A}$ with $\mu(E^C) = 0$ (this is what your statement actually says). We need to show that $g$ is also $\mu-$measurable. Indeed, we may write $g = f \chi_E + h \chi_{E^C},$ where $\chi_E$ denotes the indicator function of the set $E.$ Consider $U \subseteq \overline{\mathbb{R}}$ open. Then $g^{-1}(U) = (f^{-1}(U) \cap E) \cup (h^{-1}(U) \cap E^C).$ Since $f$ is measurable and $E \in \mathcal{A},$ we deduce that $f^{-1}(U) \cap E \in \mathcal{A}.$ Moreover, since the measure space we are working with is complete and $h^{-1}(U) \cap E^C \subseteq E^C,$ $\mu(E^C) = 0,$ we also get $h^{-1}(U) \cap E^C \in \mathcal{A},$ so that finally $g^{-1}(U) \in \mathcal{A}.$ Thus, we have showed that $g$ is also measurable.