Halmos "Measure theory" exercise on limit of sequence of sets Problem statement.
This is an exercise from chapter 1, section 4 (problem 13) from Halmos textbook:
If $\{E_n\}$ is a sequence of sets, write
$D_1=E_1, D_2=D_1 \triangle E_2, D_3=D_2 \triangle E_3$
and, in general,
$D_{n+1}=D_n \triangle E_{n+1}$
Prove that the limit of the sequence $\{D_n\}$ exists if and only if $\lim_n E_n=\emptyset$.
The attempt at a solution.
I am pretty stuck with this problem, I have to use the definition of limit for a sequence of sets, that is, if $\lim$ sup=$\lim$ inf, then, this is the limit
of the sequence. In this particular case, suppose I want to show the right implication:
$\rightarrow$ Suppose $\lim_{n \to \infty} D_n$ exists, I want to show that $\lim_{n \to \infty} E_n=\emptyset$. I thought that maybe the easier way to prove this is by the absurd:
Suppose that $\lim_{n \to \infty} E_n \neq\emptyset$. This means that 
(1) there is $x \in \bigcap_{n \in \mathbb N} (\bigcup_{k \geq n} E_k)$, 
or, equivalently, 
(2) $x \in \bigcup_{n \in \mathbb N} (\bigcap_{k\geq n} E_k)$ 
If I consider (2), (2) implies there is $x \in \bigcap_{k\geq n_0} E_k$ for some $n_0 \in \mathbb N$. 
From here I have no idea how to conclude that the limit of $(D_n)_{n \in \mathbb N}$ doesn't exist. 
$\leftarrow$ The assumption is that $\lim_{n \to \infty} E_n=\emptyset$, I want to show that this implies that $\lim_{n \to \infty} D_n$ exists.
But $\lim_{n \to \infty} E_n=\emptyset \implies \bigcap_{n \in \mathbb N} (\bigcup_{k \geq n} E_k)=\emptyset= \bigcup_{n \in \mathbb N} (\bigcap_{k\geq n} E_k)$. Again, I am confused on how could I use this information to prove the existence of the limit of $(D_n)_{n \in \mathbb N}$.
I would appreciate some help and suggestions on how could I continue the problem to prove both implications.
 A: Be careful on the forward direction. You can't assume $\lim_{n\to\infty} E_n$ exists, so (1) and (2) may not be equivalent a priori. If $\limsup_{n\to\infty} E_n = \emptyset$, then we can say $\lim_{n\to\infty} E_n = \emptyset$. Note that the limit of $\{D_n\}$ exists if and only if $\limsup_{n\to\infty} D_n - \liminf_{n\to\infty} D_n = \emptyset$. Therefore it is sufficient to show 
$$
\limsup_{n\to\infty} E_n = \limsup_{n\to\infty} D_n - \liminf_{n\to\infty} D_n.
$$
For example, imagine that $x \in \bigcap_{n \ge 1} E_n$. Notice that $x \in D_1,D_3,D_5,\ldots$, which implies $x \in \limsup_{n\to\infty}D_n$. However, $x \notin D_2,D_4,D_6,\ldots$, so $x \notin \liminf_{n\to\infty}D_n$. I'll give a sketch of how to proceed in general.
Suppose $x \in \limsup_{n\to\infty} E_n$. Let $\{E_{n_k}\}_{k \ge 1}$ be the subsequence of all sets in $\{E_n\}$ that contain the point $x$. It's easy to see that $x \in D_{n_1}$ but $x \notin D_{n_2}$. The above example might lead you to believe that $x \in D_{n_{k}}$ for odd $k$ and $x \notin D_{n_{k}}$ for even $k$. This turns out to be the case. Proceed by induction and use the fact that for every $k \ge 2$, 
$$
D_{n_k} = D_{n_k-1} \triangle E_{n_k} = D_{n_{k-1}} \triangle (E_{n_{k-1}+1} \triangle \cdots \triangle E_{n_k}).
$$
 This proves that $x \in \limsup_{n\to\infty} D_n - \liminf_{n\to\infty} D_n$.
Now suppose that $x \in \limsup_{n\to\infty} D_n - \liminf_{n\to\infty} D_n$. Then there are infinitely many $D_n$ that contain $x$, and infinitely many $D_n$ that do not contain $x$. So we may choose a subsequence $\{D_{n_k}\}$ such that $x \in D_{n_k}$ but $x \notin D_{n_k-1}$. Now try to show that $x \in E_{n_k}$ for all $k$ using the definition of $D_{n_k}$. This completes the proof.
