Are we allowed to assume $X \ge E(X)$ in general? Given a random variable $X : \Omega \rightarrow \Bbb R$ and its expected value $E(X)$, it's safe to assume there are certain $\omega \in \Omega$ such that $X(\omega) \ge E(X)$. This technique is often used in probabilistic proofs. But are we also allowed to assume that $X \ge E(X)$ in general, without taking a certain $\omega$ into account? 
 A: If $E[X]$ exist it necessarily holds that $$P( X \ge E[X] ) > 0$$ but $\{X \ge E[X]\}$ is just a shorten notation for $$\left\{\omega \in \Omega: X(\omega) \ge E[X]\right\}$$ so actually we have $$P( \left\{\omega \in \Omega: X(\omega) \ge E[X]\right\} ) > 0$$ so there is necessarily an $\omega \in \Omega$ s.t. $$X(\omega) \ge E[X]$$ But ofc the assumption $X \ge E[X]$ does not hold in general even more it just holds for real-valued $X$ iff $X$ is almost surely constant. 
E.g. take $X$ s.t. $$P(X = -1) = P(X=1) = \frac{1}{2}$$ then $E[X] = 0$ but obviously $X \not\ge E[X]$.
A: I am certainly no expert for stochastics, but I will try to give what I understand as a counterexample.

Let $\Omega=\Bbb N$ (without zero) and $X=\mathrm{id}$. All you need is to choose the probabilities $p_i$ in such a way so that
$$\sum_{i=1}^\infty p_i=1\qquad\text{and}\qquad E[X]= \sum_{i=1}^\infty ip_i =\infty.$$
For example, choose
$$p_i=\begin{cases}2^{-k} & \text{if $i=2^k$ for some $k\in\Bbb N$}\\0&\text{otherwise}\end{cases}.$$
Then $X(n)=n<\infty=E[X]$ for all $n\in\Omega$. All assuming that you consider $E[X]=\infty$ as an existing but infinite expectation value.
A: Take 20 people and compute their average height. There will be some people higher than the average (i.e., exists $\omega \in\Omega$ such that $X(\omega) \ge E(X)$. 
But clearly you cannot expect that everyone is higher than the average (that is, $X \ge E(X)$). 
Actually, unless $X$ is constant, $X\ge E(X)$ is never true (i.e. there will be someone below the average). This is a direct consequence of the fact that $X\ge Y \implies E(X) \ge E(Y)$
