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Earlier today in my stochastic processes lecture, the prof mentioned that there does not exist a probability measure P(A) defined for all subsets of [0,1] which would satisfy all the 3 Kolmogorov axioms. If I understand correctly, the Kolmogorov axioms are supposed to lay out the basis of any given probability measure so I find the statement a bit unintuitive and can’t fully understand how a probability measure wouldn’t satisfy the axioms it’s built upon. Am I missing something critical? I’d appreciate any tips or further readings.

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  • $\begingroup$ The assertion is false. Pick a number $\xi\in[0,1].$ The function $\mathbb{P}:\mathscr{P}([0,1])\to[0,1]$ defined by $\mathscr{P}(\mathcal E) = \mathcal{I}_{\mathcal E}(\xi)$ (trivially) satisfies all three axioms. ($\mathscr{P}$ is the power set and $\mathcal I$ is the indicator function.) $\mathbb{P}$ is called the atom at $\xi.$ It is the probability measure determined by any random variable $X:\Omega\to\mathbb R$ (for any probability space $\Omega$) that is almost surely equal to $\xi.$ $\endgroup$
    – whuber
    Sep 14, 2022 at 16:34

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the prof mentioned that there does not exist a probability measure P(A) defined for all subsets of [0,1] which would satisfy all the 3 Kolmogorov axioms

At the very outset, please note whuber's comment. The statement is then overreaching (perhaps misinterpreted by OP) and problematic.

Common counter-example: Dirac measure $\delta_\omega$ for a point $\omega \in [0, 1].$

With that stated, define for $a,b\in[0,1],$ $$\mathbf P([a, b])= \mathbf P((a, b])=\mathbf P([a, b)) = \mathbf P((a, b)) = b-a .\tag 1$$

A related question can be addressed: can $\mathbf P(A) $ be plausible? Is it possible for each $A\subseteq [0, 1]? $

The concern stems from the fact that there exists a non-Lebesgue measurable set in $[0,1].$

A brief sketch:

We will use an equivalence relation (check): $$x\sim y\iff y-x~\text{rational}.\tag 2$$

As a result of the equivalence, it partitions $[0, 1].$ Now, the important aspect: invoke Axiom of Choice to create a subset $T$ consisting of exactly one element from each equivalence class. Assume $0\notin T. $

Define $r$-shift of $A\subseteq[0, 1]$ as

$$A\oplus r := \{a+r; ~a\in A, ~a+r\leq 1\}\cup\{a+r-1;~a\in A, ~a+r> 1\}\tag 3.$$ This is the $r$-translate of $A$ modulo $1.$

Then (check) $$\mathbf P(A\oplus r) = \mathbf P(A).\tag 4 $$

Now consider the union of rational shifts in $[0, 1)$ of $T$ i.e. $$S:= \bigcup_{r\in[0, 1), ~r~\text{rational}}(T\oplus r) .\tag 5$$ It is easy to check $(0, 1]$ is contained in $S$ and that all $T \oplus r$ are disjoint.

Countable additivity then allows

$$\mathbf P((0, 1]) =\sum_{r\in[0, 1), ~r~\text{rational}}\mathbf P(T\oplus r) .\tag 6$$

Therefore, by dint of $(4) ,$

$$\underbrace{\mathbf P((0, 1])}_{=1}=\sum_{r\in[0, 1), ~r~\text{rational}}\mathbf P(T) .\tag{ 6.I}$$

What is $\rm (6.I) $ saying? Countably infinite sum of a same quantity is equal to 1. Can it be possible? If $\mathbf P(T) = 0,$ then $0=1$ and if $\mathbf P(T) > 0,$ then $1=\infty.$


Reference:

A First Look at Rigorous Probability Theory, Jeffrey S. Rosenthal, World Scientific, $2019.$

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    $\begingroup$ This doesn't appear to address the question that was stated, because it only analyzes a particular measure (Borel measure). The assertion in the question is that there is no measure, Borel or not, defined on the power set $\mathscr{P}([0,1])$ that satisfies all three Kolmogorov axioms. That's problematic because the assertion is wrong. $\endgroup$
    – whuber
    Sep 14, 2022 at 15:43
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    $\begingroup$ You are correct @whuber. I missed that no measure. Let me leave a message in my post. Thanks for noticing. $\endgroup$ Sep 14, 2022 at 15:48
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    $\begingroup$ Because you have made a significant contribution, I would suggest adding an introduction that discusses two possible interpretations of the question: as asked and as you think it was intended. You can dismiss the question as asked with a simple counterexample and then move on to what you already wrote. $\endgroup$
    – whuber
    Sep 14, 2022 at 16:09
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    $\begingroup$ Good suggestion. Done @whuber. $\endgroup$ Sep 14, 2022 at 17:20

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