Show that there exists a measurable set $A\in\mathcal{A}$ such that $\chi_{A_n}\to\chi_{A}$ $\mu$-a.e.

Question

I'm studying for measure theory and I am wondering how to understand a line in the proof, as well as how the problem could be solved without referencing the fact that $L_p$ spaces are Banach:

Question

Let $(X,\mathcal{A},\mu)$ be a finite measure space, and $A_n \in \mathcal{A}$, $n\geq 1$, be a collection of measurable sets such that $\mu( (A_n \setminus A_{n+1})\cup (A_{n+1} \setminus A_{n})) < \frac{1}{2^n} $

Show that there exists a measurable set $A\in\mathcal{A}$ such that the indicator functions converge: $\chi_{A_n}\to\chi_{A}$ $\mu$-a.e.

My Concerns

$1)$ The proof claims that for $m<n$, $$\sum_{k=m}^{n-1} \mu( (A_k \setminus A_{k+1})\cup (A_{k+1} \setminus A_{k})) < \sum_{k=m}^{n-1}\frac{1}{2^k} < \frac{1}{2^{m-1}}$$ but how is that last inequality derived? I can't seem to work it out

$2)$ How can you prove this question directly without using the fact that $L_p$ spaces are Banach?

Answer 1

1) The inequality is $$ \sum_{k=m}^{n-1}\frac1{2^k} < \sum_{k=m}^\infty\frac1{2^k} = \frac1{2^{m-1}}. $$

2) Let $$ E_n = \left(\bigcap_{k=n}^\infty A_k\right)\cup\left(\bigcap_{k=n}^\infty A_k^c\right). $$ From the property of $A_n$ you get $\mu(E_n)\geq 1-2^{-n+1}$. Define $$ A = \{x: x\in A_n \text{ eventually or } x\in A_n^c \text{ eventually}\}. $$ Then $A=\bigcup_{n=1}^\infty E_n$ and therefore $A$ has full measure.

Edit: sorry, I misread "How can you prove this question directly using the fact that $L^p$ spaces are Banach?".

Old answer for 2):

2) The functions $f_n=\chi_{A_n}$ are in $L^1$ and the property of the sets $A_n$ is equivalent to $\|f_n+1-f_n\|<2^{-n}$. This implies that the sequence $(f_n)_n$ is Cauchy in $L^1$, therefore $f_n\to f\in L^1$. Since you have a.e. pointwise convergence up to a subsequence, you get that also $f$ is an indicator function ($f$ has to be $0$ or $1$ at any point).