Assigning Probability to each and every subset of $\mathbb{R}$ Supposing $\Omega = \mathbb{R}$ and $\mathcal{F} = 2^\Omega$. Then, I have read somewhere that it is not possible to define a set function $P:\mathcal{F} \to [0,1]$ such that $P$ satisfies $P(\cup_{i=0}^\infty A_i) = \sum_{i=0}^\infty P(A_i)$ whenever $A_i$ are pairwise disjoint and $P(\Omega) = 1$. Can somebody explain or point me to a good reference which proves this.
Any help is much appreciated.
EDIT: As ArturoMagidin pointed out, it is indeed possible to define such a $P$. I was trying to understand the need for defining a $\sigma$-algebra of sets and
defining $P$ on those sets. I might have to put more conditions on $P$ but I 
dont know which ones.
Thanks,
Phanindra
 A: The "Hard measure problem on $\mathbb{R}$" is to find a nonnegative set theoretic function $\rho$ whose domain are subsets of $\mathbb{R}$, such that:


*

*$\rho(E)$ is defined for all $E\subseteq\mathbb{R}$;

*$\rho(I) = \mathrm{length}(I)$ for all intervals $I\subseteq\mathbb{R}$;

*$\rho$ is $\sigma$-additive;

*$\rho$ is invariant under isometries; i.e., if $j\colon\mathbb{R}\to\mathbb{R}$ is an isometry, then $\rho(j(E)) = \rho(E)$ for all $E\subseteq $\mathbb{R}$.


If we assume the Axiom of Choice, then the hard measure problem on $\mathbb{R}$ cannot be solved, as shown by the construction of Vitali sets. On the other hand, Solovay proved that in there is a model of Set Theory without the Axiom of Choice in which the hard problem can be solved.
The "Easy measure problem on $\mathbb{R}$" requires a measure that satisfies 1, 2, and 4 above, and replaced 3 with simple finite additivity. Banach proved that the problem can be solved (non-uniquely) in $\mathbb{R}$, and in $\mathbb{R}^2$ (where 2 replaces interval and length with rectangle and area), and Hausdorff proved that the easy measure problem cannot be solved in $\mathbb{R}^n$ with $n\geq 3$.
The does not directly address your question, because a measure that satisfies 2 above cannot have $\rho(\mathbb{R})=1$. H
A probability measure is just a function that takes values on $[0,1]$, is $\sigma$-additive, and satisfies $\rho(\emptyset)=0$ and $\rho(\mathbb{R})=1$. There are plenty of probability measures on all of $\mathbb{R}$ that are defined on every subset: you can define atomic measures in which there are countably many atoms and their measures add up to $1$, for instance. However, usually you want some sort of "uniformity" to the measure (e.g., there is no uniform probability measure on $\mathbb{N}$, or on any countable set). So one needs to specify exactly what conditions you want to place on the measure in order to determine whether or not one can construct such a probability measure.
