(1) Your proof that $\liminf A_n\subseteq \limsup A_n$ is correct.
Exercise 1: Under what conditions does equality hold in the above inclusion, i.e., under what conditions is it true that $\liminf A_n=\limsup A_n$?
(2) Note that $x\in \liminf A_n$ if and only if there exists a positive integer $N$ such that $x\in A_n$ for all $n\geq N$ if and only if there exists a positive integer $N$ such that $x\in \bigcap_{k=N}^{\infty} A_k$ if and only if $x\in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$.
I will leave the remaining questions as easy exercises (with similar solutions).
Exercise 2: Let $\{A_n\}$ be a sequence of measurable subsets of a measurable space $(X,\mu)$. Assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. Let $A=\limsup A_n$. Prove that $\mu(A)=0$. (Hint: Use the fourth assertion in your question, namely, use the characterization of $\limsup A_n$ in your question.)
Exercise 3: Give an example (in the context of Exercise 2) where $\lim_{n\to\infty} \mu(A_n)=0$ but that $\mu(A)>0$. Do not assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. (Hint: let $\{A_n\}$ be an appropriate sequence of intervals in $[0,1]$, for example.)
You cannot guarantee that $\langle a_n:n\in\Bbb N\rangle$ is a decreasing sequence: it might start out $$\left\langle 1,2,\frac1{10},\pi,\frac12,\frac13,\frac1{2^{100}},4,\frac1{100},\dots\right\rangle\;,$$ for instance. All you know is that all of its terms are positive, and for each $\epsilon>0$ there is some $n_a(\epsilon)\in\Bbb N$ such that $0<a_n<\epsilon$ whenever $n\ge n_a(\epsilon)$.
Similarly, all you can say for sure about $\langle b_n:n\in\Bbb N\rangle$ is that for every $\epsilon>0$ there is some $n_b(\epsilon)$ such that $1<b_n<1+\epsilon$ whenever $n\ge n_b(\epsilon)$. And above all you cannot assign specific values to the numbers $a_n$ and $b_n$: that’s changing the problem. (Of course, you can do so to look at an example or two in order to get a better idea of what’s going on, but that’s a different matter altogether.)
Now let’s take a look at $\liminf_n A_n$, where $A_n=[a_n,b_n)$: we want to determine which real numbers are eventually in the sets $A_n$, i.e., which are in all $A_n$’s from some point on. Here’s where those numbers $n_a(\epsilon)$ and $n_b(\epsilon)$ come in handy. For $\epsilon>0$ let $n(\epsilon)=\max\{n_a(\epsilon),n_b(\epsilon)\}$; then $0<a_n<\epsilon$ and $1<b_n<1+\epsilon$, and hence $$[\epsilon,1]\subseteq[a_n,b_n)\subseteq(0,1+\epsilon)\;.$$ for all $n\ge n(\epsilon)$. In other words, $[\epsilon,1]\subseteq A_n$ for all $n\ge n(\epsilon)$, and we conclude that for each $\epsilon>0$, $[\epsilon,1]\subseteq\liminf_n A_n$. It follows that $$\liminf_n A_n\supseteq\bigcup_{\epsilon>0}[\epsilon,1]=(0,1]\;.$$ On the other hand, if $x>1$, let $\epsilon=x-1$: for every $n\ge n(\epsilon)$ we have $1<b_n<1+\epsilon=x$, so $x\notin A_n$ whenever $n\ge n(\epsilon)$. This shows that $x$ isn’t even in infinitely many of the $A_n$’s, let alone in a tail of them, so $x\notin\limsup_n A_n$, and therefore certainly $x\notin\liminf_n A_n$. It’s also clear that no $x\le 0$ belongs to any of the $A_n$’s, so we’ve established that $\liminf_n A_n=(0,1]$, as you thought.
Along the way we’ve also seen that $\limsup_n A_n\subseteq(0,1]$, so $$\liminf_n A_n\subseteq\limsup_n A_n\subseteq(0,1]=\limsup_n A_n\;,$$ and it follows that $\limsup_n A_n=(0,1]$ as well, also as you thought.
It appears that you’re getting the concepts but might have a bit of difficulty actually writing down an argument to justify your reckoning of $\liminf$ or $\limsup$ of a sequence of sets.
Best Answer
Note that $A_n=\left(-\frac1n,2-\frac1n\right]=-\frac1n+(-1,2]$. So, if $N\in\mathbb N$,$$\bigcup_{n\geqslant N}A_n=\left(-\frac1N,2\right)$$and therefore$$\limsup_nA_n=\bigcap_{N\in\mathbb N}\left(-\frac1N,2\right)=[0,2).$$On the other hand, if $N\in\mathbb N$,$$\bigcap_{n\geqslant N}A_n=\left[0,2-\frac1N\right]$$and therefore$$\liminf_n=\bigcup_{N\in\mathbb N}\left[0,2-\frac1N\right]=[0,2)$$too.