TITLE: Show that every finite family of random variables is tight.
QUESTION [5 upvotes]: A family of random variables $(X_i)_{i \in \mathcal{I}}$ is said to be $tight$ if for every $\epsilon>0$ there exist a compact set $K_\epsilon$ such that 
$$\displaystyle\sup_{i \in \mathcal{I}}\mathbb{P}(X_i\notin K_{\epsilon})<\epsilon$$
Show that every finite family of random variables is tight.
So I tried to star with only one random variable, say $X$. So, I have that 
$\mathbb{P}(X\notin K_\epsilon)=\mathbb{P}(\{\omega : X(\omega) \notin K_\epsilon \})$
And I get stuck right there. I don't know how to construct my compact set $K_\epsilon$ because I don't know how to work with the event $\{\omega : X(\omega) \notin K_\epsilon \}$. Any help?

REPLY [3 votes]: This isn't true in general; some (fairly light) assumptions are necessary on  the topology of your probability space $\Omega$, and your family of events $\mathcal{F}$.  
For instance, suppose I let $\Omega = [0,1]$ with the discrete topology, $\mathcal{F}$ to be Lebesgue measurable sets,  and $\mathbb{P}$ to be the Lebesgue measure.  Under the discrete topology, only finite sets are compact, so for any compact set $K$, $\mathbb{P}(X \notin K) = 1$, implying that $\mathbb{P}$ isn't tight.  
Typically, we assume that $\Omega$ is $\sigma$-compact or (sometimes) Polish and take $\mathcal{F}$ to be the Borel sets.  Most commonly, $\Omega$ is $\sigma$-compact and $\mathcal{F}$ is the Borel sets, in which case we can use the solution David posted.