Nesting of different Asymptotic operators Is it possible to nest big-oh notation with omega-notation? I came across this here, while doing calculations on an exercise: 
$$
f(x) \in O(\Omega(\log x))
$$
I'm really unsure on how to properly read this. Either it means, that $f(x)$ is unbounded, as $\Omega(\log x)$ is anything growing as fast or faster than $\log x$, and applying big-O is then somewhat meaningless and can be omitted. Or, it results in $\Theta(\log x)$, but I think this would be wrong. 
Am I therefore correct that the big-O can be omitted here? 
Edit: 
Okay, I think I have made a mistake on my part: 
I have: 
$$\mathbb{E}[X] \in \Omega(\log x)$$
And as I have to use Chernoff bounds, I wanted to plug it into the first chernoff inequality:
$$
P\left[ |X - \mathbb{E}[X]| \in O\left(\log n + \sqrt{\mathbb{E}[X]\log n}\right)\right] > 1 - \frac{1}{n^c} for\, a\, selectable\, constant\,\, c
$$
So I shamelessly plugged in $\mathbb{E}[X]$ which resulted in the first equation. But I guess that was the first real mistake - is that correct?
If so, how would I properly do that? Note, that one can assume $n$ to be large as its all about asymptotics. 
 A: Yes, it seems quite strange, and no, it (probably) cannot be omitted.
Typically, when used in a contexts like: $f(x) = x + O(\log x)$, it normally means that there is some element $g$ in the set $O(\log x)$ which appears as $f(x) = x + g(x)$.
In $f \in O(\Omega(\log x))$), If we take it to mean that there is some $g \in \Omega(\log x)$ such that $f \in O(g)$, then this means that $f$ is bounded asymptotically above by a function $g$, which itself is bounded below asymptotically by $\log x$, which is kind of a strange way of writing things, can probably drop both $\Omega$ and $O$.
If you interpret it to mean for every $g \in \Omega(\log x)$, $f \in O(g)$, then $f = O(\log x)$ and you can drop the $\Omega$ (not the $O$), and it is yet another strange way of writing things.
Can you provide more context?
A: You are probably right. We can write
$$
g(x)= \Omega(\log x) \geq c_1 \log x
$$
hence
$$
\lim_{x \to \infty}\frac{f(x)}{g(x)} \leq \frac{1}{c_1} \lim_{x \to \infty} \frac{f(x)}{\log x}\leq \frac{c_{2}}{c_1}
$$
hence $f(x) = O(\log x)$
