Let $(X, \mathcal{M}, \mu)$ be a measured space. Let $f : X \rightarrow \mathbb{C}$ be a measurable function. Let $A \in \mathcal{M}$ and $t>0$, then we define $\forall n \in \mathbb{N}$: $$A_n = \{ x \in A ; |f(x)| > tn \},\text{ } \text{ } \text{ } \text{ } B_n = A_n -A_{n+1} $$ Show that $B_n$ are disjoint, and that $$A_n = \bigcup_{k \geq n}B_n$$ and $$\sum_{n=0}^{\infty} \mu (A_n) = \sum^{\infty}_{n=0}(n+1)\mu (B_n)$$
Tho show that the $\{ B_n \} $ are disjoint, I simply show that $\forall n\in \mathbb{N}$, we have $B_n \cap B_{n+1} = (A_n - A_{n+1}) \cap (A_{n+1} - A_{n+2})$.
As $ A_{n+1} - A_{n+2} \subset A_{n+1}$, we have $B_n \cap B_{n+1} = (A_n - A_{n+1}) \cap (A_{n+1} - A_{n+2}) \subset (A_n -A_{n+1}) \cap A_{n+1} = \emptyset.$
But I have difficulties showing that $A_n = \bigcup_{k \geq n}B_n$, and the equality with the sum. Can someone help me out?
By definition we have $$B_n:=A_n\setminus A_{n+1}=\{x\in A: t(n+1)\geqslant |f(x)|>tn\}$$ It is clear the $\{B_n\}$ are disjoint since for $k\neq n$ we have $$B_k\cap B_n=\{x\in A: t(k+1)\geqslant |f(x)|>tk\}\cap\{x\in A: t(n+1)\geqslant |f(x)|>tn\}\\=\{x\in A:t(k+1)\geqslant |f(x)|>tk\hspace{0.2cm}\& \hspace{0.2cm}t(n+1)\geqslant |f(x)|>tn\}=\emptyset$$ otherwise there would be an $x$ such that $f(x)$ gave two different values violating thus the definition of a function. Now $$\bigcup_{k\geqslant n}B_k=\bigcup_{k\geqslant n}(A_k\setminus A_{k+1})\\=\bigcup_{k\geqslant n} \{x\in A: t(k+1)\geqslant |f(x)|>tk\}=\{x\in A: |f(x)|>tn\}=A_n$$ Next $$\mu(A_n)=\mu\Big(\bigcup_{k\geqslant n}B_k\Big)=\sum_{k\geqslant n}\mu(B_k)$$ where equality follows since $\{B_k\}$ are disjoint. Therefore $$\sum_{n=0}^{\infty}\mu(A_n)=\sum_{n=0}^{\infty}\sum_{k\geqslant n}\mu(B_k)=\sum_{n=0}^{\infty}(n+1)\mu(B_n)$$