I came up with an alternative proof of Egoroff's theorem, which looks simpler than the original one in Follan's book. I suspect it might be wrong. Can anyone give some comments?
To prove: if $\mu(X) < \infty$ and $f_n$ converges to $f$ a.e., then for any $\varepsilon > 0$, there exists $E \subset X$ such that $\mu(E) < \varepsilon$ and $f_n$ converges to $f$ uniformly on $E^c$.
Here is my proof:
for any $\varepsilon > 0$, define $E_{n,\varepsilon} = \bigcup_{m \geq n}\left\{ x \colon \left\vert f_{m}(x)-f(x) \right\vert \geq \varepsilon \right\} $.
By $\mu(X) < \infty$, convergence a.e. is equivalent to:
$\forall \varepsilon > 0$, $\lim_{n \to \infty}\mu\left(E_{n,\varepsilon}\right) = 0$
Then, there exists $N \in \mathbb{N}$, such that $\mu\left( E_{n,\varepsilon} \right) < \varepsilon, \quad \forall n \geq N$.
Consider set $E_{N,\varepsilon}^{c} := \bigcap_{m \geq N}\left\{ x \colon \left\vert f_{m}(x)-f(x) \right\vert < \varepsilon \right\} $.
Then $f_{n} \rightrightarrows f$ on $E_{N,\varepsilon}^{c}$:
$\forall \varepsilon > 0,$ $\exists N \in \mathbb{N}, \text{s.t.} \left\vert f_{m}(x) - f(x) \right\vert < \varepsilon, \quad \forall x \in E_{N,\varepsilon}^{c} ~\text{and}~n \geq N.$
Done!
My concern is that the construction of set $E_{N,\varepsilon}^{c}$ depends on $N$, while it does not in Folland's proof.
For uniform convergence you have to find a set $E$ such that $\sup_{x \in E^{c}}|f_n(x)-f(x)|$ can be made less than any given number $\eta$ by choosing $n$ sufficiently large. It is no enough to take $\eta=\epsilon$. The set $E$ itself should not to depend on $\eta$.