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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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3 Answers 3

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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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  • $\begingroup$ Interesting (and helpful), I had not thought of this. $\endgroup$
    – Justin
    May 21, 2012 at 3:28
  • $\begingroup$ @Nate Eldredge why is $Q$ countable? $\endgroup$
    – Dave ddd
    Apr 29, 2016 at 11:40
  • $\begingroup$ @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/… $\endgroup$ Apr 29, 2016 at 13:43
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    $\begingroup$ @Anoldmaninthesea.: Sure, you can apply the same technique to get a one-to-one function from $D$ to the set $\mathbb{I}$ of irrationals. That will prove the cardinality of $D$ is no larger than that of $\mathbb{I}$, i.e. $|D| \le |\mathbb{I}|$. That is completely true, but not useful; you already knew it. (Hint: I never said the function would be onto...) $\endgroup$ Jun 7, 2017 at 14:27
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    $\begingroup$ @Atom: Certainly, if you like. Enumerate the rationals in your favorite way, and for each $x$ let $q_x$ be the lowest-numbered rational in the interval $(F(x+), F(x-))$. $\endgroup$ Feb 17, 2022 at 8:16
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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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  • $\begingroup$ Thank you for the response! $\endgroup$
    – Justin
    May 21, 2012 at 3:29
  • $\begingroup$ passerby, can you elaborate on the 3rd point? Why to be finite the number of nonzero terms must be countable? Thanks ;) $\endgroup$ Jun 7, 2017 at 14:42
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    $\begingroup$ $$ \sum_{|x| < N} [F(x+) - F(x-)] \le F(-N) - F(N) < \infty $$. Hi community. I didn't undestand why this sum is countable. Is there any way to write this in step by step? $\endgroup$
    – Silvinha
    Mar 18, 2021 at 1:48
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    $\begingroup$ Me too. I have the same doubt as @Silvinha $\endgroup$
    – user901100
    Mar 18, 2021 at 1:50
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    $\begingroup$ @Silvinha: This is probably a lemma in Folland where uncountable sums are defined. Let $A$ be the set of all nonzero terms, and $A_n$ the set of terms that are greater than $1/n$, so $A = \bigcup_{n=1}^\infty A_n$. A countable union of finite sets would be countable, so if $A$ is uncountable then some $A_n$ is infinite. If you sum infinitely many terms, all greater than $1/n$, you get infinity. $\endgroup$ Feb 17, 2022 at 8:20
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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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  • $\begingroup$ Thank you for your help. I was closer than I thought. $\endgroup$
    – Justin
    May 21, 2012 at 3:28
  • $\begingroup$ @Justin $~~~~~$ $\endgroup$
    – MathMan
    Oct 10, 2019 at 17:07
  • $\begingroup$ Could you please explain why $\textrm{The above limits exist and discontinuity points occur where}\\ F(\varphi^-)\neq F(\varphi)=F(\varphi^+)\\$ $\endgroup$
    – MathMan
    Oct 10, 2019 at 17:08

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