Property of cumulative distribution function I was taking the course on random variables , where I faced below property of cumulative distribution function:
$$\lim_{x\rightarrow a^+}F_X(x)=F_X(a^+)=F_X(a)\qquad\qquad a^+=\lim_{0<\epsilon\rightarrow0}a+\epsilon$$
with single addition remark that above property indicates that cdf is continuous on the right.
However this is not making any sense to me. What does that property mean? Maybe that $a^+=\lim_{0<\epsilon\rightarrow0}a+\epsilon$ needs more explanation. 
 A: In fact $\lim\limits_{\varepsilon\,\downarrow\,0} (a+\varepsilon) =a$, so there is no separate number $a^+$ that is equal to $\lim\limits_{\varepsilon \, \downarrow\,0} (a+\varepsilon) =a$.  The notation $F(a^+)$ does not mean the value of $F$ at a number called $a^+$; rather it is just a shorthand for $\lim\limits_{\varepsilon\,\downarrow\,0}F(a+\varepsilon)$, also denoted $\lim\limits_{x\,\downarrow\,a}F(x)$.
Cumulative distribution functions can be shown to satisfy the relation
$$
\lim_{x\,\downarrow\,a} F_X(x) = F_X(a)
$$
by using countable additivity of probability.  (Notice that I am distinguishing between capital $X$, the random variable, and lower-case $x$, the number that is approaching $a$ from above. Only by doing that can one understand the meaning of the expression $\Pr(X\le x)$.)
If $x>a$ then
$$
F_X(x) = \Pr(X\le x) = \Pr(X\le a\text{ or } a<X\le x)
= \Pr(X\le a)+\Pr(a<X\le x).
$$
So the problem is to prove $\lim\limits_{x\,\downarrow\,a}\Pr(a<X\le x)=0$.
Notice that $\lim\limits_{x\,\downarrow\,a}\Pr(a<X\le x)$ must exist since as $x$ decreases to $a$, $F_X(x)$ decreases and is bounded below.  Consequently that limit is the same as
$$
\lim_{n\to\infty} \Pr\left(a<X\le\frac 1 n\right).
$$
Now use countable additivity:
\begin{align}
1\ge \Pr\left( a<X\le 1 \right) & = \Pr\left( \bigcup_{n=1}^\infty a+\frac 1{n+1}<X\le a + \frac 1 n \right) \\[10pt]
& = \sum_{n=1}^\infty \Pr\left( a+\frac 1{n+1}<X\le a + \frac 1 n \right).
\end{align}
Since the series converges, its tails have to approach $0$:
$$
\Pr\left( a < X \le \frac 1 N\right) = \sum_{n=N}^\infty \Pr\left( a+\frac 1 {n+1} < X\le a +\frac 1 n \right) \to 0\text{ as }N\to\infty.
$$
A: Hopefully you're familiar with the following result: if $f(x)$ is continuous at $y$, then $\lim_{n\rightarrow\infty }f(y_n)=f(\lim_{n\rightarrow\infty} y_n)=f(y)$ for all sequences $y_n\rightarrow y$. 
It looks like this is silly notation for a similar result involving right continuous functions. Essentially the above holds for $y_n\rightarrow y$ with $y_n\geq y$ for all $n$. 
So $a^+$ is limit of any sequence going to $a$ from the right. 
A: CDF continuous. If the random variable $X$ is 'continuous' then the CDF $F_X(x)$ is a continuous function. 
Examples: (1) The CDF of $X \sim Unif(0,1)$ is $F_X(x) = x,$ for
$0 \le x \le 1$ (and $0$ for smaller values of $x,$ and $1$ for larger values of $x$).
For a continuous random variable $X$, we have $P(X = a) =0.$
Positive probability is assigned to intervals and to events
made up of countable unions and intersections of intervals. 
So $P(a < X \le b) = F_X(b) - F_X(a)$ may be positive.
In particular, if $X \sim Unif(0,1),$ then 
$$P(1/3 < X \le 2/3) = P(X \le 2/3) - F(X \le 1/3)\\ = F_X(2/3) - F_X(1/3) = 2/3 - 1/3 = 1/3.$$
(2) The CDF of the standard normal distribution is known to be
continuous, even though it cannot be written as a 'closed-form'
function.
If $X$ is standard normal, then one can determine by
numerical integration that $$P(1/3 < X \le 2/3) = 0.7475075 - 0.6305587 = 0.1169488.$$
CDF discontinuous. The issue of the value of $F_X(x)$ at a particular value $x = a$ arises only when there is a discontinuous jump at $a$. For example, if $X \sim Binom(2, 1/2),$ then $P(X = 0) = 1/4,$ and $F_X$ jumps
upward by $1/4$ at $x = 0$ (as one 'moves from left to right'). By convention, we say that $F_X(0) = 1/4$ and $F_X(-\epsilon) = 0$, for small $\epsilon > 0$.
Whether CDFs are defined to be continuous from the right or from the left is an arbitrary convention. In some countries, especially in parts of Eastern Europe and Asia,
the convention is that CDFs are continuous from the left. Then one writes $F_X(x) = P(X < x)$ instead of $F_X(x) = P(X \le x)$.
