obvious property of big O, big Omega, and big Theta
EDIT: Now that I have fresh eyes in the morning, I see that I have thoroughly misread the question. The answer below applies to “if $f(n)\ne O(g(n))$, then $f(n)=\Omega(g(n))$”. As noted in comments above, the question as actually stated has a trivially true answer for all functions, that is, $f(n)=O(g(n))$ is equivalent to $g(n)=\Omega(f(n))$ immediately from the definition.



The answer is positive if $f$ and $g$ belong to a “well-behaved” family of functions which ensures that they are asymptotically comparable; in particular, if they are (restrictions to $\mathbb N$ of) functions taken from a Hardy field. For example, this holds if both functions are first-order definable (with parameters) in the real exponential field $(\mathbb R,+,\cdot,\exp)$, or in any o-minimal expansion of the real field for that matter.