On the proof of right-continuity of the distribution function In today's statistics class, we saw properties of the distribution function, i.e. defined by $F(x) = P(X\leq x)$ for a random variable $X$. One of these was:

$F(x)$ is right continuous.

The proof was:
Let $E_n$ be a decreasing sequence of events s.t. $\cap_{i=1}^{\infty} E_i = E$, with $E = \{X\leq x_0\}$ Then
$$
F(x_0) = P(X\leq x_0) = P(E) = P(\cap_i E_n) = \lim_i P(E_i) = F(x_0^{+}).
$$
Surely I must be missing something since I don't see how she jumped from the third equality to the next one. Could you tell me why those equalities in the proof are true?

First of all, let's write the desired result mathematically: $F$ is right continuous-means really $F(x) = F(x^+)$ for all $x$, where
$$
F(x^{+}) = \lim_{y\to x^+} F(y).
$$
I'm going to assume you are familiar with the axioms of probability.
To prove this result you need a couple of preliminary properties of the probability measure. For, let $(\Omega, \mathcal F, P)$ be the usual probability triple.
Property 1. For any two given sets $E_i, E_j$, s.t. $E_i\subseteq E_j$, then $P(E_j\cap E_i^c) = P(E_j) - P(E_i)$.
Proof. Since $E_j$ includes $E_i$, then $E_j = E_i \cup (E_j\cap E_i^c)$. This union forms a partition of $E_j$, thus $P(E_j) = P(E_i) + P(E_j\cap E_i^c)$ and the result follows.
$\blacksquare$
Property 2. If $E_1,E_2,\ldots$ is an expanding sequence of sets in $\mathcal F$, that is $E_n\subseteq E_{n+1}$ for all $n$, and $E = \cup_{n=1}^\infty E_n$, then $P(E) = \lim_{n\to\infty} P(A_n)$
Proof. We can write $E = E_1 \cup (E_2\cap E_1^c)\cup (E_3\cap E_2^c)\cup\cdots(E_n\cap E_{n-1}^c)\cdots$, which is again a disjoint union (I leave this as an exercise to you). By one of the axioms of probability,
\begin{align*}
P(E) &= P(E_1) + P(E_2\cap E_1^c) + P(E_3\cap E_2^c) + \cdots\\
&= P(E_1) + P(E_2)- P(E_1) + P(E_3) - P(E_2)+\cdots (\text{using Property 1})\\
&= \lim_n P(E_n).
\end{align*}
$\blacksquare$
Property 3. If $E_1,E_2,\ldots$ is a contracting sequence of sets in $\mathcal F$, that is $E_{n+1}\subseteq E_n$, for all $n$, and $E = \cap_n E_n$, then $P(E) = \lim_n P(E_n).$
Proof. If $E = \cap_{n=1}^\infty E_n$, then on of the DeMorgan laws imply $E^c = \cup_n E_n^c$. Now $E_{n+1}\subseteq E_n$, hence $E_n^c\subseteq E_{n+1}^c$. Thus the sets $E_n^c$ form an expanding sequence so by Property 2 $\lim_n P(E_n^c) = P(E^c)$, that is $1-\lim_n P(E_n) = 1 - P(E)$ and the result follows.
$\blacksquare$
Now the proof of the desired result is essentially there.
Proof. Let $x_1,x_2,\ldots$ be a sequence of real numbers striclty decreasing to $x_0$; that is $x_n$ approach $x_0$ from above. Furthermore, let $E_n = \{X\leq x_n\}$. The $A_n$ form a contracting sequence whose limit, i.e. the intersection is $A =\{X\leq x_0\}$. To show that $\cap A_n = \{X\leq x_0\}$ reason as follows. If $\omega\in \{X\leq x_n\}$ for all $n$, then since $x_n\to x_0$, $\omega\in \{X\leq x_0\}$. Conversely, if $\omega\in \{X\leq x_0\}$, then since $x_0\leq x_n$ for all $n$, $\omega\in \{X\leq x_n\}$ for all $n$. Thus by Property 2,
$$
F(x) = P(E) = \lim_n P(E_n) = \lim_n F(x_n) = F(x_0).
$$
$\blacksquare$
