Proof of Monotone convergence theorem.

$X=\cup_n A_n$ is false if you take $\alpha=1$. For example take $f$ to be a simple function and $\phi =f$. If $f_n$ is strictly increasing to $f$ then $A_n$ is empty.

Proof of the fact that $X=\cup_n A_n$: If $\phi (x)=0$ then $x \in \cup_n A_n$ because $f_n(x) \geq 0$. If $\phi (x) >0$ then $f_n(x) \to f(x)$ and $f(x)\geq \phi (x) >\alpha \phi(x)$ so there exisst $n_0$ such that $f_n(x) >\alpha \phi(x)$ for all $n \geq n_0$. It follows that $x \in A_{n_0}$.

$\alpha$ is there to obtain the result $X=\bigcup_{n=1}^{\infty} A_{n}$.

The reason is as follows.

Since $\ \lim_{n \to \infty}f_n(x)=f(x), \ $ we have $$\lim_{n \to \infty} A_n = \{ x:\alpha \phi(x) \le \lim_{n \to \infty}f_n(x) \}$$ $$ \ \ \ \ =\{ x:\alpha \phi(x) \le f(x) \}.$$

But, for all $x\in X$, we have $f(x) \ge \phi (x)$, so since $\alpha <1, \ $ $f(x) > \alpha \phi (x). \ $ Hence

$$\lim_{n \to \infty} A_n = \{ x:\alpha \phi(x) <f(x) \}.$$

Now, because this is true for all $x\in X$, it follows that $\ \lim_{n \to \infty} A_n = X$.

And since $\ A_1 \subseteq A_2 \subseteq A_3 \subseteq... \ $, we have $\ \bigcup_{n=1}^{\infty} A_{n}= \lim_{n \to \infty} A_n = X$.

If $\alpha\in(0,1)$ is missing in this proof then it fails because there is no guarantee that in that case $X=\bigcup_{n=1}^{\infty}A_n$.

It is quite well possible that for some $x$ we have: $$\forall n [\phi(x)>f_{n}(x)] $$