Why is a CDF right-continuous at "a" in [a,b), when property Pr(ac.f. wikipedia:Cumulative distribution function properties
"Every cumulative distribution function F is (not necessarily strictly) monotone non-decreasing (see monotone increasing) and right-continuous. "
...
"If the CDF F of X is continuous, then X is a continuous random variable; if furthermore F is absolutely continuous, then there exists a Lebesgue-integrable function f(x) such that
$$F(b)-F(a) = \operatorname{P}(a< X\leq b) = \int_a^b f(x)\,dx$$
for all real numbers a and b. The function f is equal to the derivative of F "almost everywhere", and it is called the probability density function of the distribution of X."
 A: I'm not sure I fully understand the question.
You have the CDF $F(\alpha) = P \{ \omega | X(\omega) \in (-\infty,\alpha] \}$, and since $(-\infty, a] \cup (a,b] = (-\infty, b]$, we have $ P \{ \omega | X(\omega) \in (-\infty,a] \} + P \{ \omega | X(\omega) \in (a,b] \} = P \{ \omega | X(\omega) \in (-\infty,b] \}$, from which we get $P \{ \omega | X(\omega) \in (a,b] \} = F(b)-F(a)$.
There is no choice in the matter once you let $F(\alpha) = P (X^{-1}(-\infty,\alpha])$.
As in the comment above, if $\alpha_n \downarrow \alpha$, then $(-\infty, \alpha] = \cap_n (-\infty, \alpha_n]$. It follows that $X^{-1}(-\infty, \alpha] = \cap_n X^{-1}(-\infty, \alpha_n]$, and since $P$ is a probability measure (hence bounded above), we have $P(X^{-1}(-\infty, \alpha]) = \lim_n P (X^{-1}(-\infty, \alpha_n])$. Or, in other words, $F(\alpha) = \lim_n F(\alpha_n)$.
If $\beta_n \uparrow \beta$, we have $F(\beta_n)$ is non-decreasing and bounded above by $F(\beta)$, hence $\lim_{n} F(\beta_n)$ exists (and is $\le F(\beta)$, of course).
Suppose $a<b$. Choose any $c \in [a,b)$, then $[a,b) = [a,c]\cup (c,b)$, from which we get 
$[a,b) = (\cap_n (a-\frac{1}{n},c]) \cup (\cup_n (c,b-\frac{1}{n}]) $. We have $P(X^{-1} (a-\frac{1}{n},c]) = F(c)-F(a-\frac{1}{n})$, hence $P(X^{-1} (a,c]) = F(c)-\lim_n F(a-\frac{1}{n})$, and similarly, $P(X^{-1} (c,b-\frac{1}{n}]) = \lim_n F(b-\frac{1}{n}) -F(c)$, from which it follows that $P(X^{-1}[a,b)) = \lim_n F(b-\frac{1}{n}) - \lim_n F(a-\frac{1}{n})= \lim_{b_n \downarrow b} F(b_n) - \lim_{a_n \downarrow a} F(a_n)$. If $a=b$ then $[a,b) = \emptyset$, and it is easy to see that the formula agrees in this case as well.
Again, there is no choice in the matter.
Note that the formula in the question is incorrect, as the CDF $F = 1_{[0,\infty)}$ shows, with $a=0, b=1$. Then $P(X^{-1}[0,1) ) = 1$, whereas the formula in the question above would result in zero.
