Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $D$ be the set of all functions $F: \mathbb{R} \rightarrow \mathbb{R}$ which are nondecreasing, left-hand-side continuous and $\lim_{x \rightarrow -\infty} F(x)=0$ and $\lim_{x \rightarrow \infty} F(x)=1$. Let $d$ be a Lévy metric in $D$, that is:

$$d(F,G)=\inf \{ e >0: G(x-e)-e \leq F(x) \leq G(x+e)+e\text{ for }x\in \mathbb{R} \}\;.$$

How to prove completeness and separability of $(D, d)$ ?

I know that a sequence $(F_n)$ from $D$ is convergent to $F$ from $D$ iff

$\lim_{n\rightarrow \infty} F_n(x)=F(x)$ in each $x \in \mathbb{R}$ in which $F$ is continuous.

share|improve this question
    
What have you tried? What else do you know about weak convergence? –  Nate Eldredge Jun 28 '12 at 16:35
    
I want to do it elementary, not using probability. –  L.T Jun 28 '12 at 16:38

1 Answer 1

up vote 2 down vote accepted

Here is a sketch.

For separability: try functions of the form $$F(x) = \sum_{i=1}^n a_i 1_{(b_i, \infty)}$$ where $a_i, b_i$ are rational. (This corresponds to measures which put a rational amount of mass at each of a finite set of rational points.)

For completeness: It suffices to show every Cauchy sequence has a convergent subsequence. Given a Cauchy sequence $\{F_n\}$, use the compactness of $[0,1]$ and a diagonalization argument to extract a subsequence $\{F_{n_k}\}$ such that $F_{n_k}(x)$ converges for every rational $x$. (Equivalently, this is the compactness of $[0,1]^{\mathbb{Q}}$.) Call the limit $G : \mathbb{Q} \to [0,1]$. $G$ is nondecreasing so it corresponds to a unique nondecreasing, left-continuous $F : \mathbb{R} \to [0,1]$. (Make this correspondence precise.) Check that $F \in D$ (this will require using that the sequence is Cauchy). Now show that $F_{n_k} \to F$ in the $d$ metric.

There is a fair amount of work involved in filling in the details, but I leave it to you.

share|improve this answer
1  
The OP's space $D$ consists of left-continuous functions :) –  Byron Schmuland Jun 29 '12 at 12:25
    
Thanks for answer. I understand completness but I have still a problem why the given set of functions is dense in $D$. –  L.T Jun 29 '12 at 14:57
    
@ByronSchmuland: Oops. Fixed. –  Nate Eldredge Jun 29 '12 at 15:00
1  
@L.T: Let's call my set $E$. If you take an $F \in D$ which is continuous, it should be pretty easy to find a $G \in E$ such that $|F(x)-G(x)| < \epsilon$ for every $x$, i.e. $F,G$ are uniformly close. This implies they are also close in the $d$ metric. Now if $F$ is not continuous, the problem is that it may have jumps at irrational $x$. But you can choose $G$ to have a jump at a rational that is less than $\epsilon$ away, and such a function will be $d$-close to $F$. –  Nate Eldredge Jun 29 '12 at 15:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.