So I'm looking at the examples and one thing that threw me off was this  f( n ) = n * log n = Ω( n ).

This is the example I'm looking at:

1. T(n) = 3 * T(n / 4) + n * log n

Answer: a = 3, b = 4, f(n) = n * log n, n(logba) = O( n0.793)
Since f( n ) = n * log n = O( n ) and the regularity condition holds, namely, a * f(n/b) = ¾ * n * log n = c * f(n) for c = ¾, we can apply case 3, and find T(n) = T(n * log n)

How did they obtain Ω(n)?