Kolmogorov axioms consequences 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.


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, $2019.$
