Conditions for applying Case 3 of Master theorem
Yes, your sharp observation is completely correct.

To be compatible with the highly strict style shown at section 4.6, Proof of the master theorem of Introduction to Algorithms, here is the complete proposition and a slightly more rigorous proof. It seems that the proof in the question ignores the requirement that $f$ is defined only on exact powers of $b$.
(Regularity implies lower-bounded by a greater-exponent polynomial.) Let $a\geq1$, $b&gt;1$ and $f$ be a nonnegative function defined on exact powers of $b$. Suppose $af(\frac nb)\leq cf(n)$ for some fixed $c&lt;1$ and for all sufficiently large $n$. Furthermore, $0 &lt; f(n)$ for all sufficiently large $n$. Then $f(n) = \Omega(n^{log_ba +\epsilon})$ for some fixed $\epsilon&gt;0$.
Proof. There exists some $n_0&gt;0$ such that $af(\frac nb)\leq cf(n)$ and $0 &lt; f(n)$ for all $n\ge n_0$. We can assume $n_0$ is an exact power of $b$ since, otherwise, we can replace $n_0$ by $b^{\lceil\log_b{n_0}\rceil}$.
Let $n\ge n_0$ be an exact power of $b$. So $n = n_0b^m$, where $m=\log_b\frac n{n_0}$ is an integer since both $n$ and $n_0$ are exact powers of $b$.
Applying $af(k/b)\leq cf(k)$ multiple times, we get
$$f(n) \ge \frac acf(\frac nb) \ge (\frac ac)^2f(\frac n{b^2})\ge \cdots \ge (\frac ac)^mf(\frac n{b^m})=(\frac ac)^mf(n_0)$$
Since
$$(\frac ac)^m=(\frac ac)^{\log_b\frac n{n_0}} =(\frac n{n_0})^{\log_b\frac ac}=(\frac n{n_0})^{\log_ba-\log_bc}=c_0n^{log_ba+\epsilon}$$
where $\epsilon=-\log_bc &gt; 0$ and $c_0=(\frac1{n_0})^{log_ba +\epsilon}$ are two constants, we have
$$f(n) \ge c_0f(n_0)n^{log_ba +\epsilon}.$$
So,
$$f(n)=\Omega(n^{log_ba +\epsilon}).\quad \checkmark$$

What happens if $n$ is not necessarily an exact power of b?  The same result will hold if we replace $\frac nb$ by $\lfloor \frac nb\rfloor$ or $\lceil \frac nb\rceil$.  The following is a version when $\lfloor \frac nb\rfloor$ is used. 
Let $a\ge1$, $b&gt;1$ and $f$ be a nonnegative function defined on positive integers. Suppose $af(\lfloor \frac nb\rfloor)\leq cf(n)$ for some fixed $c&lt;1$ and for all sufficiently large $n$. Furthermore, $0 &lt; f(n)$ for all sufficiently large $n$. Then $f(n) = \Omega(n^{log_ba +\epsilon})$ for some fixed $\epsilon&gt;0$.