Sum of a function Θ(g) with a function that is not O(g)

Consider g a function of n: $$g(n)$$.

Knowing that the function $$f(n) \in Θ(g(n))$$ and the function $$h(n) \notin O(g(n))$$, could we conclude anything, related to it's asymptotic behaviour, about $$f(n) + h(n)$$ with respect to $$g(n)$$?

Given that $$f(n)$$ is $$Θ(g(n))$$, I have that $$f(n)$$ grows the same way as $$g(n)$$.
Given that $$h(n)$$ is not $$O(g(n))$$, I have that $$h(n)$$ grows more rapidly compared to $$g(n)$$.

In my mind, at least, it seems like these two above observations are sufficient to conclude that $$f(n) + h(n)$$ grows (just like $$h(n)$$) more rapidly than $$g(n)$$ and, thus, $$f(n) + h(n) \in O(g(n))$$, but I'm afraid it's a counterintuitive case or I'm missing something.

Suppose $$f\in \Theta(g)$$ and $$h\notin \mathcal{O}(g)$$. That means: $$\exists A>0, B>0, \forall n\geqslant 0, Ag(n) \leqslant f(n) \leqslant Bg(n)$$ and $$\forall C>0, \exists n\geqslant 0, h(n) > Cg(n)$$
Therefore, for all $$C>0$$, there exists $$n\geqslant 0$$ such that $$h(n) > (C-A)g(n)$$. We conclude that $$f(n) + h(n) > Ag(n) + (C-A)g(n) = Cg(n)$$.
We just proved that $$f +h\notin\mathcal{O}(g)$$.