Convergence of distribution functions

Question

I am trying to prove the equivalence of the following statements:

(i) $F_n(z) \to F(z)$ as $n \to \infty$ for all $z$ at which $F$ is continuous.

(ii) $F_n(z) \to F(z)$ as $n \to \infty$ for all $z$ in a dense subset $D$ of $\mathbb{R}$

where $F_n$, $n \in \mathbb{N}$, and $F$ are distributions functions on $\mathbb{R}$.

I have found 2 questions, namely this and this, which look related to my topic.

One approach mentioned there might come in handy here, I guess.

We introduce $C:=\left\{ x\in\mathbb{R}\mid F\text{ continuous at }x\right\}$.

And if I prove that $C$ is dense in $\mathbb{R}$ I am done with one direction already. I think, one idea to start is to find for every $x\in\mathbb{R}$ a sequence $(x_k)_{k\in\mathbb{N}}$ that converges to $x$. I feel that the definition of a dense subset is now somewhere close, but I still cannot catch it.

Any help on the matter would be highly appreciated.

Answer 1

There can only be a countable number of atoms (points at which the CDF is not continuous). An atom is a point $x$ where $P(X=x)>0$. And $P(X=x)$ is called the mass of the atom at $x.$

Observe that there cannot be more than $n$ atoms with mass greater than $1/n.$ So the set $\{\mbox{atoms with mass >1/n}\}$ is finite. The set of all atoms is $\bigcup_n\{\mbox{atoms with mass >1/n}\},$ which is a countable union of finite sets and thus countable. Thus the set of non-atoms (continuity pts) has countable complement and is thus dense.

Hint for the other direction: note that a function's values on a dense subset determines its value at each point of continuity.