Formulating the master theorem with Little-O- and Little-Omega notation
I am wondering if this means that we can write this first case of the theorem as $f \in o(n^{\log_b{a}})$ (and following the same logic, $f \in \omega(n^{\log_b{a}})$ for the third case), rather than the more convoluted alternative?

That is indeed a natural attempt to understand that &quot;convoluted&quot; condition in simple terms. Unfortunately, it is not correct.
Here is a simple counterexample. Let $a=1$ and $b=2$, and let $T : \mathbb{N} \rightarrow \mathbb{R}$ where $T(n) = T(\frac{n}{2}) + f(n)$ and $T(1)=0$, where $f(n)=\frac1{\ln n} \in o(1)=o(n^{\log_21})$.
If the first case of master theorem still holds, we should have $T(n) \in \Theta(n^{\log_21})=\Theta(1)$. However,
$$T(2^m)=T(2^{m-1})+\frac1m=T(2^{m-2})+\frac1m+\frac1{m-1}=\cdots=\frac1m+\frac1{m-1}+\cdots+1.$$
Letting $m$ goes to infinity, we see that $T(n)$ is not bounded because the harmonic series diverges.

For the same or a variety of other reasons, the usage of an arbitrarily small positive constant, which is usually denoted by $\epsilon$ pops up constantly in different places, especially in asymptotic estimate and complexity analysis.  You may want to get used to it or even get addicted to it.

Exercise 1. Verify that for any $\epsilon&gt;0$, it is not true that $\frac1{\ln n}\in o(n^{-\epsilon})$ although  $\frac1{\ln n}\in o(n^0).$
Exercise 2. Construct a counterexample for the master theorem where $a=4$, $b=2$ for

the first case where $f \in o(n^{\log_b{a}})$ instead of $ f \in O(n^{log_b{a}-\epsilon})$ for some  $\epsilon &gt; 0$ and
the third case where $f \in \omega(n^{\log_b{a}})$ instead of $f \in \Omega(n^{\log_b{a}+\epsilon})$ for some $\epsilon &gt; 0$

respectively.