In this proof of MCT, I don't understand the the use of the number $\alpha$. When i ignore it and read the proof I don't see where it fails. 

Thanks for help.

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Proof. 

Let $(X,\mathbb{X},\mu)$ be a measure space.

Then since $f\_{n} \leq f$ for all $n$ it follows that $\lim \_{n \rightarrow \infty} \int f\_{n} \mathrm{d} \mu \leq$
$\int f \mathrm{d} \mu .$ 

To show the inequality in the other direction we fix a simple function
$\phi \leq f, 0<\alpha<1$ and let
$$
A\_{n}=\left\{x: \alpha \phi(x) \leq f\_{n}(x)\right\}
$$
Note that for each $n$, $A\_{n} \in \mathbb{X}$ and $A\_{n} \leq A\_{n+1} .$ We also have that $X=$
$\cup\_{n=1}^{\infty} A\_{n} .$ By Lemma 4.5 we can define a measure $\nu$ on $(X, \mathbb{X})$ by
$$
\nu(A)=\int\_{A} \phi \mathrm{d} \mu(x)
$$

for all $A \in \mathbb{X} .$ It follows from Lemma 3.4 that
$$
\int \phi \mathrm{d} \mu=\nu(X)=\nu\left(\cup\_{n=1}^{\infty} A\_{n}\right)=\lim \_{n \rightarrow \infty} \nu\left(A\_{n}\right)
$$
However for all $n \in \mathbb{N}$
$$
\int\_{X} f\_{n} \mathrm{d} \mu \geq \int\_{A\_n} f\_{n} \mathrm{d} \mu \geq \alpha \int\_{A\_{n}} \phi \mathrm{d} \mu=\alpha \nu\left(A\_{n}\right)
$$
and thus $\alpha \int \phi \mathrm{d} \mu \leq \lim \_{n \rightarrow \infty} \int f\_{n} \mathrm{d} \mu .$ 

Since this holds for all $0<\alpha<1$ and
simple functions $\phi \leq f$ it follows that
$$
\int f \mathrm{d} \mu \leq \lim \_{n \rightarrow \infty} \int f\_{n} \mathrm{d} \mu
.$$