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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.

Community
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Justin
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3 Answers3

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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 $D$.

Nate Eldredge
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  • Interesting (and helpful), I had not thought of this. – Justin May 21 '12 at 03:28
  • @Nate Eldredge why is $Q$ countable? – Dave ddd Apr 29 '16 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 http://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set – Nate Eldredge Apr 29 '16 at 13:43
  • Nate, I've read this answer of yours, but somehow I think there's something missing. Couldn't I do this also for the irrationals? And the irrationals are not countable... – An old man in the sea. Jun 07 '17 at 13:03
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    @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...) – Nate Eldredge Jun 07 '17 at 14:27
  • @NateEldredge Thanks. I get it now. ;) – An old man in the sea. Jun 07 '17 at 14:41
  • @NateEldredge $\bf{Q}$ is countable because all elements of $\bf{Q}$ can be written as $\frac{p}{q}$, where $p,q \epsilon \bf{Z}$ and $Z$ is counatable – Jayanth Kumar Sep 13 '17 at 01:49
  • Hi, I've just read your solution, but I got confused with one point. $F$ is non-decreasing, not increasing? right? (or this depends on the terminology?) also, why $F$ being increasing implies if $x\neq y$, $q_{x}\neq q_{y}$? – JacobsonRadical Oct 04 '19 at 17:54
  • @JacobsonRadical: Sure, "non-decreasing" if you like. I guess just saying "increasing" may be ambiguous, but $F$ is certainly allowed to have flat spots. This means that if, say, $x < y$, we must have $F(x+) \le F(y-)$ and therefore $q_x < F(x+) \le F(y-) < q_y$, so $q_x < q_y$ and they are not equal. – Nate Eldredge Oct 04 '19 at 21:42
  • @NateEldredge oh! Thank you so much! Brilliant answer by the way :) – JacobsonRadical Oct 05 '19 at 15:55
  • Can we get away without axiom of choice in some way? – Atom Feb 16 '22 at 10:56
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    @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 $(F(x+), F(x-))$. – Nate Eldredge Feb 17 '22 at 08:16
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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.)

passerby51
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  • Thank you for the response! – Justin May 21 '12 at 03:29
  • passerby, can you elaborate on the 3rd point? Why to be finite the number of nonzero terms must be countable? Thanks ;) – An old man in the sea. Jun 07 '17 at 14:42
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    $$ \sum_{|x| < N} [F(x+) - F(x-)] \le F(-N) - F(N) < \infty $$. Hi community. I didn't undestand why this sum is countable. Is there any way to write this in step by step? – Silvinha Mar 18 '21 at 01:48
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    Me too. I have the same doubt as @Silvinha –  Mar 18 '21 at 01:50
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    @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. – Nate Eldredge Feb 17 '22 at 08:20
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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.

Alex Becker
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