Proving a set-theoretic identity Context: Measure theory.
Reason: Just curious.
Question: Given $\{A_k\}$ with $A_k$ not disjoint, $B_1=A_1$ and $B_n = A_n - \bigcup\limits_{k=1}^{n-1} A_k$ for $n \in \mathbb{N}-\{1\}$ and $k \in \mathbb{N}$, how can I show that $$\bigcup\limits_{n=2}^{\infty}A_n = \bigcup\limits_{n=2}^{\infty}B_n?$$
Attempt: $$\bigcup\limits_{n=2}^{\infty}B_n=\bigcup\limits_{n=2}^{\infty}\left(A_n \cap (\bigcup\limits_{k=1}^{n-1} A_k)^c\right)=\bigcup\limits_{n=2}^{\infty}A_n \bigcap \bigcup\limits_{n=2}^{\infty}\left(\bigcup\limits_{k=1}^{n-1} A_k\right)^c = \cdots$$
Where do I go from here?

Try showing the double inclusion. One side is easy as $B_i \subseteq A_i$ for all $i$. For the other side, think of $x \in \bigcup A_i$, and let $A_k$ the first $k$ such that $x \in A_k$, what can you say about $x$ and $B_k$?
