Let $f$ be a real valued function and let
1) $$\limsup_{r \to \infty}f(r) = \mu$$ 2) $$\liminf_{r \to \infty}f(r) = \phi$$
Does
1) $\iff$ $\exists\ r_n$ increasing, $r_n \to \infty$ such that for $\mu'<\mu,\ f(r_n)>\mu'$
2) $\iff$ $\exists\ r_n$ increasing, $r_n \to \infty$ such that for $\phi'>\phi,\ f(r_n)<\phi'$
where $\mu'$ and $\phi'$ are finite?
I've noticed them being used in the proofs for the limit superior and inferior of real-valued functions in my complex analysis textbook (Markushevich). Since I know nothing about them, I have to make up some conjectures and see if I am correct.
P.S. I only know extremely little about the limit superior and inferior of functions, so I would prefer a simple explanation. Thank you.
Let's first understand the concept in English language and then translate it to symbols.
Meaning of $\limsup\limits_{r \to \infty}f(r) = \mu$: This means if we take any number $\mu' > \mu$ then for all sufficiently large values of $r$ we have $f(r) < \mu'$. At the same time if we take any $\mu' < \mu$ then there are infinitely many values of $r$ without any bound for which $f(r) > \mu'$.
If we translate the above into symbols it means that if $\mu' > \mu$ then there is a positive number $N$ such that $f(r) < \mu'$ for all values of $r > N$. And if $\mu'< \mu$ and $N$ is any positive number then there exists a value $r > N$ such that $f(r) > \mu'$.
Clearly this means that your statement 1) is true. We can make the sequence $r_{n}$ as follows. By definition above we can choose a number $r_{1}$ such that $f(r_{1}) > \mu'$. Next choose $r_{2} > r_{1} + 1$ such that $f(r_{2}) > \mu'$. Then choose $r_{3} > r_{2} + 2$ with $f(r_{3}) > \mu'$. Similarly choose $r_{n + 1} > r_{n} + n$ with $f(r_{n + 1}) > \mu'$. Clearly then $r_{n} \to \infty$.
Your statement 2 is also true and I hope you can provide similar explanation as above for the case of limit inferior.