R&C Analysis - Monotone Convergence Theorem

Rudin skips some nice background information that should precede Lesbesgue integration. Reader of the book will be wondering what’s so special about Lesbegue integration. First observation is that every Riemann integral function is Lesbegue integration. Every Lesbegue integral function is not Riemann integrable.
Ex: Let $X=[0,1]$, $f(x)=r$ for each $r\in Q\subset X$ and $f(x)=0$ otherwise. Riemann Integral lower limit is $0$ and largest value is $1$. However, using measurable set as $Q\cap X$, the Lesbegue integral has value and it is zero.
Thus Lesbegue Integral deals with larger classes of functions.
The Leabesgue Integral has the following nice property while Riemann Integral doesn’t have.
Lesbesgue’s Montone Convergence Theorem: Let $\{f_n\}$ be be a sequence of measurable functions on $X$ and suppose that Then $f$ is measurable, and $${\int }_X{f}_{n}d\mu \to {\int }_{X}fd\mu$$ PROOF: Since $\int f_n\le \int f_{n+1}$, there exists $\alpha \in [0,\mathrm{\infty }]$ such that $${\int }_X{f}_{n}d\mu \to \alpha \text{ as }n\to \mathrm{\infty }$$ Because $sup(f_n)$ and $limsup\{f_n\}$ as $n\to \mathrm{\infty }$ are both measurable, $f$ is measurable. Since $f_n\le f$, we have $\int f_n\le \int f$ for every $n$. Hence, $$\alpha \le {\int }_{E}fd\mu$$ All that needs to be shown is $\alpha \ge {\int }_{E}fd\mu$. Let $s$ be any simple measurable functions such that $0\le s\le f$, let $c$ be a constant with $0<c<1$, and define $$E_n=\{x:f_n(x)\ge cf(x)\}n=1,2,3,\cdots$$ Each $E_n$ us neasurable, $E_1\subset E_2\subset E_3\subset \cdots$ and $X=\cup E_n$. Also $${\int }_E{f}_{n}d\mu \ge {\int }_{E_n}{f}_{n}d\mu \ge c{\int }_{E_n}sd\mu n=1,2,\cdots$$ As $n\to \mathrm{\infty }$ Then, $$\alpha \ge c{\int }_{X}sd\mu$$ Since above equation holds for every $c<1$, we have $$\alpha \ge {\int }_{X}sd\mu$$ for every measurable $s$ satisfying $0\le s\le f$, so that $$\alpha \ge {\int }_{X}fd\mu$$ Conclusion follows.