Let $(E,\mathcal{A},\mu)$ be a finite measure space and $\{f\_n\}\_n$ be a sequence of simple functions such that:
\begin{align\*}
f\_n1\_{\{\lvert f\_n\rvert\leq k\}}\overset{\sigma(L^2,L^2)}{\underset{n}{\longrightarrow}} u\_k, &\qquad\forall k\geq 1 \\
\|u\_k\|\_2\leq 2\|f\_n1\_{\{\lvert f\_n\rvert\leq k\}}\|\_2,&\qquad \forall n\geq k.
\end{align\*}
Put $\epsilon\_k=\frac{1}{2^k}$ $(k\geq 1)$. Why does there exist, for each $k\geq 1$, a simple function $v\_k$ such that:
$$
\|u\_k-v\_k\|\_2\leq \min \bigg(\inf\_{n\geq k}{\|f\_n1\_{\{\lvert f\_n\rvert\leq k\}}\|\_2},\frac{\epsilon\_{k-1}}{4k}\bigg)
$$



$ \|u_k-v_k\|_2\leq \min \bigg(\inf_{n\geq k}{\|f_n1_{\{|f_n|\leq k\}}\|_2},\frac{\epsilon_{k-1}}{4k}\bigg) $
$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$  

Fix any natural $k$. Let $$\de\_k:=\inf\_{n\ge k}\|f\_n 1\_{\{|f\_n|\le k\}}\|\_2$$ and $$\eta\_k:=\ep\_{k-1}/(4k).$$ 
We have 
$$\|u\_k\|\_2\le2\de\_k \tag{1}$$ 
and 
$$\eta\_k>0.$$ 
We want to show that there exists a simple function $v\_k$ such that
$$
\|u\_k-v\_k\|\_2\le\al\_k:=\min(\de\_k,\eta\_k). \tag{2}
$$
If $\de\_k=0$, then, by (1), $\|u\_k\|\_2=0$, and hence we may let $v\_k:=0$, to have (2). 


Otherwise, $\al\_k>0$. By the condition $f\_n 1\_{\{|f\_n|\le k\}}
\overset{\sigma(L^2,L^2)}{\underset n\longrightarrow} u\_k$, we have $|u\_k|\le k$ $\mu$-almost everywhere ($\mu$-a.e.). So, we can find a simple function $v\_k$ such that $|u\_k-v\_k|\le\al\_k/\sqrt{1+\mu(E)}$ $\mu$-a.e. Then we will have the inequality in (2), as desired. 


