Can I understand Egorov's theorem in this way? Egorov's Theorem:

Suppose $\{f_k\}_{k=1}^{\infty}$ is a sequence of measurable functions defined on a measurable set $E$ with $m(E)<\infty$, and assume that $f_k \rightarrow f$ a.e. on $E$. Given $\epsilon > 0$, we can find a closed set $A_\epsilon \subset E$ such that $m(E-A_\epsilon)\leq \epsilon$ and $f_k \rightarrow f$ uniformly on $A_\epsilon$

Because $\epsilon$ is arbitrary, can I say that except on a set of measure zero, $f_k \rightarrow f$ uniformly on $E\;$? Is this understanding right? Thanks very much.
 A: No, this is not understood correctly, because when you find an such $A_\varepsilon$ to apply Ergorov's theorem you also find an $n(\varepsilon)$, so if you try to exaust the set $E$ by this $A_\varepsilon$ your $n(\varepsilon)$ blows up, and then you don't have uniformly on $E$ except on a set of measure zero, because you don't have an upper limit for $n(\varepsilon)$  , instead of upper limited your $n(\varepsilon) \to \infty$ as $\varepsilon \to 0$, and you can not take a $N_0$ that satisfies the hypotheses of the desired uniformly convergency property. I hope this interpretation helps you!
A: Here is an argument (built on Martin's comment) why we are not guaranteed almost everywhere uniform convergence by Egorov's Theorem:
Let $\forall n: f_n:[0.1]\to[0,1], f_n(x):=x^n, f:[0,1]\to[0,1],f(x):=\chi_{\{1\}}(x)$. Then $\forall x\in[0,1]:f_n(x)\to f(x)$.
Claim: Let $F\subseteq[0,1]$. If $1\in F'$, then $f_n\stackrel{u.}{\not\to} f$ on $F$.
Proof of Claim: Suppose $1\in F'$, let $\{x_m\}_m\subseteq F-\{1\}: x_m\uparrow1$. Then
$$\forall n: \sup_{x\in F}|f_n(x)-f(x)|\geq \sup_{m}|f_n(x_m)-f(x_m)|=\sup_{m}|f_n(x_m)|=1,$$
so that any member of $\{f_n\}_n$ is bounded away from $f$ in supremum norm (over $F$), which implies that $f_n\stackrel{u.}{\not\to} f$ on $F. \checkmark$
Thus if $F$ is a subset on which we have uniform convergence, $f\subseteq[0,1-\delta]$ for some $\delta\in]0,1]$, that is to say, we cannot have uniform convergence on a subset with full measure.
