In the answer given in the book for the proof of $\lg n=\Omega(n\lg n)$ there are several steps which I don't understand .
$$\lg n!=\lg n+\lg(n-1)+\lg(n-2)+ ....+\lg(2)+\lg 1$$
Then it says that
$$\lg n!=\lg n+\lg(n-1)+\lg(n-2)+....+\lg(2)+\lg 1\\\ge \lg n +\lg(n-1)+\lg(n-2)+...+\lg\lceil \frac n2\rceil$$
Here in the right hand side of the inequality is the log terms considered upto the $n\over2$ term. How is it decided that up to only $n\over2$ terms should be considered.
$$\lg n+\lg(n-1)+\lg(n-2)+...+\lg({n\over2})\ge \lg\lceil \frac n2 \rceil+\lg\lceil \frac n2 \rceil+...+\lg\lceil \frac n2 \rceil=[ \lceil \frac {n+1}2 \rceil \lg\lceil \frac n2 \rceil]$$
Here why is the ceiling function being used? Why can't the floor be used?
We want to estimate a sum from below. An easy lower bound is: the number of terms times the smallest term in the sum.
Problem with the above: the smallest term is too small. Solution: drop the smallest half of terms and estimate the rest, as above. Why $1/2$? Because why not. Could be $1/3$ or $2/3$, or $14/73$, but none of those choices are more appealing than $1/2$.
First of all: using the ceiling does not mean that the floor could not be used. When deriving asymptotic estimates, there are plenty of ways that lead to the same end result. I think the ceiling was used so that later on we can get rid of it with an inequality that points in the desired direction, $\ge$:
$$ \left\lceil \frac {n+1}2 \right\rceil \lg\left\lceil \frac n2 \right\rceil \ge \frac n2 \lg \frac n2 = \Omega(n\lg n)$$