How do I show that the sum of two random variables is random variable? How do I prove the following?

If $X$ and $Y$ are random variables on a probability space $(\Omega, F, \mathbb P)$, then so is $X+Y$.

The definition of a random variable is a function $X: \Omega \to \mathbb R$, with the property that $\{\omega\in\Omega: X(\omega)\leq x\}\in F$, for each $x\in\mathbb R$.

Furthermore, how to approach $X+Y$ and $\min\{X, Y\}$?
 A: There are a number of ways to do it.  A standard trick for proving things like this is by noticing that $\{X+Y\leq x\}^c=\{X + Y > x\} = \displaystyle \bigcup_{r \in \mathbb{Q}} \{X > r\} \cap \{Y > x - r\} $, and that showing that this is in $F$ is enough to show that $X + Y$ is measurable.  Then use the properties of $X$, $Y$, and $\sigma-$algebras to deduce that this set is measurable.
A: In fact, a random variable is a measurable function from $\Omega$ to $\mathbb{R}$.
\begin{align}
& \{X+Y>x\}=\{X>x-Y\}=\bigcup_{q\in \mathbb{Q}}\{X>q>x-Y\} \\[10pt]
= {} & \bigcup_{q\in \mathbb{Q}} (\{X>q\ \}\cap\{Y>x-q\}) = \bigcup_{q\in \mathbb{Q}}(\{X\le q\ \}^c \cap \{Y\le x-q\}^c)
\end{align}
Since $ \{X\le q\ \}^c\in \mathcal{F}\ ,\  \{Y\le x-q\}^c\in \mathcal{F} $, $\mathbb{Q}$ is countable and $\mathcal{F}$ is a $\sigma$-field, we obtain $\{X+Y>x\}=\{X+Y\le x\}^c\in \mathcal{F}$. Then
$$ \{\omega:X(\omega)+Y(\omega)\le x\}\in \mathcal{F}$$
$$ \{\omega:\min\{X(\omega),Y(\omega)\}\le x\}=\{\omega:X(\omega)\le x\}\bigcup\{\omega:Y(\omega)\le x\}\in \mathcal{F}$$
By definition, $X+Y$, and $\min\{X,Y\}$ are both random variables.
