In part of some proof, I have a doubt in the following step:
Let $(\Omega, \mathcal{A}, \mu)$ be a measure space. Let $M\in\mathcal{A}$, $(M_n)_{n\in\mathbb{N}}$ be a sequence of disjoint set of $\mathcal{A}$ and $M:=\bigcup_{n}M_n$. For $f\in L^2(\Omega, \mu)$, set $f_k := \sum_{n=1}^{k} \chi_{M_n}f$.
Why do we have $|\chi_{M}f-f_k|^2 \leq 4|f|^2$ on $\Omega$? If $x\in M$ but $x \notin M_n$ $\forall n \leq k$, then $\chi_{M}f=f, f_k=0$ thus $|\chi_{M}f-f_k|^2 \leq |f|^2$, and for other cases, the RHS of the equality is $0$. So, where does the 4 in front of $|f|^2$ come from?
I think that the proof would simply be
$|\chi_Mf-f_k| \leq |\chi_Mf|+|f_k| \Rightarrow |\chi_Mf-f_k|^2 \leq |\chi_Mf|^2+|f_k|^2+2|\chi_Mf||f_k| \leq |f|^2+|f|^2+2|f|^2 = 4|f|^2$
since $|\chi_Mf| \leq |f|$ and $|f_k| \leq |f|$ on $\Omega$.