EDIT: Can cdf have a set of continuity points which is a.s. different from $\emptyset$?

Question

It can be easily proven that a cumulative distribution function (cdf) $$F(x)=\mathbb{P}\left(X\le x\right)\hspace{0.3cm}\text{for }-\infty<x<+\infty$$ is right-continuous $\left(\text{that is }F(x)=F(x^+)\right)$, but NOT left-continuous $\left(\text{that is }F(x)\neq F(x^-)\right)$.

Given this, could it be even possible that there exists a set of continuity points of $F$ $$D=\{x: F(x^-)=F(x)\}$$ which is a.s. different from $\emptyset$? (That is, considering $\mathbb{P}$ as the probability measure corresponding to the cdf $F$, $\mathbb{P}(D\neq\emptyset)=0$).
Clearly, for $D$ it would be sufficient to specify the left-continuity condition, since, as above-stated, right-continuity is already granted by definition of cdf.


However, the question is: could one have that such a D is a.s. different from $\emptyset$ if cumulative distribution functions are known to be not left-continous at any points at all?

Answer 1

The set $D^{\complement}$ of discontinuity points is a countable set.

This becomes clear if we write $D^{\complement}=\bigcup_{n=1}^{\infty}E_n$ where $E_n=\{x\in\mathbb R\mid F(x)-F(x^-)>\frac1{n}\}$.

Observe that for every $n$ the fact $P(E_n)\leq1$ tells us that the set $E_n$ has less than $n$ elements.

Answer 2

$D$ is precisely the set of points at which $F$ is continuous.

Any monotone function $F:\mathbb R \to \mathbb R$ has at most countable number of discontinuity points. Hence $D^{c}$ is necessarily at most countable. Every countable set has Lebesgue measure $0$. Hence $D^{c}$ has measure $0$. Since the measure of the real line is $\infty$ this implies that the measure of $D$ is necessarily $\infty$.