Continuity of a multivariate CDF at a point I want to show that if $(X_1,X_2,\dots,X_n)$ is an $n$-variate random variable, then its CDF is continuous at a point $\vec a$ iff $$P\left(\bigcup_{k=1}^n\{X_k= a_k, X_j\le a_j\:\forall\:j\ne k\}\right)=0$$
I don't understand how to show this result. If I use the usual definition of continuity, then it would be very difficult to handle balls (because they are spherical, and we are working with CDFs). If we consider rectangles instead of spherical balls, then $F$ is continuous at $\vec a$ iff for every $2\varepsilon$ sided square centered at $F(\vec a)$, we have a $2\delta$ sided square centered at $\vec a$ whose $F$-image lies in the former square.
And I don't know what to do then.
 A: Write $\mathbf{X}=(X_1,\ldots, X_n)$ for an arbitrary random variable with distribution function $F.$  By definition, the value of $F$ at any point $\mathbf a = (a_1,a_2,\ldots, a_n)$ is
$$F(\mathbf a ) = \Pr(X_1\le a_1, X_2\le a_2, \ldots, X_n\le a_n) = \Pr(\max_i(X_i-a_i)\le 0).$$
For any point $\mathbf{a}\in\mathbb{R}^n$ and any number $\delta$ define the events
$$\mathcal{M}(\mathbf a, \delta) = \{\mathbf X\mid \max_i(X_i - a_i) \le \delta\}.$$

Here, in $n=2$ dimensions, is shown $\mathcal{M}(\mathbf a, 0)$ relative to $\mathbf a.$  It is a buffer of width $\delta$ around all points to the left and below $\mathbf a.$
The demonstration below will analyze the region between two such buffers: an outer buffer and an inner buffer, as shown.

These buffers are relevant to the question because the boundary of $\mathcal{M}(\mathbf a, 0),$ written $\partial\mathcal{M}(\mathbf a, 0),$ is the event described in the question and their set difference is a buffer of radius $\delta$ around that event.

Let
$$F_{\mathbf a}(\delta) = \Pr(\mathcal{M}(\mathbf a, \delta)) = F\left(\mathbf a + \delta(1,1,\ldots, 1)\right).$$
Since
$$\mathcal{M}(0) = \bigcap_{n=1}^\infty \mathcal{M}\left(\mathbf a, \frac{1}{n}\right),$$
computing probabilities gives
$$F(\mathbf a) = \Pr(\mathcal{M}(0)) = \inf_n\, \Pr(\mathcal{M}\left(\mathbf a, \frac{1}{n}\right) = \inf_n\, F_{\mathbf{a}}\left(\frac{1}{n}\right) = \lim_{n\to \infty} F_{\mathbf a}\left(\frac{1}{n}\right).\tag{*}$$
Similarly, we can fill up the interior of $\mathcal{M}(\mathbf a, 0)$ with a union of sets that are disjoint from the boundary, so that their probabilities (of the interior and of the boundary) add:
$$\mathcal{M}(\mathbf a, 0)  = \partial \mathcal{M}(\mathbf a, 0)\, \cup\, \bigcup_{n=1}^\infty \mathcal{M}\left(\mathbf a, -\frac{1}{n}\right)$$
implies
$$F(\mathbf a) = \Pr\left(\partial \mathcal{M}(\mathbf a, 0)\right) + \lim_{n\to \infty} F\left(\mathbf a, -\frac{1}{n}\right).\tag{**}$$
In these terms, the statement to be proven is that

$F$ is continuous (as a function) at $\mathbf a$ if and only if $\Pr\left(\partial \mathcal{M}(\mathbf a, 0)\right) = 0.$

Forward direction
Suppose $F$ is continuous at $\mathbf a.$  This means that for any $\epsilon \gt 0,$ there is some ball (say of radius $R(\epsilon)$ around $\mathbf{a},$ written $B(\mathbf a, R(\epsilon))$) for which the image of $F$ is close to $F(\mathbf a):$ namely,
$$F\left(B(\mathbf a, R(\epsilon))\right) \subseteq [F(\mathbf{a})-\epsilon/2, F(\mathbf{a})+\epsilon/2].$$
When $\delta \le R(\epsilon)/\sqrt{n},$ the rectangle $[a_1-\delta,a_1+\delta]\times \cdots \times [a_n-\delta,a_n+\delta]$ lies entirely within this ball.  But, because $F$ cannot decrease when any of its arguments are increased, the largest value attained by $F$ on this rectangle is $F_{\mathbf a}(\delta)$ while the smallest is $F_{\mathbf a}(-\delta).$  Since
$$\partial\mathcal{M}(\mathbf a, 0) \subset \mathcal{M}(\mathbf a, \delta) \setminus \mathcal{M}(\mathbf a, -\delta),$$
we deduce
$$\Pr\left(\partial\mathcal{M}(\mathbf a, 0) \right) \le F_{\mathbf a}(\delta) - F_{\mathbf a}(-\delta) = F(\mathbf a) + \epsilon/2 - (F(\mathbf a) - \epsilon/2) = \epsilon.$$
The only probability less than or equal to all positive numbers $\epsilon$ is $0,$ which completes the demonstration of the forward direction.
Reverse direction
Given $\epsilon \gt 0,$ $(*)$ shows there exists an integer $n_+\gt 0$ for which $F_\mathbf{a}\left(\frac{1}{n_+}\right) - F(\mathbf a) \lt \epsilon.$  Similarly, assuming $\Pr\left(\partial\mathcal{M}(\mathbf a, 0) \right),$ $(**)$ shows there exists an integer $n_{-} \gt 0$ for which $F(\mathbf a) - F_\mathbf{a}\left(-\frac{1}{n_{-}}\right)  \lt \epsilon.$
Choosing any $\delta \gt 0$ smaller than $1/n_+$ and $1/n_{-}$  assures the values of $F$ on $B(\mathbf a, \delta)$ are within $\epsilon$ of $F(\mathbf{a}).$  Since $\epsilon\gt 0$ was arbitrary, this establishes the continuity of $F$ at $\mathbf a,$ QED.
