I have the following problem:
Let $\{ f_{n}\}$ be a sequence of measurable functions on $E$ that converges to the real-valued $f$ pointwise on $E$. Show that $E = E_{0} \cup \left( \cup_{k=1}^{\infty}E_{k} \right)$, where all sets $E_{0}$, $E_{1}, \cdots$ are measurable, and $\{f_{n} \}$ converges uniformly to $f$ on each $E_{k}$ if $k>0$, and $m(E_{0}) = 0$.
Now, the way to approach this problem is by a repeated application of Egoroff's Theorem.
So, if $\{f_{n}\}$ is a sequence of measurable functions on $E$ that converges pointwise on $E$ to the real-valued function $f$, then by Egoroff's Theorem, $\forall \epsilon > 0$, $\exists$ a closed set $E_{1}$ contained in $E$ on which $f_{n} \to f$ uniformly, and such that $m(E\backslash E_{1})<\frac{\epsilon}{2}$. Further, since all closed sets are measurable, $E_{1}$ is measurable.
Since uniform convergence implies pointwise convergence, $\{f_{n}\}$ also converges pointwise to $f$ on $E_{1}$. Thus, $\exists$ a closed set $E_{2}\subset E_{1}$ such that $f_{n} \to f$ uniformly on $E_{2}$ and $m(E_{1} \backslash E_{2}) < \frac{\epsilon}{4}$.
If we keep proceeding in this manner, eventually, we get that $E = (\cup_{k=1}^{\infty} E_{k}) \cup E_{0}$, and I'm trying to figure out what $E_{0}$ is and how to show that its measure is zero.
My professor gave me the hint that the complement of their union is going to have measure 0 and I will need to explain why. However, I'm not sure what he means by the complement of their union. The complement of which union?? The complement of the union of the $E_{k-1}\backslash E_{k}$'s? The complement of $\cup_{k=1}^{\infty} E_{k}$? That part is unclear to me. And in any case, I still am unsure how to show the measure of this part equals 0.
Please help!!! I need this explained to me thoroughly and detailed as though you were explaining it to a completely clueless person. A complete solution would be best, since I already have most of it except for this part.