Q: Is the set of continuity points of any high dimensional distribution function dense in R^n? Given any distribution function $F(x_1,...,x_n): R^n \rightarrow [0,1]$, we know it's increasing in each component and right continuous.
My ultimate goal is: trying to deduce that the set of continuity points of $F$ is dense in $R^n$.
My attempt: to this end, I think it suffices to deduce that the discontinuity set of $F$ has Lebesgue measure $0$, so that, by a basic result, its complement (i.e. the continuity set of $F$) is dense in $R^n$. To proceed with this route, I read this post. However, I do not understand why he sets $\delta$ to $0$ and also the last step: "a discontinuity of $g$ is a discontinuity of $\phi_t$" (I think the result is based on a fixed $t$, so I don't know how he generalized it). Even if his approach is correct, I also do not know if it can be generalized to $R^n$.
I much appreciated any effort in continuing with my attempt or showing any other good approaches.

A: Let $F_{X,Y},F_X, F_Y $ be the distribution function of random variable $ (X,Y),X,Y $ respectively and $ C_{X,Y},C_X,C_Y $ be the set of continuous points of $F_{X,Y},F_X, F_Y $ respectively. Firstly, introduce following inequlities:
\begin{align*}
 |F_{X,Y}(x_1,y_1)&-F_{X,Y}(x_2,y_2)|\le F_{X,Y}(x_M,y_M)-F_{X,Y}(x_m,y_m)\\
 &\le F_X(x_M)-F_X(x_m)+ F_Y(y_M)-F_Y(y_m), \tag{1}\\
  x_M&=\max(x_1,x_2),\qquad x_m=\min(x_1,x_2). 
\end{align*}
From (1),
\begin{equation*}
  C_X\times C_Y \subset C_{X,Y}  .
\end{equation*}
Furthermore, for $\overline{C_{X,Y}} $, the closure of $C_{X,Y}$,
\begin{equation*}
 \overline{C_{X,Y}} \supset \overline{C_X\times C_Y} 
                        =\overline{C_X}\times \overline{C_Y}
                       = \mathbb{R}\times \mathbb{R}.  
\end{equation*}
This means $C_{X,Y}$ is dense in the $\mathbb{R}^2$.

A: This is good to know—thanks for your hint.
Assume the fact @JGWang mentioned, we get: if $F_{X,Y}$ is discontinuous at $(x_0, y_0)$, then either $F_X$ is discontinuous at $x_0$, or $F_Y$ is discontinuous at $y_0$  (we cannot directly conclude from here that the discontinuity set of $F_{X,Y}$ is countable because $F_{X,Y}$ may be discontinuous at $(x_0, y)$ for the same $x_0$ but with uncountably many $y$'s at which $F_Y$ is continuous).
However, we may use it to show $\lambda (D)=0$ ($D$ being the discontinuity set of $F_{X,Y}$; $\lambda$ being the Lebesgue measure on $R^2$): By definition and Fubini's theorem,
$$ \lambda(D)=\int \int 1_{D}(x,y) dxdy. $$
For any continuity point $y$ of $F_Y$, $1_{D}(x,y)\equiv 1_{D_y}(x)=1$ iff $F_X$ is discontinuous at $x$, so $1_{D_y}(x)=0$ Lebesgue almost everywhere due to the countability of such points. Hence,
$$ \lambda(D)=\int_{y: F_Y\, \text{discontinuous at}\, y} \int 1_{D}(x,y) dxdy. $$
Then we may apply Tonelli's theorem to swap the order of integrals. After that, the inner integral will be $0$ due to countability of $\{y: F_Y\, \text{discontinuous at}\, y\}$. Hence, $\lambda(D)=0$.
The same logic applies generally to $R^n$.
