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?

  • $\begingroup$ $X$ is a fixed random variable, not the limit of a sequence of random variables. $\endgroup$ Oct 4, 2013 at 3:02
  • $\begingroup$ It doesn't make sense to say $X$ is defined as $\lim_{n\to\infty}X_n$ where $X_n$ has a certain distribution depending on $n$. It takes more than specifying the distributions of the random variables for an expression like $\lim_{n\to\infty}X_n$ to mean something. $\endgroup$ Oct 4, 2013 at 3:39
  • $\begingroup$ @MichaelHardy may i ask, if "It takes more than specifying the distributions of the random variables for an expression like limn→∞Xn to mean something", then what prerequisite condition of a probability definition i missed in the one above? $\endgroup$
    – athos
    Oct 4, 2013 at 6:49
  • $\begingroup$ When you write $X=\lim_{n\to\infty} X_n$, where $X_n$ are random variables, then you seem to be trying to define a random variable called $X$, rather than trying to specify a probability distribution. If you do define a random variable, then it has a probability distribution, but if you are simply trying to define a probablity distribution, then writing $X=\lim_{n\to\infty}X_n$ does not make that clear. If you mean you want the limiting distribution, rather than the limit of the sequence of random variables, that's what you should say. $\endgroup$ Oct 4, 2013 at 13:36
  • $\begingroup$ @athos : I've posted a proper answer to your question below. It appears that you are not sufficiently attending to the differences between different kinds of convergence, nor to the specifics of the details of the definition of one of them. $\endgroup$ Oct 4, 2013 at 14:06

2 Answers 2


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

  • 1
    $\begingroup$ thanks a lot. I didn't reply or accept the answer actually is because i felt quite headache reading the book and gave up. however recently i pick it up and start reading from the first page again. now i see what you are saying. $A_1, A_2, \dots \in \mathscr{F} \Rightarrow A:=\cup_i A_i \in \mathscr{F}$ indirectly allows the limit to be done inside the P be taken out. $\endgroup$
    – athos
    Apr 11, 2014 at 2:22

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 $\lim_{n\to\infty}F_n(x)=F(x)$.


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