In the proof of the Beppo Levi theorem in Schilling the following claim is made: $$f\leq u\Rightarrow I_\mu(f)\leq\sup_{j\in\mathbb{N}}\int u_j d\mu.$$
The proof of this claim is not hard but there is one point that I have a hard time seeing of why it is done, which is why is the parameter $\alpha$ being introduced.
Let $(X,\mathcal{A},\mu)$ be a measure space, $f=\sum\limits_{k=0}^My_k\mathbf{1}_{A_k}$ a standard representation of a non-negative measurable simple function, $u_i:X\rightarrow [0,\infty]$ an increasing sequence of measurable functions and $u=\sup\limits_{i\in\mathbb{N}}u_i$. Define,
$I_\mu(f):=\sum\limits_{k=0}^My_k\mu(A_k)$ and for any non-negative measurable function $w$,
$\int wd\mu:=\sup\{I_\mu(g):g\leq w$ & $g\geq 0$ measurable simple function$\}.$
The proof of the claim from Schilling is as follows. For a simple non-negative $f$ such that $f\leq u$, we can see that $$(\forall\alpha\in(0,1))(\forall x\in X)(\exists N(\alpha,x))(\forall j\in \mathbb{N})(j\geq N(\alpha,x)\Rightarrow \alpha f(x)\leq u_j(x))\space\space \space -(\dagger)$$
Thus the (measurable) sets $B_j:=\{\alpha f \leq u_j\}\uparrow X$ as $j\uparrow\infty$. Also we clearly have, $$\alpha f\mathbf{1}_{B_j}f\leq \mathbf{1}_{B_j}u_j\leq u_j.$$ So then, $$\sum_{k=0}^M\alpha y_k\mu(A_k\cap B_j)=I_\mu\bigg(\sum_{k=0}^M\alpha y_k\mathbf{1}_{A_k}\mathbf{1}_{B_j}\bigg)=I_\mu(\alpha\mathbf{1}_{B_j}f)\leq\int u_jd\mu\leq\sup_{j\in\mathbb{N}}\int u_jd\mu.$$
Then by $\sigma$-additivity of $\mu$, $\mu(A_k\cap B_j)\uparrow \mu(A_k)$ as $B_j\uparrow X$, $j\uparrow \infty$.
Thus $$\alpha I_\mu(f)=\alpha\sum\limits_{k=0}^My_k\mu(A_k)\leq\sup_{j\in\mathbb{N}}\int u_jd\mu$$ and letting $\alpha\rightarrow 1$, the claim follows.
Now my problem is with $(\dagger)$ and why the $\alpha$ is needed? Now as $\alpha\in(0,1)$ and $f\geq 0$, we obviously have $\alpha f\leq f$. So if $f\leq u$ we then have $\alpha f\leq sup_{i\in\mathbb{N}}u_i$, and consequently for a $x\in X$ there is a $N(\alpha,x)\in\mathbb{N}$, such that $\alpha f(x)\leq u_{N(\alpha,c)}(x)$ and as the $u_i$'s are increasing we have that $\alpha f(x)\leq u_j(x)$, for all $j\geq N(\alpha,x)$.
But from this reasoning and the one above it will seem to work if the $\alpha$ was never introduced, so I would appreciate any help to explain to me why I am wrong in thinking the $\alpha$ is not needed.
Thanks in advance.