TITLE: Discontinuity points of a Distribution function
QUESTION [17 upvotes]: Possible Duplicate:
Distribution Functions of Measures and Countable Sets 

The question at hand is:
Let F be a distribution function on $\mathbb{R}$. Prove that F has at most countably many discontinuities.
My attempt at a solution:
$\textrm{F is non-decreasing by assumption}\\
F(\varphi ^-)=\lim_{t \uparrow \varphi}F(t),F(\varphi ^+)=\lim_{t \downarrow \varphi}F(t)\\
\textrm{The above limits exist and discontinuity points occur where}\\
F(\varphi^-)\neq F(\varphi)=F(\varphi^+)\\
\textrm{let (a,b] be a ﬁnite interval with n discontinuity points such that: }
\\
a<\varphi_1<...< \varphi_n < b \Rightarrow \sum_{\varphi =1}^{n}P(\varphi_k) \leq F(b)-F(a)\\
\textrm{therefore the number of discontinuity points is at most: } \frac{1}{\varepsilon }F(b)-F(a)$
As is (painfully) evident, I am just learning these concepts on my own and have little background in rigorous proof writing. I think all I have done is restrict the # of discontinuities of size $\frac{1}{\epsilon}$, and I'm not sure this does much for me.
Any help would be greatly appreciated, as always.

REPLY [40 votes]: Another approach: let $D$ be the set of points of discontinuity.  For each $x \in D$ we have $F(x-) < F(x+)$ so we can choose a rational $q_x$ with $F(x-) < q_x < F(x+)$.  Since $F$ is increasing we can check that if $x \ne y$ then $q_x \ne q_y$.  So $x \mapsto q_x$ is a one-to-one function from $D$ to $\mathbb{Q}$, and $\mathbb{Q}$ is countable, hence so is $D$.