Theorem 1.4 (Bounded convergence theorem) Suppose that $\{f_n\}$ is a sequence of measurable functions that are all bounded by $M$, are supported on a set $E$ of finite measure, and $f_n(x) \to f(x)$ a.e. $x$ as $n \to \infty$. Then $f$ is measurable, bounded, supported on $E$ for a.e. $x$, and$$\int |f_n \to f| \to 0 \text{ as } n \to \infty.$$Consequently,$$\int f_n \to \int f \text{ as } n \to \infty.$$
Proof. From the assumptions one sees at once that $f$ is bounded by $M$ almost everywhere and vanishes outside $E$, except for possibly on a set of measure zero. Clearly, the triangle inequality for the integral implies that it suffices to prove that $\int |f_n - f| \to 0$ as $n$ tends to infinity.
The proof is a reprise of the argument in Lemma 1.2. Given $\epsilon > 0$, we may find, by Egorov's theorem, a measurable subset $A_\epsilon$ of $E$ such that $m(E - A_\epsilon) \le \epsilon$ and $f_n \to f$ uniformly on $A_\epsilon$. Then, we know that for all sufficiently large $n$ we have $|f_n(x) - f(x)| \le \epsilon$ for all $x \in A_\epsilon$. Putting these facts together yields\begin{align*} \int |f_n - f(x)|\,dx & \le \int_{A_\epsilon} |f_n(x) - f(x)|\,dx + \int_{E - A_\epsilon} |f_n(x) - f(x)|\,dx \\ & \le \epsilon m(E) + 2M\,m(E - A_\epsilon)\end{align*}for all large $n$. Since $\epsilon$ is arbitrary, the proof of the theorem is complete.$$\tag*{$\square$}$$
For reference, we include the statement of Lemma 1.2 here.
Lemma 1.2 Let $f$ be a bounded function supported on a set $E$ of finite measure. If $\{\varphi_n\}_{n = 1}^\infty$ is any sequence of simple functions bounded by $M$, supported on $E$, and with $\varphi_n(x) \to f(x)$ for a.e. $x$, then:
(i) The limit $\lim_{n \to \infty} \int \varphi_n$ exists.
(ii) If $f = 0$ a.e., then the limit $\lim_{n \to \infty} \int \varphi_n$ equals $0$.
My question is, could anybody supply me their intuitions behind the proof of the bounded convergence theorem here? What are the key steps I should distill the proof into as to be able to recreate it from scratch?
To remember the proof, maybe it is best to keep a particular example in mind.
Let $E = [0,1]$, the closed unit interval on the line. Let $f_n(x) = x^n$, which is bounded by $M=1$. Then $f_n \rightarrow 0$ almost everywhere on $E$ but not uniformly.
But we can exclude the bits where uniform convergence fails (this is Egorov's theorem). In this particular case, we can take $A_\epsilon = [0, 1-\epsilon]$. Then $f_n \rightarrow 0$ uniformly on $A_\epsilon$, i.e. for large enough $n$ we have that $|f_n(x) - 0| < \epsilon$ on $A_\epsilon$.
Now add up the two pieces and let $\epsilon$ get arbitrarily small.