Show that $\lim\limits_{n\to\infty}D_n$ exists if and only if $\lim\limits_{n\to\infty}E_n=\emptyset$

Question

This is one of the exercises of Halmos's measure theory book.
I know it has been already treated but I've been asked to compute it by going via indicator functions.

Exercises goes as follows:

Let $(E_n)_{n\in\mathbb{N}}$ sequence of sets. Define the sequence $(D_n)_{n\in\mathbb{N}}$ as follows: $$D_1=E_1\;,\;\;D_2=D_1\triangle E_2\;,\\D_n=D_{n-1}\triangle E_n\;,\;\;\forall n\in\mathbb{N}_{\geq 3}\,.$$ Prove that the sequence $(D_n)$ converge to some set, say $A$, if and only if $\lim\limits_{n\to\infty}E_n=\emptyset$.

I have to calculate the "easiest arrow" so assuming that the $\lim\limits_{n\to\infty}D_n$ exists implies that $\lim\limits_{n\to\infty}E_n =\emptyset$, but I don't know how solve it with the indicators.

I know that:

$$D_n=D_{n-1}\triangle E_n=1_{D_{n-1}\triangle E_n}\\=\left|1_{D_{n-1}}+1_{E_n}\right|\mod2$$

but I don't know how to go forward.

Answer 1

Exactly how you do it using indicator functions will depend on what you already know about such functions, but here is one way. Let $X$ be the underlying set, so that all indicator functions have domain $X$.

The sequence $\langle D_k:k\in\Bbb N\rangle$ converges if and only if for each $x\in X$ the sequence $\left\langle\mathbf{1}_{D_k}(x):k\in\Bbb N\right\rangle$ is eventually constant, i.e., if and only if for each $x\in X$ there is an $n_x\in\Bbb N$ such that either $\mathbf{1}_{D_k}(x)=0$ for all $k\ge n_x$, or $\mathbf{1}_{D_k}(x)=1$ for all $k\ge n_x$.

Note that for each $k\in\Bbb N$ we have

$$\begin{align*} \mathbf{1}_{E_{k+1}}&=\mathbf{1}_{\varnothing\mathop{\triangle}E_{k+1}}\\ &=\mathbf{1}_\varnothing\oplus\mathbf{1}_{E_{k+1}}\\ &=\mathbf{1}_{D_k\mathop{\triangle}D_k}\oplus\mathbf{1}_{E_{k+1}}\\ &=\left(\mathbf{1}_{D_k}\oplus\mathbf{1}_{D_k}\right)\oplus\mathbf{1}_{E_{k+1}}\\ &=\mathbf{1}_{D_k}\oplus\left(\mathbf{1}_{D_k}\oplus\mathbf{1}_{E_{k+1}}\right)\\ &=\mathbf{1}_{D_k}\oplus\mathbf{1}_{D_k\mathop{\triangle}E_{k+1}}\\ &=\mathbf{1}_{D_k}\oplus\mathbf{1}_{D_{k+1}}\,, \end{align*}$$

where $\oplus$ is pointwise addition mod $2$.

In case you don’t immediately see where this calculation came from, it’s just the observation that

$$D_k\mathop{\triangle}D_{k+1}=D_k\mathop{\triangle}(D_k\mathop{\triangle}E_{k+1})=(D_k\mathop{\triangle}D_k)\mathop{\triangle}E_{k+1}=\varnothing\mathop{\triangle}E_{k+1}=E_{k+1}$$

expressed in terms of indicator functions.

Suppose first that $\langle D_k:k\in\Bbb N\rangle$ converges, and let $x\in X$; then there is an $n_x\in\Bbb N$ such that $\mathbf{1}_{D_k}(x)=\mathbf{1}_{D_{k+1}}(x)$ for all $k\ge n_x$, and hence $\mathbf{1}_{E_{k+1}}(x)=0=\mathbf{1}_\varnothing(x)$ for all $k\ge n_x$. This is the case for all $x\in X$, so $\lim\limits_{k\to\infty}\mathbf{1}_{E_k}=\mathbf{1}_\varnothing$, and hence $\lim\limits_{k\to\infty}E_{k}=\varnothing$.

For the reverse implication just run the argument backwards. Start by assuming that $\lim\limits_{k\to\infty}E_{k}=\varnothing$, so that for each $x\in X$ there is an $n_x\in\Bbb N$ such that $\mathbf{1}_{E_{k+1}}(x)=0$ for all $k\ge n_x$; conclude that $\mathbf{1}_{D_k}(x)=\mathbf{1}_{D_{k+1}}(x)$ for all $k\ge n_x$ and hence that either $\mathbf{1}_{D_k}(x)=0$ for all $k\ge n_x$, or $\mathbf{1}_{D_k}(x)=1$ for all $k\ge n_x$, i.e., that $\langle D_k:k\in\Bbb N\rangle$ converges.