why distribution function is right continuous?

Question

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?

Answer 1

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, $$F_X(x) = \lim_{n\to \infty} F_X(x+x_n)~\text{where}~ x_n \downarrow 0 ~ \text{as}~ n \to \infty.$$

Answer 2

The question as posted is at best unclear. It superficially appeared that the question was supposed to be about the distribution of a random variable $X$ defined by $X=\lim_{n\to\infty}X_n$, when there was not enough information given about the joint distribution of $(X_n : n=1,2,3,\ldots)$ for that to define a random variable $X$. But in comments under the question, it appears that the question was intended to be about the limit of distributions rather than about a limit of random variables; thus $\lim_{n\to\infty} F_{X_n}(x)$ rather than $F_{\lim_{n\to\infty}X_n}(x)$.

In the expression $\lim_{n\to\infty}X_n$ one would probably mean almost sure convergence, although one might have in mind convergence in probability. In the expression $F_{\lim_{n\to\infty}X_n}(x)$, one would normally mean convergence in distribution.

To say that a sequence of probability distributions on the reals converges to a particular distribution is equivalent to saying that the sequence of cumulative distribution functions converges EXCEPT at points where the c.d.f. of the limiting distribution is discontinuous. For example, suppose the $n$th distribution, with c.d.f. $F_n$, assigns probability $1$ to the set containing just the one point $1/n$. Then the limiting distribution $F$ assigns probability $1$ to $0$, and has a discontinuity at $0$. Notice that $F_n(0)=0$ but $F(0)=1$. Thus $\lim_{n\to\infty}F_n(0)\ne F(0)$, so $\lim_{n\to\infty}F_n\ne F$ pointwise, but $\lim_{n\to\infty}F_n = F$ in the relevant sense. For every value of $x$ except the one where $F$ is discontinuous, it is true that $\lim_{n\to\infty}F_n(x)=F(x)$.