# Discontinuity points of a Distribution function [duplicate]

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.

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## marked as duplicate by Nate Eldredge, t.b., LVK, tomasz, WilliamAug 21 '12 at 8:44

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 $D$.

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Interesting (and helpful), I had not thought of this. – Justin May 21 '12 at 3:28
@Nate Eldredge why is $Q$ countable? – Dave ddd Apr 29 at 11:40
@Daveddd: $\mathbb{Q}$ is the set of rational numbers. I suppose you have seen a proof that this is countable, as it's a fundamental fact; if somehow not, see math.stackexchange.com/questions/659302/… – Nate Eldredge Apr 29 at 13:43

You can finish off this argument by noting that any discontinuity is of "size" at least $1/n$ for some $n\in\mathbb N$, and thus the set of discontinuities is a countable union of finite sets, which is itself countable.

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Thank you for your help. I was closer than I thought. – Justin May 21 '12 at 3:28

Here is the proof from Folland Real Analysis Theorem 3.23 (p. 101):

• $F$ is increasing, hence intervals $(F(x-),F(x+))$, $x \in \mathbb{R}$ are mutually disjoint.
• For $|x| < N$, the interval $(F(x-),F(x+))$ lies in $(F(-N),F(N))$.
• These two imply $$\sum_{|x| < N} [F(x+) - F(x-)] \le F(-N) - F(N) < \infty$$ This sum over an uncountable set is defined on p. 11. (Essentially it is the supremum of the sum over all of its finite partial sums.) Being finite implies that the number of nonzero terms of the sum is countable.
• This means that the set $D_N := \{ x \in (-N,N) : F(x-) \neq F(x+) \}$ is countable.
• The whole set of discontiuities is $\bigcup_{N \in \mathbb{N}} D_N$ which is countable (since it is the countable union of countable sets.)

The interesting trick here is that you map the discontinuities to a collection of disjoint intervals, and then bound the total length of those intervals. (Very useful in general in proving that a set is countable.)

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Thank you for the response! – Justin May 21 '12 at 3:29