Why should the hyperplane define an at most countable set of discontinuity points?

Question

In Durrett's book, 4th edition, on page 173, we have the following sentence:

Let $H^i_c = \{x : x_i = c\}$ be the hyperplane where the $i$th coordinate is $c$. For each $i$, the $H^i_c$ are disjoint so $D^i = \{c : P(X \in H^i_c) > 0\}$ is at most countable. It is easy to see that if $x$ has $x_i \notin D^i$ for all $i$ then $F$ is continuous at $x$. This gives us more than enough points to reconstruct $F$.

(Original image here.)

I don't get why $D^i$ is at most countable and supposedly has the discontinuity points on the $i$th-axis for the distribution function $F(x)=P(X\leq x)$, where $x \in \mathbb{R}^d$.

Edit: For example what's the problem with $P(X\in H^i_c)=1/(b-a)$ and zero otherwise. $P(X\in R^d)\geq \int_{[b,a]} 1/(b-a) \ d c = 1$

Answer 1

The author claims that at most countably many of the hyperplanes have positive probability. Suppose, by way of contradiction, that there are uncountably many such. For $n=1,2,3,\dots$ let $S_n^i = \{H_c^i|P(H_c^i)>1/n\}.$ Each of the uncountably many $H_c^i$ with positive probability belongs to at least one of the $S_n^i.$ It cannot be the case that all the $S_n^i$ are finite, because there are countably many $S_n^i$ and the countable union of finite sets is countable. But then we have infinitely many disjoint sets with probability greater than $1/n$ for some $n,$ and this is absurd, so there are at most countably many $H_c^i$ with positive probability.

The set of such hyperplanes is in one-to-one correspondence with $D^i,$ so $D_i$ is at most countable.

Answer 2

There's an entire "genre" of results about probability distributions which state that probability distributions can only have countably many of certain kinds of "discontinuities". All of them ultimately come from a general (and simple) mathematical principle which every student should have seen at least once: "an uncountable sum always diverges". I've posted a simple explanation and proof of this principle here.

From this we get the classic fact that a probability distribution has at most countably many "atoms" (singletons with positive probability), since otherwise it would have finite sets of arbitrarily large probability - not possible when the probability of the whole set is supposed to be $1$.

You can use the principle to prove your statement directly. If there were uncountably many of these disjoint hyperplanes of positive probability, we would be able to find a finite family of them whose union had arbitrarily large probability (notice we use the disjointness to be able to say that the probability of the union is the sum of the probabilities). In fact, more generally, a probability distribution cannot have an uncountable family of disjoint sets of positive probability.

In this case, however, since you're dealing with particularly simple sets (hyperplanes), you can also use another proof. Consider the random variable $X_i$, the $i$-th coordinate of $X$. The set $H^i_c$ is exactly the set $\{X_i = c\}$. So $P(X\in H^i_c)=P(X_i=c)$. Since the distribution of $X_i$ can only have countably many singletons of positive probability, only countably many of the $H_i$ can have positive probability.

The problem with your attempted counterexample is that when you write $P(X\in H^i_c)=\frac1{b-a}$, what you really mean is that the probability density function of $X_i$ is $\frac1{b-a}$ at $c$. We would still have $P(X\in H^i_c)=0$. Your idea is obviously based on the uniform distribution, and notice that even for a regular old uniform random variable $U$, we don't have $P(U=c)=\frac1{b-a}$, we have $P(U=c)=0$.