discontinuity at an isolated point of support of distribution Let $\text F$ be a distribution function(i.e. $\text F$ is non decreasing, right continuous and $\lim_{x\rightarrow \infty}\text F(x)=1$ and $\lim_{x\rightarrow{-}\infty}\text F(x)=0$ and the support of the distribution function $\text F$ is the set $\text S_F=\{x\in \mathbb R$ such that $\text F(x+\epsilon)-\text F(x-\epsilon)>0$ for all $\epsilon >0\}$.
We need to show that if $x$ is an isolated point of $\text S_F$ then $F$ must be discontinuous at $x$.
I am not able to do it. Any type of help will be appreciated. Thanks in advance.
 A: Let $x_0$ be an isolated point of $S_F$ . Then there exists a deleted neighbourhood of $x_0$ ,$ B(x_0,\eta) = (x_0-\eta, x_0+\eta ) \setminus \{x_0\} $, such that  $ B(x_0,\eta) \subset S_F^c $. 
Let $ y \in B(x_0,\eta) $. Assume $y < x_0$ , and $F(y) = c_0$.
Since $y \not\in  S_F$ ,  there must exist a $\delta_y > 0 $ , such that $ F(y +\delta_y) - F(y  - \delta_y ) = 0 $.
In other words , $ F $ is constant on the interval $ ( y - \delta_y , y + \delta_y ) $. Clearly , $ y +\delta_y  \leq x_0 $, else $x_0$ would not belong to $S_F$. 
Hence $x_0 \ge  s := sup\{ x : F(x) = c_0 \} $.
But if $s < x_0$, F has a jump at $s$, which contradicts the fact that $s \not\in S_F$.
Hence,
$$ x_0 = sup\{ x : F(x) = c_0 \} $$
Let $ z \in B(x_0,\eta) $. Assume $ z > x_0$ , and $F(z) = c_1$. Using a similar argument as above , we can show that 
$$ x_0 = inf\{ x : F(x) = c_1 \} $$
By right continuity of $F$ , the infimum is attained , and $ F(x_0) = c_1$.
If $c_1 = c_0 $, then $F$ would be constant in a neighbourhood of $x_0$ , and this would contradict the fact that $x_0 \in S_F$.Hence $F$ is discontinuous at $x_0$.
A: Note that the complement of $S_F$ consists of points $x$ such that $F$ is constant on some neighborhood of $x$. [There exists $\epsilon > 0$ such that $F(x+\epsilon) = F(x- \epsilon)$. Then since $F$ is non-decreasing, $F$ is constant on $[x-\epsilon, x+\epsilon]$.]
If $x$ is an isolated point of $S_F$, then there exists $\delta > 0$
such that $(x-\delta, x+\delta) \setminus \{x\} \subseteq S_F^c$.
I believe this implies that $F$ is constant on the two intervals $(x-\delta, x)$ and $(x, x+\delta)$. Combining this with the fact that $x \in S_F$ should imply that $F$ is discontinuous at $x$.
