# Completeness and separability of Lévy's metric

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.

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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

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.

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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
@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