Find a sequence $(p_n)_{n\in N}$ of discrete density Let $F:\mathbb{R}\to [0,1]$ be a repartition function and $D=\{x\in \mathbb{R}: F(x)\neq F(x^-)\}\subset \mathbb{R}$. I have to find a sequence $(p_n)_n$ of discrete density in $\mathbb{R}$ such that $\lim_{n\to\infty}F_n(x)=F(x)\ \forall x\in\mathbb{R}\setminus D$, where $F_n$ is the repartition function associated to $p_n$, (i.e. $F_n(x)=\sum_{y\in\mathbb{R}:y\le x}p_n(y)$).

I use the fact that $\{X\le x\}=\bigcap_{n\in N}\{X\le x_n\}$. Indeed if $X(\omega)\le x$, we necessarily have $X(\omega)\le x_n$ because $x\le x_n$ and vice versa, since the sequence of events $\{X\le x_n\}$ is decreasing, $\{X\le x_{n+1}\}\subseteq\{X\le x_n\}$, so $F(x)=P(X\le x)=\lim_{n\to\infty}P(X\le x_n)=\lim_{n\to\infty}F(x_n)$. But this is a sequence of events and not of discrete density. So my question is: is it possible to use this fact to find the density sequence?
Let $F_n (t)=P\{[nX] \leq nt\}$ where $[x]$ is the largest integer not exceeding $x$ and $X$ is random variable whose distribution is $F$. Since $\frac {[nX]} n \to X$ it follows that $F_n(t) \to F(t)$ at all continuity points $t$ of $F$, i.e., at all points of $\mathbb R \setminus D$. Also the distribution of $\frac {[nX]} n $ is discrete since it takes only countable many values.
Let $F_n (t)=P\{[nX] \leq nt\}$ where $[x]$ is the largest integer not exceeding $x$ and $X$ is random variable whose distribution is $F$. Since $\frac {[nX]} n \to X$ it follows that $F_n(t) \to F(t)$ at all continuity points $t$ of $F$, i.e., at all points of $\mathbb R \setminus D$. Also the distribution of $\frac {[nX]} n $ is discrete since it takes only countable many values.