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.
Thanks in advance.
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 $\mathbb{Q}$.
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 $\mathbb{R}$.