# lim inf, lim sup in measure theory

I am struggling with this question from an old measure theory question set, mainly because I'm confused by the notation.

Determine $$\liminf A_n$$ and $$\limsup A_n$$ for the sequence $$(A_n)_{n∈\mathbb{N}}$$ of sets given by: $$A_n = \begin{cases} (-\frac{1}{n}-2,1) & n \text{ odd}\\ (0,3+\frac{1}{n}) & n \text{ even}. \end{cases}$$

Clearly the odd sequence is converging to $$(-2,1)$$ and the even series is converging to $$(0,3)$$, but I don't know what $$\limsup$$ and $$\liminf$$ mean on ordered pairs like this.

Edit: Clearly I was over-thinking this - I thought I was dealing with ordered pairs from $$\mathbb{R}^2$$, when clearly these are open intervals on $$\mathbb{R}$$. Thank you to the people who set me straight.

• They are open intervals. For a family $F=\{F_n:n\in \Bbb N\}$ of sets we have (i) $x\in \lim \inf F$ iff $\{n\in \Bbb N: x\not \in F_n\}$ is finite, and (ii) $x\in \lim\sup F$ iff $\{n\in \Bbb N: x\in F_n\}$ is infinite. Commented Apr 17, 2020 at 1:58

If $$0 the $$x$$ clearly belongs to all of these sets. Suppose $$y$$ belongs to $$A_n$$ for all $$n >n_0$$ for some integer $$n_0$$. Then $$-2-\frac 1 n for $$n >n_0$$ which implies $$-2 \leq y <1$$. Similarly we get $$0 . Combining these we get $$0. Hence $$\lim \inf A_n =(0,1)$$. Can you now verify that a point $$x$$ belongs to infinitely many of the set $$A_n$$ iff $$x \in [-2,1) \cup (0,3]$$. This part is easy so I will let you handle this.
• Ah! The sets were intervals of $\mathbb{R}$! Now it all makes sense. Thank you. Commented Apr 17, 2020 at 0:21