Prove $\min\{X(\omega)|\,\omega\in\Omega\} \le \mathbb{ E}[X] \le \max \{X(\omega)|\,\omega\in\Omega\}$ I'm not sure how to prove that 
 $$\min\{X(\omega)|\,\omega\in\Omega\} \le \mathbb{ E}[X] \le \max \{X(\omega)|\,\omega\in\Omega\}$$
With $\mathbb{E}[X]$ as the expected value of $X$and $$\sum_{i=1}^k r_i \mathbb{P}(X = r_i)$$
I know $$\mathbb{E}[X] = r_1\mathbb{P}(X = r_1) + r_2\mathbb{P}(X = r_2) +\cdots + r_k\mathbb{P}(X = r_k)$$ based on the above theorem, and thus $$r_2 \ge r_1\,\,,\,\, r_3 \ge r_1,\cdots,\,\, r_k \ge r_1$$ and that $\sum_{i=1}^k \mathbb{P}(X = r_i) = 1$. 
 A: How about
$$\mathbb{E}[X] = \sum_{i=1}^k r_i \mathbb{P}(X = r_i) \leq \sum_{i=1}^k \max_{\omega \in \Omega}\{X(w)\}\cdot \mathbb{P}(X = r_i) $$
$$= \max_{\omega \in \Omega}\{X(w)\}\sum_{i=1}^k \mathbb{P}(X = r_i) = \max_{\omega \in \Omega}\{X(w)\} $$
And the same approach for the other inequality.
A: Let $r_{min}, r_{max}$ be the minimal/maximal values respectively. Then we have that:
$$
\mathbb{E}(X)=\sum_{i=1}^n r_i \mathbb{P}(X=r_i) \leq \sum_{i=1}^n r_{max} \mathbb{P}(X=r_i)= \\
r_{max} \sum_{i=1}^n \mathbb{P}(X=r_i)=r_{max}
$$
Can you try the other part for the minimum?
A good way to think of the expected value is as some sort of average. And the average is in between the minimal and maximal value.
A: $\min\{X(\omega)|\,\omega\in\Omega\}$ and $\max \{X(\omega)|\,\omega\in\Omega\}$ are just numbers. They're not like $\min\{X_1, X_2\}$ whose value depends on $\omega$. In $\min\{X(\omega)|\,\omega\in\Omega\}$, the $\omega$'s are all given. 
Anyway,
Pf:
$\forall \omega \in \Omega$, 
$$\min \{X(\omega)|\,\omega\in\Omega\} \le X(\omega) \le \max \{X(\omega)|\,\omega\in\Omega\}$$
By monotonicity of expectation, we have
$$E[\min \{X(\omega)|\,\omega\in\Omega\}] \le E[X(\omega)] \le E[\max \{X(\omega)|\,\omega\in\Omega\}]$$
$$\to \min\{X(\omega)|\,\omega\in\Omega\} \le E[X(\omega)] \le \max \{X(\omega)|\,\omega\in\Omega\}$$
QED

We can even extend this to $A \in \mathscr F$
$$\min \{X(\omega)|\,\omega\in A\} \le X(\omega)1_A(\omega) \le \max \{X(\omega)|\,\omega\in A\}$$
where $1_A(\omega) = 1$ for $\omega \in A$ and 0 otherwise.
By monotonicity of expectation, we have
$$E[\min \{X(\omega)|\,\omega\in A\}] \le E[X(\omega)1_A(\omega)] \le E[\max \{X(\omega)|\,\omega\in A\}]$$
$$\to \min\{X(\omega)|\,\omega\in A\} \le E[X(\omega)1_A(\omega)] \le \max \{X(\omega)|\,\omega\in A\}$$
