2.33 Egoroff's Theorem - Suppose that $\mu(X) < \infty$, and $f_1,f_2,\ldots$ and $f$ are measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e. Then for every $\epsilon > 0$ there exists a set $E\subset X$ such that $\mu(E) < \epsilon$ and $f_n\rightarrow f$ and $f_n\rightarrow f$ uniformly on $E^c$.
Proof - Let $\mu(X) < \infty$ and $\{f_j\}_{j\in\mathbb{N}}$ and $f$ be measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e. Then, since $f_n\rightarrow f$ a,e, there exists a set $E\in M$ such that $\mu(E) = 0$ and for all $x\in E^c$, $f_n(x) \rightarrow f(x)$. So we have $f_n\chi_{E_c}\rightarrow fX_{E^c}$ and $f\chi_{E^c} = f$ a.e.
Now without loss of generality assume $f_n\rightarrow f$ pointwise. For $n,k$ define $E_n(k) = \bigcup_{m=k}^{\infty}\{|f_m - f|\geq 1/k\}$. Then clearly $E_n(k)$ is decreasing since $E_{n+1}(k)\subseteq E_n(k)$ and $\bigcap_{n=1}^{\infty}E_n = \emptyset$ by pointwise converence. So by continuity from above $\lim_{n\rightarrow \infty}\mu(E_n(k)) = 0$.
Let $\epsilon > 0$, for each $k$, choose $n_k$ such that $\mu(E_{n_k}) < \epsilon 2^{-k}$. Set $E = \bigcup_{1}^{\infty}E_{n_k}(k)$ then $\mu(E) < \epsilon$. So we have uniform convergence on $X\setminus E$.
I tried to follow what Folland did but I am not sure if this is correct completely and I honestly don't really like his approach. If there is an alternative way of proving this theorem any suggestions would be greatly appreciated. Also note that the first paragraph is not used, although it is still true.