Let $(\Omega, \mathcal{F}, \mathbb P)$ and $c \in \mathbb R$ so that $\mathbb P(X = c) > 0$
Show that the distribution function $F_{X}$ of $X$ has a point of discontinuity at $c$.
My ideas:
Since $F_{X}$ is left-continuous by definition we need to look at the right-hand side. Let $\epsilon > 0$
Note that $F_{X}(c+ \epsilon)=\mathbb P(X \leq c + \epsilon)=\mathbb P(X<c+\epsilon)+\mathbb P(X=c+\epsilon)$ and then I want to introduce $\delta \in (0, \epsilon)$ so that $\mathbb P(X \leq c + \epsilon)\geq\mathbb P(X<c+\delta)+\mathbb P(X=c+\epsilon)$ and then letting $\epsilon \to 0$ we have $P(X<c+\delta)+\mathbb P(X=c+\epsilon)\to P(X < c + \delta)+P(X=c) >P(X < c + \delta)\geq P(X\leq c)$
I am unsure about the above, because since I have let $\epsilon \to 0$, does that mean by definition of $\delta$ that $\delta \to 0$?
I am also struggling to find a case where the converse does not hold, i.e. if $F_{X}$ has a point of discontinuity at $c$ then $\mathbb P (X=c)\leq0$.
You have that $F_X(x) = \Bbb P(X \leq x)$. So $F_X$ is actually right-continuous instead of left-continuous, since if $x_n \downarrow x$ then $[X \leq x] = \bigcap_{n} [X \leq x_n]$ implies that $$\Bbb P(X \leq x) = \Bbb P\left(\bigcap_n [X \leq x_n]\right) = \lim_{n \to +\infty} \Bbb P(X \leq x_n) = \lim_{n \to +\infty} F_X(x_n).$$It follows that $F_X(c^+) \doteq \lim_{x \to c^+}F_X(x) = F_X(c)$, but you can't really say anything about the other limit. It follows from this that $\Bbb P(a< X \leq b) = F_X(b)-F_X(a)$, and so $$\Bbb P(X = c) = F_X(c) - \lim_{x \to c^-}F_X(x).$$Then $\Bbb P(X = c)=0$ if and only if $F_X(c) = \lim_{x\to c^-}F_X(x)$, which in view of the above, is equivalent to $F_X$ being continuous at $c$.