It is well known that a cdf of a random variable in $\mathbb{R}$ has at most countably many discontinuities. How does this generalize to $n$-dimensions from the $1$-dimensional case? That is, for the $1$-dimensional case we show the following:
If $F_X$ is the cdf of a random variable $X$ in $\mathbb{R}$, then we can let $B_n := \{x : F(x^{+}) - F(x^{-}) \geq 1/n\}$. Further, since $F$ is increasing, $\lim_{x \to -\infty} F(x) = 0$, and $\lim_{x \to \infty} F(x) = 1$ we can conclude that $|B_n| \leq n$. Therefore, the set of (jump) discontinuities is $B = \bigcup_{n \geq 1} B_n$, and is at most countable.
For $n$-dimensions we define the cdf as follows: $$F_X(x_1, \dots, x_n) := \mathbb{P}(X_1 \leq x_1, \dots, X_n \leq x_n),$$
where $x_1, \dots, x_n \in \mathbb{R}$. I believe we could approach the $n$-dimensional case in either/or ways:
1. If we assume that the cdf is equipped with the Lebesgue measure $\lambda$, we could some how apply Lebesgue's Differentiation (Density) Theorem: That is, if $A \subset \mathbb{R}^n$, then a.e. point of $A$ is either $1$ or $0$, where the set of discontinuities $\mathcal{B} \subset A$ are those points of density not $0$ or $1$ (countable).
2. The set of discontinuities in $n$-dimensions is $\mathcal{B} := \mathcal{B}_1 \times \dots \times \mathcal{B}_n$. Presumably, each $\mathcal{B}_i$ are countable and have measure $0$. Hence, if we show any $\mathcal{B}_i$ has measure $0$ the entire $\mathcal{B}$ does too.
I'm not sure how to go about doing this formally though, and whether (1) or (2) is the more viable/correct approach?
