# It is possible to define our intuitive notion for probability in subsets of $[0,1]$

I've always heard and read the sentence:

If you pick a real number $x\in[0,1]$ at random, the probability to obtain a rational number is $0$.

What is the meaning for that? Is this the "real" probability?

First, we need to define accurately what means at random, I'll assume a base $10$. Then all numbers in $[0,1]$ begins with $0$ (for $1$ we discard this notation and we will use $0.9999\ldots$).

Then I'll give a classical approach defining random meaning, if we pick a real number $x$ in $[0,1]$ at random we mean that the probability to get a digit in certain position is $1/10$ (extending for $[0,1]$ the Laplace classical definition for probability), roughly speaking each digit has the same "probability" to be in $x$.

Strictly speaking we are building a notion of probability in $\{0,1,2,3,4,5,6,7,8,9\}^\omega$ since numbers with a $0$ periodic has two representations. We want to pass the intuitive notion of probability to this set, but if there is one, this must be unique.

The Lebesgue measure $m$ rescues some of these probabilistic notions for example $m(\emptyset)=0$, $m(\{x\})=0$, and $m(\text{numbers starting with the given digit$k$})=\frac{1}{10}$, but intuitively a probability notion must be defined in all subsets (Edit: sos440 in comments says that is natural this phenomenon (not defined in all subsets) since we can't describe some sets without using additional axioms like "Zorn Lemma") apart of having some properties like (finite or $\sigma$)-additivity.

Lebesgue measure also satisfies $m(\mathbb{Q\cap [0,1]})=0$. Does our intuitive notion of probability tell us that the probability to get a rational number is $0$?

Other functions("measures") in $[0,1]$ are defined in whole $\mathcal{P}([0,1])$ but doesn't coincides with $m$ and we loss other notions for our intuitive probability meaning.

Does Lebesgue measure deserve to be called a "real" probability in $[0,1]$ (or $\{0,1,2,3,4,5,6,7,8,9\}^\omega$)? Why?

This was completely answered by Michael Greinecker ♦ The answer is a big Yes.

Is possible to define a probability notion for $[0,1]$ and for whole $\mathbb{R}$?

Fully answered in comments and answers. The answer is no for $\mathbb{R}$, and for $[0,1]$ is yes.

Is there any function(measure) that is the best representative for our concept of probability? Why?

This was completely answered by Michael Greinecker ♦: The Lebesgue measure.

Second Edit: People in comments and answers talk me about importance of additional axioms like "Axiom of Choice" and "Continuum Hypothesis". How acceptance or denial of these axioms affects the answers to my questions here?

This was fully answered by hot_queen.

Third Edit: I marked the questions were completely answered and offered a bounty for intuitive answer and essencially the first question about $\mathbb{Q}$.

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Unfortunately we cannot give the meaning of the probability to every subset of $\Omega = \{1, \cdots, 9\}^{\omega}$. This phenomenon is in some sense natural, since there are plenty of subsets of $\Omega$ such that we cannot describe the set explicitly and can only prove their existence by some axioms (such as AC). Intuitively, such sets lack information. Thus it is more natural to put them aside and choose those subsets which have information of which we can measure. –  sos440 May 10 '13 at 0:44
@sos440: Can't we? If there is an rvm below $|\mathbb{R}|$, surely we can (relative to a witnessing measure). –  hot_queen May 10 '13 at 0:53
what do you mean? @hot_queen –  Gastón Burrull May 10 '13 at 1:00
I meant math is funny and full of surprises. –  hot_queen May 10 '13 at 1:11
hot_queen means to say that under certain large cardinal axioms it is consistent that there is a measure extending the Lebesgue measure which measures every set. This extension is not translation invariant, though. –  Asaf Karagila May 10 '13 at 1:11

First, under the continuum hypothesis, every probability measure on the powerset of $[0,1]$ is concentrated on a countable set of numbers, so it would give a highly asymmetrical notion of "at random".

Second, $\mathbb{Q}\cap[0,1]$ is countable and any measure that does not put positive mass on some numbers will put zero measure on countable sets. Every probability measure that is invariant under all permutations of countably many points will put zero probability on $\mathbb{Q}\cap[0,1]$.

Third, Lebesgue measure is in some sense the natural notion of a uniform distribution. Let $X_n$ be a random variable with uniform distribution on the set $\{0,1/n,2/n,\ldots,n/n\}$. Then the sequence $(X_n)$ converges in distribution to a random variable with Lebesgue measure as its distribution.

Ultimately, a probability measure deserves to be called random if it doesn't discriminate between events we consider equivalent. If we think of $[0,1]$ as being wrapped around a circle and think that shifting a set along the circle should not change its probability, Lebesgue measure is the only measure (on the Borel sets) satisfying this criterion. There is no similar probability measure for all of $\mathbb{R}$.

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what do you mean by "a probability measure is concentrated on a countable set of numbers"? –  Gastón Burrull May 10 '13 at 0:34
From second point. Since the fact "lebesgue measure of a point is 0" coincides with our probability intuition then the fact "lebesgue measure of countable sets of real line is $0$" respects our intuition? –  Gastón Burrull May 10 '13 at 0:44
@Gaston That the probability measure is concentrated on a countable set of numbers means there is a countable set $C$ such that $\mu(C)=1$. In that case, there is a family $(p_x)_{x\in C}$ of nonnegative numbers such that $\sum_{x\in C}p_x=1$ and such that for every set $A$, we have $\mu(A)=\sum_{x\in A} p_x$. –  Michael Greinecker May 10 '13 at 0:48
How can any probability measure be concentrated in countable sets? Each measure that assigns $0$ to single points and satisfies $\sigma$-subadditivity must assings $0$ to countable sets. In particular lebesgue measure is not concentrated on a countable set and is a probability measure. –  Gastón Burrull May 10 '13 at 0:55
@GastónBurrull Exactly. And under the continuum hypothesis, such a measure defined on all subsets off $[0,1]$ does not exist. –  Michael Greinecker May 10 '13 at 0:56

I wans't planning to write a "rant" but it turns out that my beer is pretty good.

Mr. Greinecker talked about a scenario (CH) where it is impossible to measure all sets of reals (this was shown by the Banach and Kuratowski a long time ago late 1920s maybe). Let me add that if you do not require translation invariance then it is conceivable that all sets of reals could be assigned a measure. One way of "justifying" this is to say that people have worked in the theory ZFC + "there is such a total measure" for quite some time now and have never obtained a contradiction. The other way of justifying this is that they "feel" that in their set theoretic universe ZFC + "a measurable cardinal exists" holds and hence by Solovay's work the former assumption must be consistent too.

Of course the moral is "it is difficult to compute the truth content of infinitary statements" (Nelson believes that PA is inconsistent) and once formalized in ANY decent system, not only these questions but their independence becomes independent (over your decent system).

This is a day-to-day phenomenon for logicians (set theorists, in particular) and maybe someday all mathematicians will ask not just for a yes or no answer to their conjectures but also an independence result (ZFC in my view exhausts my intuition of truth and knowable).

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No problems. I did not mean any disrespect to anyone anyway - I am a hot queen. –  hot_queen May 10 '13 at 2:04
Dear Asaf, I am not sure. Just let me say this (and I am not going to take this back - Hail my beer). I have given a talk where your advisor was asleep. –  hot_queen May 10 '13 at 2:12
Epilogue: Say A is a positive measure set of reals. Must A contain a similar copy of $\{1, 1/2, 1/4, 1/8, \dots \}$? –  hot_queen May 10 '13 at 3:22
It's remarkably funny to come and read this conversation only AFTER half of it has been deleted and just the HOT QUEEN's side of the conversation remains. –  Quinn Culver May 10 '13 at 3:36
Math + Beer + Bike = Buenos –  Quinn Culver May 10 '13 at 3:41
If the intuitive notion of probability in the interval is that subintervals of equal length contain the sample point with equal probability, then it is intuitive that the rationals have measure 0. Take a small, small interval around 1/2, say of width $\epsilon$, then one half that big around 1/3, etc. and go on until you've surrounded every one of the countably many rational points by a collection of intervals whose total length is 2$\epsilon$. By our intuitive notion, this means that the probability of getting a rational point is less than $2\epsilon$ for any positive $\epsilon$, so it must be 0.