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?
 A: 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,
$$
F(x) = P(E) = \lim_n P(E_n) = \lim_n F(x_n) = F(x_0).
$$
$\blacksquare$
