I'm reading Bernt Oksendal's "Stochastic Differential Equations" and this is one of the exercise: 2.2 a) (iii).
Let $X:\Omega \rightarrow \mathbb{R}$ be a randome variable. The distribution function $F$ of $X$ is defined by
$$F(x) = P[X\leq x]$$
a) prove that $F$ has the following properties:
(iii) $F$ is right-continuous, i.e. $F(x) = \lim_{\substack{h\rightarrow 0\\h>0}}F(x+h)$.
I'm wondering, if $X$ is defined as:
$$X = \lim_{n\rightarrow \infty}X_n,$$
where for $X_n$ $$\text{F}_{X_n}(x) = \left\{
\begin{array}{l l}
0 & \quad \text{if } x \leq 0 \\
n\cdot x & \quad \text{if } 0<x<1/n\\
1 & \quad \text{if } 1/n \leq x
\end{array} \right.$$
then wouldn't that $$F(0) = 0,$$ and $\forall \epsilon > 0$, $\exists n_\epsilon$, s.t. $\forall n>n_\epsilon$, $$F_{X_n}(\epsilon) =1 \text{ ?}$$
added after reading @MichaelHardy 's comment.
I think the question was not put clearly, due to my superficial understanding of probability theory.
Now let me try again. My real doubt is:
The assertion "distribution function $F$ is right-continuous" from "Stochastic Differential Equations" exercise 2.2 a) (iii) actually means:
it's not possible to define a random variable $ X:\Omega \rightarrow \mathbb{R}$, such that its distribution function fulfills:
$$F_X(x) = \left\{
\begin{array}{l l}
0 & \quad \text{if } x \leq 0 \\
1 & \quad \text{if } x > 0
\end{array} \right.$$
I'd like to ask: why?
Best Answer
The third axiom of probability implies that if $A_1 \supset A_2 \supset A_3 \supset \cdots $ is a telescoping countable sequence of sets, then $$P\left(\lim_{n\to \infty} A_n \right) = \lim_{n\to \infty} P(A_n).$$ Apply this to the sequence of sets $A_n = \{X \leq x + x_n\}$ where $x_1>x_2>x_3 > \cdots$ is a monotone decreasing positive sequence with limit $0$ as $n \to \infty$. The limit of the $A_n$'s is thus $\{X \leq x\}$ and so we have that $$P\left(\lim_{n\to \infty} A_n \right) = P\{X \leq x\} = F_X(x) = \lim_{n\to \infty} P\{X \leq x + x_n\} = \lim_{n\to \infty} F_X(x+x_n),$$ that is, $$F_X(x) = \lim_{n\to \infty} F_X(x+x_n)~\text{where}~ x_n \downarrow 0 ~ \text{as}~ n \to \infty.$$