Missing part of the proof of Master Theorem's case 2 (with ceilings and floors) in CLRS?
As you said, the book does not include a proof for $g(n) = \Omega(n^{\log_b a}{\log_bn})$. There is no need to.

Recall that part of proof is to show the upper bound of $f(n)$, when $f(n)$ is defined by the following recurrence relation using the ceiling function, which is the formula 4.25 in CLRS.
$$T(n) = aT(\lceil\frac nb\rceil) + f(n)$$
Noting that the lower bound for $T(n)$ "is routine, since we can push through the bound $\lceil\frac nb\rceil\ge\frac nb$", the book proceed to prove an upper bound for the above $T(n)$, where we only need to show an upper bound for $g(n)$.



What you meant might be the corresponding situation in the case when $f(n)$ is defined using the floor function as the formula 4.26 in CLRS,
$$T(n) = aT(\lfloor\frac nb\rfloor) + f(n)$$
in which case, we will defined $h(n) = \sum_{j=0}^{\lfloor \log_b n \rfloor-1} a^jf(n_j)$ and $n_j = \begin{cases}
n,  &amp; \text{if j = 0} \\
\lfloor n_{j-1}/b \rfloor, &amp; \text{if j &gt; 0}
\end{cases}$.

Note that the definition of $h(n)$ is not the same as that of $g(n)$, although the two are almost the same.

I will leave as an exercise for you to prove a lower bound for the above $T(n)$ using $h(n)$. In particular, you will need to show that in case 2, i.e., when $f(n)=\Theta(n^{\log_b a})$, 
$$h(n) = \Omega(n^{\log_b a}{\log_bn}).$$

Hint, as said in the last statement of section 4.6.2 and by Yuval's comment, the proof of the lower bounds is similar.