Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers 3

up vote 8 down vote accepted

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 $D$.

share|improve this answer
    
Interesting (and helpful), I had not thought of this. –  Justin May 21 '12 at 3:28

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.

share|improve this answer
    
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.)

share|improve this answer
    
Thank you for the response! –  Justin May 21 '12 at 3:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.